question_id
stringlengths 8
35
| subject
stringclasses 1
value | chapter
stringclasses 32
values | topic
stringclasses 178
values | question
stringlengths 26
9.64k
| options
stringlengths 2
1.63k
| correct_option
stringclasses 5
values | answer
stringclasses 293
values | explanation
stringlengths 13
9.38k
| question_type
stringclasses 3
values | paper_id
stringclasses 149
values |
|---|---|---|---|---|---|---|---|---|---|---|
GEjRKxAiRms6EYbD
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
Let $$f\left( x \right)$$ be a polynomial function of second degree. If $$f\left( 1 \right) = f\left( { - 1} \right)$$ and $$a,b,c$$ are in $$A.P, $$ then $$f'\left( a \right),f'\left( b \right),f'\left( c \right)$$ are in
|
[{"identifier": "A", "content": "Arithmetic -Geometric Progression "}, {"identifier": "B", "content": "$$A.P$$"}, {"identifier": "C", "content": "$$G.P$$"}, {"identifier": "D", "content": "$$H.P$$ "}]
|
["B"]
| null |
$$f\left( x \right) = a{x^2} + bx + c$$
<br><br>$$f\left( 1 \right) = f\left( { - 1} \right)$$
<br><br>$$ \Rightarrow a + b + c = a - b + c$$
<br><br>or $$b = 0$$
<br><br>$$\therefore$$ $$f\left( x \right) = a{x^2} + c$$
<br><br>or $$f'\left( x \right) = 2ax$$
<br><br>Now $$f'\left( a \right);f'\left( b \right);$$
<br><br>and $$f'\left( c \right)$$ are $$2a\left( a \right);2a\left( b \right);2a\left( c \right)$$
<br><br>i.e.$$\,2{a^2},\,2ab,\,2ac.$$
<br><br>$$ \Rightarrow $$ If $$a,b,c$$ are in $$A.P.$$ then
<br><br>$$f'\left( a \right);f'\left( b \right)$$ and
<br><br>$$f'\left( c \right)$$ are also in $$A.P.$$
|
mcq
|
aieee-2003
|
lYKxaNHOCAaR6Yxj
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
Let $$y$$ be an implicit function of $$x$$ defined by $${x^{2x}} - 2{x^x}\cot \,y - 1 = 0$$. Then $$y'(1)$$ equals
|
[{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$\\log \\,2$$"}, {"identifier": "C", "content": "$$-\\log \\,2$$ "}, {"identifier": "D", "content": "$$-1$$"}]
|
["D"]
| null |
$${x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0$$
<br><br>$$ \Rightarrow 2\,\cot \,y = {x^x} - {x^{ - x}}$$
<br><br>$$ \Rightarrow 2\,\cot \,y\, = u - {1 \over u}$$
<br><br>where $$u = {x^x}$$
<br><br>Differentiating both sides with respect to $$x,$$
<br><br>we get $$ \Rightarrow - 2\cos e{c^2}y{{dy} \over {dx}}$$
<br><br>$$ = \left( {1 + {1 \over {{u^2}}}} \right){{du} \over {dx}}$$
<br><br>where $$u = {x^x} \Rightarrow \log \,u = x\,\log \,x$$
<br><br>$$ \Rightarrow {1 \over u}{{du} \over {dx}} = 1 + \log \,x$$
<br><br>$$ \Rightarrow {{du} \over {dx}} = {x^x}\left( {1 + \log \,x} \right)$$
<br><br>$$\therefore$$ We get $$ - 2\cos e{c^2}y{{dy} \over {dx}}$$
<br><br>$$ = \left( {1 + {x^{ - 2x}}} \right){x^x}\left( {1 + \log \,x} \right)$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {{\left( {{x^x} + {x^{ - x}}} \right)\left( {1 + \log x} \right)} \over { - 2\left( {1 + {{\cot }^2}y} \right)}}\,\,\,\,\,\,...\left( i \right)$$
<br><br>Now when
<br><br>$$x=1,$$ $${x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0,$$
<br><br>gives $$1 - 2\,\cot y - 1 = 0$$
<br><br>$$ \Rightarrow \,\,\cot y\, = 0$$
<br><br>$$\therefore$$ From equation $$(i),$$ at $$x=1$$
<br><br>and $$\cot \,y = 0,$$ we get
<br><br>$$y'\left( 1 \right) = {{\left( {1 + 1} \right)\left( {1 + 0} \right)} \over { - 2\left( {1 + 0} \right)}} = - 1$$
|
mcq
|
aieee-2009
|
a7PCNBHZY1NukfT9vwNCx
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
If y = $${\left[ {x + \sqrt {{x^2} - 1} } \right]^{15}} + {\left[ {x - \sqrt {{x^2} - 1} } \right]^{15}},$$
<br/><br/> then (x<sup>2</sup> $$-$$ 1) $${{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is equal to :
|
[{"identifier": "A", "content": "125 y"}, {"identifier": "B", "content": "124 y<sup>2</sup>"}, {"identifier": "C", "content": "225 y<sup>2</sup>"}, {"identifier": "D", "content": "225 y"}]
|
["D"]
| null |
<p>The given equation is</p>
<p>$$y = {({x^2} + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}}$$</p>
<p>Differentiating w.r.t. x, we get</p>
<p>$${{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) + 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {1 - {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)$$</p>
<p>Here, we have used the standard differentiatials</p>
<p>$${d \over {dx}}{x^n} = n\,{x^{n - 1}}$$</p>
<p>That is, $${d \over {dx}}(\sqrt {f(x)} ) = {1 \over {2\sqrt {f(x)} }} \times {d \over {dx}}(f(x))$$</p>
<p>Therefore</p>
<p>$${{dy} \over {dx}} = {{15{{(x + \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} + {{15{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}$$</p>
<p>$$ \Rightarrow \sqrt {{x^2} - 1} {{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{15}} - 15{(x - \sqrt {{x^2} - 1} )^{15}}$$</p>
<p>Differentiating w.r.t. x, we get</p>
<p>$${{1(2x)} \over {2\sqrt {{x^2} + 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 15 \times 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) - 15 \times 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {{{1 - 1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)$$</p>
<p>$$ \Rightarrow {x \over {\sqrt {{x^2} - 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 225{(x + \sqrt {{x^2} - 1} )^{14}}{{(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} - {{225{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}$$</p>
<p>$$ \Rightarrow \sqrt {{x^2} - 1} \left[ {{x \over {\sqrt {{x^2} - 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}}} \right] = 225{(x + \sqrt {{x^2} - 1} )^{15}} + 225{(x - \sqrt {{x^2} - 1} )^{15}}$$</p>
<p>$$ \Rightarrow x{{dy} \over {dx}} + ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 225\left[ {{{(x + \sqrt {{x^2} - 1} )}^{15}} + {{(x - \sqrt {{x^2} - 1} )}^{15}}} \right]$$</p>
<p>Substituting $${(x + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}} = y$$, we get</p>
<p>$$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + {{x\,dy} \over {dx}} = 225y$$</p>
|
mcq
|
jee-main-2017-online-8th-april-morning-slot
|
DhiE82brjsITwGqGoJmYl
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
If $$f\left( x \right) = \left| {\matrix{
{\cos x} & x & 1 \cr
{2\sin x} & {{x^2}} & {2x} \cr
{\tan x} & x & 1 \cr
} } \right|,$$ then $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$
|
[{"identifier": "A", "content": "does not exist. "}, {"identifier": "B", "content": "exists and is equal to 2. "}, {"identifier": "C", "content": "existsand is equal to 0."}, {"identifier": "D", "content": "exists and is equal to $$-$$ 2."}]
|
["D"]
| null |
Given,<br><br>
$$f\left( x \right) = \left| {\matrix{
{\cos x} & x & 1 \cr
{2\sin x} & {{x^2}} & {2x} \cr
{\tan x} & x & 1 \cr
} } \right|$$<br><br>
= cosx(x<sup>2</sup> - 2x<sup>2</sup>) - x(2 sinx - 2x tanx) + (2x sinx - x<sup>2</sup> tanx)<br>
= x<sup>2</sup> (tanx - cosx)<br>
$$ \therefore $$ $${f^{'}}(x)$$ = 2x (tanx - cosx) + x<sup>2</sup>(sec<sup>2</sup>x + sinx)<br><br>
$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$<br><br>
= $$\mathop {\lim }\limits_{x \to o} {{2x(\tan x - \cos x) + {x^2}({{\sec }^2}x + \sin x)} \over x}$$<br><br>
= $$\mathop {\lim }\limits_{x \to o} \,\,2(\tan x - \cos x) + x({\sec ^2}x + \sin x)$$<br><br>
= 2 (0-1) + 0<br><br>
= -2
|
mcq
|
jee-main-2018-online-15th-april-morning-slot
|
C8sdKeF0ryQ9Msh5eBUCc
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
If x<sup>2</sup> + y<sup>2</sup> + sin y = 4, then the value of $${{{d^2}y} \over {d{x^2}}}$$ at the point ($$-$$2,0) is :
|
[{"identifier": "A", "content": "$$-$$ 34"}, {"identifier": "B", "content": "$$-$$ 32"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$-$$ 2"}]
|
["A"]
| null |
Given, x<sup>2</sup> + y<sup>2</sup> + sin y = 4
<br><br>After differentiating the above equation w.r.t.x we get
<br><br>2x + 2y $${{dy} \over {dx}}$$ + cos y $${{dy} \over {dx}}$$ = 0 . . . . (1)
<br><br>$$ \Rightarrow $$ 2x + (2y + cos y) $${{dy} \over {dx}}$$ = 0
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{ - 2x} \over {2y + \cos y}}$$
<br><br>At ($$-$$ 2, 0), $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = $${{ - 2x - 2} \over {2 \times 0 + \cos 0}}$$
<br><br>$$ \Rightarrow $$ $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = $${4 \over {0 + 1}}$$
<br><br>$$ \Rightarrow $$ $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = 4 . . . . .(2)
<br><br>Again differentiating equation (1) w.r.t to x, we get
<br><br>2 + 2 $${\left( {{{dy} \over {dx}}} \right)^2}$$ + 2y$${{{d^2}y} \over {d{x^2}}}$$ $$-$$ sin y $${\left( {{{dy} \over {dx}}} \right)^2}$$ + cos y $${{{d^2}y} \over {d{x^2}}}$$ = 0
<br><br>$$ \Rightarrow $$ 2 + (2 $$-$$ sin y) $${\left( {{{dy} \over {dx}}} \right)^2}$$ + (2y + cos y)$${{{d^2}y} \over {d{x^2}}}$$ = 0
<br><br>$$ \Rightarrow $$ (2y + cos y) $${{{d^2}y} \over {d{x^2}}}$$ = $$-$$ 2 $$-$$ (2 $$-$$ sin y)$${\left( {{{dy} \over {dx}}} \right)^2}$$
<br><br>$$ \Rightarrow $$ $${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - \left( {2 - \sin y} \right){{\left( {{{dy} \over {dx}}} \right)}^2}} \over {2y + \cos y}}$$
<br><br>So, at ($$-$$ 2, 0),
<br><br>$${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - \left( {2 - 0} \right) \times {4^2}} \over {2 \times 0 + 1}}$$
<br><br>$$ \Rightarrow $$ $${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - 2 \times 16} \over 1}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${{{d^2}y} \over {d{x^2}}}$$ = $$-$$ 34
|
mcq
|
jee-main-2018-online-15th-april-morning-slot
|
Ne604P2uiWuSdG4XwJ3rsa0w2w9jx5deh3j
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
If e<sup>y</sup>
+ xy = e, the ordered pair $$\left( {{{dy} \over {dx}},{{{d^2}y} \over {d{x^2}}}} \right)$$ at x = 0 is equal to :
|
[{"identifier": "A", "content": "$$\\left( {{1 \\over e}, - {1 \\over {{e^2}}}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - {1 \\over e},{1 \\over {{e^2}}}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over e}, - {1 \\over {{e^2}}}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over e},{1 \\over {{e^2}}}} \\right)$$"}]
|
["B"]
| null |
y = 1 $$ \Rightarrow $$ x = 0<br><br>
$${e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0$$<br><br>
$$ \Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e}$$<br><br>
$$ \Rightarrow {e^y}{{{d^2}y} \over {d{x^2}}} + {e^y}{\left( {{{dy} \over {dx}}} \right)^2} + x{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} = 0$$<br><br>
x = 0, y = 1<br><br>
$$ \Rightarrow e{{{d^2}y} \over {d{x^2}}} + e{\left( { - {1 \over e}} \right)^2} + 0 + 2\left( { - {1 \over e}} \right) = 0$$<br><br>
$$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {1 \over {{e^2}}}$$
|
mcq
|
jee-main-2019-online-12th-april-morning-slot
|
fd5wbl11yciwuHmLja7k9k2k5e29bcc
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
Let x<sup>k</sup> + y<sup>k</sup> = a<sup>k</sup>, (a, k > 0 ) and $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$, then k is:
|
[{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}]
|
["B"]
| null |
x<sup>k</sup> + y<sup>k</sup> = a<sup>k</sup>
<br><br>$$ \Rightarrow $$ kx<sup>k - 1</sup> + ky<sup>k - 1</sup>$${{{dy} \over {dx}}}$$ = 0
<br><br>$$ \Rightarrow $$ $${{{dy} \over {dx}} + {{\left( {{x \over y}} \right)}^{k - 1}}}$$ = 0 ...(1)
<br><br>Given $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$ ...(2)
<br><br>Comparing (1) and (2), we get
<br><br>k - 1 = $$ - {1 \over 3}$$
<br><br>$$ \Rightarrow $$ k = $${2 \over 3}$$
|
mcq
|
jee-main-2020-online-7th-january-morning-slot
|
kPxln5RHIGisQJ1WD17k9k2k5e4fsyg
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
If $$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} ,\alpha \in \left( {{{3\pi } \over 4},\pi } \right)$$<br/><br/>
$${{dy} \over {d\alpha }}\,\,at\,\alpha = {{5\pi } \over 6}is$$ :
|
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "-4"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "-$${1 \\over 4}$$"}]
|
["A"]
| null |
$$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}$$
<br><br>= $$\sqrt {2\left( {{{1 + {{\tan }^2}\alpha } \over {\tan \alpha \left( {1 + {{\tan }^2}\alpha } \right)}}} \right) + {1 \over {{{\sin }^2}\alpha }}} $$
<br><br>= $$\sqrt {2\left( {{{\cos \alpha } \over {\sin \alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} $$
<br><br>= $$\sqrt {{{\sin 2\alpha + 1} \over {{{\sin }^2}\alpha }}} $$
<br><br>= $${{\left| {\sin \alpha + \cos \alpha } \right|} \over {\left| {\sin \alpha } \right|}}$$
<br><br>At $$\alpha $$ = $${{5\pi } \over 6}$$
<br><br>$${\left| {\sin \alpha + \cos \alpha } \right|}$$ = -(sin$$\alpha $$ + cos$$\alpha $$)
<br><br>and |sin$$\alpha $$| = sin$$\alpha $$
<br><br>$$ \therefore $$ y($$\alpha $$) = $${{ - \left( {\sin \alpha + \cos \alpha } \right)} \over {\sin \alpha }}$$
<br><br>= -1 - cot$$\alpha $$
<br><br>$$ \therefore $$ $${{dy} \over {d\alpha }}$$ = cosec<sup>2</sup>$$\alpha $$
<br><br>So $${{{dy} \over {d\alpha }}}$$ at $$\alpha $$ = $${{5\pi } \over 6}$$,
<br><br>= cosec<sup>2</sup>$${{5\pi } \over 6}$$ = 4
|
mcq
|
jee-main-2020-online-7th-january-morning-slot
|
jKg7va9mIxQ9FIreIH7k9k2k5fnompn
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
Let y = y(x) be a function of x satisfying
<br/><br>$$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $$ where k is a constant and
<br/><br>$$y\left( {{1 \over 2}} \right) = - {1 \over 4}$$. Then $${{dy} \over {dx}}$$ at x = $${1 \over 2}$$, is equal to :</br></br>
|
[{"identifier": "A", "content": "$${2 \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "$$ - {{\\sqrt 5 } \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt 5 } \\over 2}$$"}, {"identifier": "D", "content": "$$ - {{\\sqrt 5 } \\over 4}$$"}]
|
["B"]
| null |
$$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $$ ....(1)
<br><br>On differentiating both side of eq. (1) w.r.t. x we
get,
<br><br>$${{dy} \over {dx}}\sqrt {1 - {x^2}} - y{{2x} \over {2\sqrt {1 - {x^2}} }}$$
<br><br>= 0 - $$\sqrt {1 - {y^2}} + {{xy} \over {\sqrt {1 - {y^2}} }}{{dy} \over {dx}}$$
<br><br>Put x = $${1 \over 2}$$ and y = $$ - {1 \over 4}$$, we get
<br><br>$${{dy} \over {dx}}{{\sqrt 3 } \over 2} - \left( { - {1 \over 4}} \right){{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}$$
<br><br>= $$ - {{\sqrt {15} } \over 4} + {{ - {1 \over 8}} \over {{{\sqrt {15} } \over 4}}}.{{dy} \over {dx}}$$
<br><br>$$ \therefore $$ $${{dy} \over {dx}} = - {{\sqrt 5 } \over 2}$$
|
mcq
|
jee-main-2020-online-7th-january-evening-slot
|
Dv3KDCMR2csVxH0QHWjgy2xukf8zrqqn
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
If $$\left( {a + \sqrt 2 b\cos x} \right)\left( {a - \sqrt 2 b\cos y} \right) = {a^2} - {b^2}$$<br/><br/>
where a > b > 0, then $${{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)$$ is :
|
[{"identifier": "A", "content": "$${{a - 2b} \\over {a + 2b}}$$"}, {"identifier": "B", "content": "$${{a - b} \\over {a + b}}$$"}, {"identifier": "C", "content": "$${{a + b} \\over {a - b}}$$"}, {"identifier": "D", "content": "$${{2a + b} \\over {2a - b}}$$"}]
|
["C"]
| null |
$$(a + \sqrt 2 b\cos x)(a - \sqrt 2 b\cos y) = {a^2} - {b^2}$$<br><br>$$ \Rightarrow {a^2} - \sqrt 2 ab\cos y + \sqrt 2 ab\cos x - 2{b^2}\cos x\cos y = {a^2} - {b^2}$$<br><br>Differentiating both sides :<br><br>$$0 - \sqrt 2 ab\left( { - \sin y{{dy} \over {dx}}} \right) + \sqrt 2 ab( - \sin x)$$<br><br>$$ - 2{b^2}\left[ {\cos x\left( { - \sin y{{dy} \over {dx}}} \right) + \cos y( - \sin x)} \right] = 0$$<br><br>At $$\left( {{\pi \over 4},{\pi \over 4}} \right)$$ :<br><br>$$ab{{dy} \over {dx}} - ab - 2{b^2}\left( { - {1 \over 2}{{dy} \over {dx}} - {1 \over 2}} \right) = 0$$<br><br>$$ \Rightarrow {{dx} \over {dy}} = {{ab + {b^2}} \over {ab - {b^2}}} = {{a + b} \over {a - b}}$$; a, b > 0
|
mcq
|
jee-main-2020-online-4th-september-morning-slot
|
1ldpt5swb
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
<p>Let $$y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)\right)$$. Then, at x = 1,</p>
|
[{"identifier": "A", "content": "$$2 y^{\\prime}+\\sqrt{3} \\pi^{2} y=0$$"}, {"identifier": "B", "content": "$$y^{\\prime}+3 \\pi^{2} y=0$$"}, {"identifier": "C", "content": "$$\\sqrt{2} y^{\\prime}-3 \\pi^{2} y=0$$"}, {"identifier": "D", "content": "$$2 y^{\\prime}+3 \\pi^{2} y=0$$"}]
|
["D"]
| null |
$f(x)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right)$
<br/><br/>$$
\begin{aligned}
& f^{\prime}(x)=3 \sin ^{2}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\
& \cos \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\
& \frac{\pi}{3}\left(-\sin \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\
& \frac{\pi}{3 \sqrt{2}} \frac{3}{2}\left(-4 x^{3}+5 x^{3}+1\right)^{1 / 2}\left(-12 x^{2}+10 x\right)
\end{aligned}
$$
<br/><br/>$f^{\prime}(1)=\frac{3 \pi^{2}}{16}$
<br/><br/>$$
\begin{aligned}
& f(1)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}} 2 \sqrt{2}\right)\right) \\\\
&=\sin ^{3}\left(-\frac{\pi}{6}\right)=\frac{-1}{8} \\\\
& \therefore 2 f^{\prime}(1)+3 \pi^{2} f(1)=0
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-31st-january-morning-shift
|
1lgoy3hxl
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
<p>Let $$f(x)=\sum_\limits{k=1}^{10} k x^{k}, x \in \mathbb{R}$$. If $$2 f(2)+f^{\prime}(2)=119(2)^{\mathrm{n}}+1$$ then $$\mathrm{n}$$ is equal to ___________</p>
|
[]
| null |
10
|
Given, $f(x)=\sum_\limits{k=1}^{10} k x^{k}$
<br/><br/>$$
\begin{aligned}
& f(x)=x+2 x^2+\ldots \ldots \ldots+10 x^{10} \\\\
& f(x) . x=x^2+2 x^3+\ldots \ldots \ldots+9 x^{10}+10 x^{11} \\\\
& f(x)(1-x)=x+x^2+x^3+\ldots \ldots \ldots+x^{10}-10 x^{11} \\\\
& \therefore f(x)=\frac{x\left(1-x^{10}\right)}{(1-x)^2}-\frac{10 x^{11}}{(1-x)}
\end{aligned}
$$
<br/><br/>$$ \Rightarrow $$ $$
f(x)=\frac{x-x^{11}-10 x^{11}+10 x^{12}}{(1-x)^2} = \frac{10 x^{12}-11 x^{11}+x}{(1-x)^2}
$$
<br/><br/>So,
<br/><br/>$$
\begin{aligned}
& f(2)=2\left(1-2^{10}\right)+10 \cdot 2^{11} \\\\
& =2+18 \cdot 2^{10}
\end{aligned}
$$
<br/><br/>$$f'\left( x \right) = {{{{\left( {1 - x} \right)}^2}\left( {120{x^{11}} - 121{x^{10}} + 1} \right) + 2\left( {1 - x} \right)\left( {10{x^{12}} - {{11.x}^{11}} + 2} \right)} \over {{{\left( {1 - x} \right)}^4}}}$$
<br/><br/>So,
<br/><br/>$$f'\left( x \right) = {{1\left( {{{120.2}^{11}} - {{121.2}^{10}} + 1} \right) + 2\left( { - 1} \right)\left( {{{10.2}^{12}} - {{11.2}^{11}} + 2} \right)} \over {{{\left( { - 1} \right)}^4}}}$$
<br/><br/>= $${2^{10}}\left( {83} \right) - 3$$
<br/><br/>Hence $2 f(2)+f^{\prime}(2)=119.2^{10}+1$
<br/><br/>$$
\Rightarrow \text { So, } \mathrm{n}=10
$$
|
integer
|
jee-main-2023-online-13th-april-evening-shift
|
1lgzxhecr
|
maths
|
differentiation
|
differentiation-of-implicit-function
|
<p>Let $$f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}, x \in[0, \pi]-\left\{\frac{\pi}{4}\right\}$$. Then $$f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right)$$ is equal to</p>
|
[{"identifier": "A", "content": "$$\\frac{2}{3 \\sqrt{3}}$$"}, {"identifier": "B", "content": "$$\\frac{2}{9}$$"}, {"identifier": "C", "content": "$$\\frac{-1}{3 \\sqrt{3}}$$"}, {"identifier": "D", "content": "$$\\frac{-2}{3}$$"}]
|
["B"]
| null |
$$f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}$$
<br/><br/>$$
\begin{aligned}
& =\frac{\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x-1}{\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x} \\\\
& =\frac{\cos \left(x-\frac{\pi}{4}\right)-1}{\sin \left(x-\frac{\pi}{4}\right)} \\\\
& =\frac{-2 \sin ^2\left(\frac{x}{2}-\frac{\pi}{8}\right)}{2 \sin \left(\frac{x}{2}-\frac{\pi}{8}\right) \cos \left(\frac{x}{2}-\frac{\pi}{8}\right)}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow f(x)=-\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \\\\
& \Rightarrow f^{\prime}(x)=-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right)
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow f^{\prime \prime}(x)=-\frac{1}{2} \cdot 2 \sec \left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \sec \left(\frac{x}{2}-\frac{\pi}{8}\right)\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \times \frac{1}{2} \\\\
& =-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \tan \left(\frac{x}{2}-\frac{\pi}{8}\right)
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Now, } f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right) \\\\
& =-\tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \frac{-1}{2} \sec ^2\left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \\\\
& =\frac{1}{2} \tan ^2\left(\frac{\pi}{6}\right) \times \sec ^2 \frac{\pi}{6} \\\\
& =\frac{1}{2} \times \frac{1}{3} \times \frac{4}{3}=\frac{2}{9}
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-8th-april-morning-shift
|
cOntCBTD82jRclEC
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
If $$y = \sec \left( {{{\tan }^{ - 1}}x} \right),$$ then $${{{dy} \over {dx}}}$$ at $$x=1$$ is equal to :
|
[{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "$${1 \\over 2}$$ "}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$$\\sqrt 2 $$ "}]
|
["A"]
| null |
Let $$y = \sec \left( {{{\tan }^{ - 1}}x} \right)$$
<br><br>and $${\tan ^{ - 1}}\,\,x = \theta .$$
<br><br>$$ \Rightarrow x = \tan \theta $$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267182/exam_images/gheykswaxc1dfck1kir0.webp" loading="lazy" alt="JEE Main 2013 (Offline) Mathematics - Differentiation Question 69 English Explanation">
<br><br>Thus, we have $$y = \sec \,\theta $$
<br><br>$$ \Rightarrow y = \sqrt {1 + {x^2}} $$
<br><br>$$\left( {\,\,} \right.$$ As $$\,\,\,\,\,\,{\sec ^2}\theta = 1 + {\tan ^2}\theta $$ $$\left. {\,\,} \right)$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {1 \over {2\sqrt {1 + {x^2}} }}.2x$$
<br><br>At $$x = 1,\,\,{{dy} \over {dx}} = {1 \over {\sqrt 2 }}.$$
|
mcq
|
jee-main-2013-offline
|
yy401t5DTBxS3W73
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
If for $$x \in \left( {0,{1 \over 4}} \right)$$, the derivatives of
<br/><br/>$${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$ is $$\sqrt x .g\left( x \right)$$, then $$g\left( x \right)$$ equals
|
[{"identifier": "A", "content": "$${{{3x\\sqrt x } \\over {1 - 9{x^3}}}}$$"}, {"identifier": "B", "content": "$${{{3x} \\over {1 - 9{x^3}}}}$$"}, {"identifier": "C", "content": "$${{3 \\over {1 + 9{x^3}}}}$$"}, {"identifier": "D", "content": "$${{9 \\over {1 + 9{x^3}}}}$$"}]
|
["D"]
| null |
Let y = $${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$
<br><br>= $${\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]$$
<br><br>= 2$${\tan ^{ - 1}}\left( {3{x^{{3 \over 2}}}} \right)$$
<br><br>$$ \therefore $$ $${{dy} \over {dx}} = 2.{1 \over {1 + {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}.3 \times {3 \over 2}{\left( x \right)^{{1 \over 2}}}$$
<br><br>= $${9 \over {1 + 9{x^3}}}.\sqrt x $$
<br><br>$$ \therefore $$ g(x) = $${9 \over {1 + 9{x^3}}}$$
|
mcq
|
jee-main-2017-offline
|
WaMVJei76O03McRXFgNSd
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
If f(x) = sin<sup>-1</sup> $$\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right),$$ then f'$$\left( { - {1 \over 2}} \right)$$ equals :
|
[{"identifier": "A", "content": "$$ - \\sqrt 3 {\\log _e}\\sqrt 3 $$"}, {"identifier": "B", "content": "$$ \\sqrt 3 {\\log _e}\\sqrt 3 $$"}, {"identifier": "C", "content": "$$ - \\sqrt 3 {\\log _e}\\, 3 $$"}, {"identifier": "D", "content": "$$ \\sqrt 3 {\\log _e}\\, 3 $$"}]
|
["B"]
| null |
Since f(x) = sin$$\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right)$$
<br><br>Suppose 3<sup>x</sup> = tan t
<br><br> $$ \Rightarrow $$ f(x) = sin<sup>$$-$$1</sup> $$\left( {{{2\tan t} \over {1 + {{\tan }^2}t}}} \right)$$ = sin<sup>$$-$$1</sup> (sin2t) = 2t
<br><br> = 2tan<sup>$$-$$1</sup> (3x)
<br><br>So, f'(x) = $${2 \over {1 + {{\left( {{3^x}} \right)}^2}}} \times {3^x}.{\log _e}3$$
<br><br>$$ \therefore $$ f '$$\left( { - {1 \over 2}} \right)$$ = $${2 \over {1 + {{\left( {{3^{ - {1 \over 2}}}} \right)}^2}}} \times {3^{ - {1 \over 2}}}$$ . log<sub>e</sub> 3
<br><br>= $${1 \over 2} \times \sqrt 3 \times {\log _e}3$$ = $$\sqrt 3 \times {\log _e}\sqrt 3 $$
|
mcq
|
jee-main-2018-online-15th-april-evening-slot
|
9onUmZNU2npIbQ1VhNA44
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
If $$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$$,
<br/><br/>x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ then $$dy \over dx$$ is equal to:
|
[{"identifier": "A", "content": "$$2x - {\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\pi \\over 6} - x$$"}, {"identifier": "C", "content": "$${\\pi \\over 3} - x$$"}, {"identifier": "D", "content": "$$x - {\\pi \\over 6}$$"}]
|
["D"]
| null |
$$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\left( {{{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 2} - {{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$$
<br><br>As x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ then
<br><br>$${{{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)}$$ = $${\left( {{\pi \over 3} + x} \right)}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 2} - \left( {{\pi \over 3} + x} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 6} - x} \right)^2}$$
<br><br>$$ \therefore $$ $$2{{dy} \over {dx}} = 2\left( {{\pi \over 6} - x} \right)\left( { - 1} \right)$$
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} = \left( {x - {\pi \over 6}} \right)$$
|
mcq
|
jee-main-2019-online-8th-april-morning-slot
|
9rwVQ5XrboppQiu5tE7k9k2k5gqvdkp
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
Let ƒ(x) = (sin(tan<sup>–1</sup>x) + sin(cot<sup>–1</sup>x))<sup>2</sup> – 1, |x| > 1.
<br/>If $${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$$ and $$y\left( {\sqrt 3 } \right) = {\pi \over 6}$$,
then y($${ - \sqrt 3 }$$) is equal to :
|
[{"identifier": "A", "content": "$${{5\\pi } \\over 6}$$"}, {"identifier": "B", "content": "$$ - {\\pi \\over 6}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${{2\\pi } \\over 3}$$"}]
|
["B"]
| null |
Given ƒ(x) = (sin(tan<sup>–1</sup>x) + sin(cot<sup>–1</sup>x))<sup>2</sup> – 1
<br><br> = (sin(tan<sup>–1</sup>x) + sin($${\pi \over 2}$$ - tan<sup>–1</sup>x))<sup>2</sup> – 1
<br><br> = (sin(tan<sup>–1</sup>x) + cos(tan<sup>–1</sup>x))<sup>2</sup> – 1
<br><br>= sin<sup>2</sup>(tan<sup>–1</sup>x) + cos<sup>2</sup>(tan<sup>–1</sup>x) + 2sin(tan<sup>–1</sup>x)cos(tan<sup>–1</sup>x) + 1
<br><br>= 1 + sin(2tan<sup>–1</sup>x) - 1
<br><br>= sin(2tan<sup>–1</sup>x)
<br><br>Also given $${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$$
<br><br> Integrating both sides we get
<br><br>y = $${1 \over 2}$$ sin<sup>-1</sup> (f(x)) + C
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (sin(2tan<sup>–1</sup>x)) + C
<br><br>Given $$y\left( {\sqrt 3 } \right) = {\pi \over 6}$$ mean x = $$\sqrt 3 $$ and y = $${\pi \over 6}$$
<br><br>$$ \therefore $$ $${\pi \over 6}$$ = $${1 \over 2}$$ sin<sup>-1</sup> (sin(2tan<sup>–1</sup>$$\sqrt 3 $$)) + C
<br><br>$$ \Rightarrow $$ $${\pi \over 6}$$ = $${1 \over 2}$$ sin<sup>-1</sup> (sin(2$$ \times $$$${\pi \over 3}$$)) + C
<br><br>$$ \Rightarrow $$ $${\pi \over 6}$$ = $${1 \over 2}$$ sin<sup>-1</sup> ($${{\sqrt 3 } \over 2}$$) + C
<br><br>$$ \Rightarrow $$ $${\pi \over 6}$$ = $${1 \over 2}$$ $$ \times $$ $${\pi \over 3}$$ + C
<br><br>$$ \Rightarrow $$ C = 0
<br><br>Now y($${ - \sqrt 3 }$$) means when x = $${ - \sqrt 3 }$$ then find y.
<br><br>y = $${1 \over 2}$$ sin<sup>-1</sup> (sin(2tan<sup>–1</sup>x))
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (sin(2tan<sup>–1</sup>($${ - \sqrt 3 }$$)))
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (sin(-2tan<sup>–1</sup>($${ \sqrt 3 }$$)))
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (sin(-2$$ \times $$$${\pi \over 3}$$))
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (-sin(2$$ \times $$$${\pi \over 3}$$))
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (-$${{\sqrt 3 } \over 2}$$)
<br><br>= $${1 \over 2}$$ $$ \times $$ -sin<sup>-1</sup> ($${{\sqrt 3 } \over 2}$$)
<br><br>= $${1 \over 2}$$ $$ \times $$ -$${\pi \over 3}$$
<br><br>= -$${\pi \over 6}$$
|
mcq
|
jee-main-2020-online-8th-january-morning-slot
|
o7hlhlRCMBt3fIKQdBjgy2xukezff9hs
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
If y = $$\sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left\{ {{3 \over 5}\cos kx - {4 \over 5}\sin kx} \right\}} $$,
<br/><br/>then $${{dy} \over {dx}}$$ at x = 0 is _______.
|
[]
| null |
91
|
Put, $$\cos \alpha = {3 \over 5},\sin \alpha = {4 \over 5}$$<br><br>
$$ \therefore {3 \over 5}\cos kx - {4 \over 5}\sin \,kx$$<br><br>
$$ = \cos \alpha .\cos kx - \sin \alpha .\sin kx$$<br><br>
$$ = \cos \left( {\alpha + kx} \right)$$<br><br>
So, $$y = \sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left( {\cos \left( {\alpha + kx} \right)} \right)} $$<br><br>
$$ = \sum\limits_{k = 1}^6 {\left( {{k^2}x + kx} \right)} $$<br><br>
$$ \Rightarrow {{dy} \over {dx}} = \sum\limits_{k = 1}^6 {{k^2}} $$<br><br>
$$ = {{6 \times 7 \times 13} \over 6} = 91$$
|
integer
|
jee-main-2020-online-2nd-september-evening-slot
|
D1eT2RJT4etOfyTR0y1kmjcgbe6
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
If $$f(x) = \sin \left( {{{\cos }^{ - 1}}\left( {{{1 - {2^{2x}}} \over {1 + {2^{2x}}}}} \right)} \right)$$ and its first derivative with respect to x is $$ - {b \over a}{\log _e}2$$ when x = 1, where a and b are integers, then the minimum value of | a<sup>2</sup> $$-$$ b<sup>2</sup> | is ____________ .
|
[]
| null |
481
|
$$f(x) = \sin {\cos ^{ - 1}}\left( {{{1 - {{({2^x})}^2}} \over {1 + {{({2^x})}^2}}}} \right)$$<br><br>$$ = \sin (2{\tan ^{ - 1}}{2^x})$$<br><br>$$f'(x) = \cos (2{\tan ^{ - 1}}{2^x}).2.{1 \over {1 + {{({2^x})}^2}}} \times {2^x}.{\log _e}2$$<br><br>$$ \therefore $$ $$f'(1) = \cos (2{\tan ^{ - 1}}2).{2 \over {1 + 4}} \times 2 \times {\log _e}2$$<br><br>$$ \Rightarrow f'(1) = \cos {\cos ^{ - 1}}\left( {{{1 - {2^2}} \over {1 + {2^2}}}} \right).{4 \over 5}{\log _e}2$$<br><br>$$ = - {{12} \over {25}}{\log _e}2$$<br><br>$$ \Rightarrow a = 25,b = 12$$<br><br>$$|{a^2} - {b^2}| = |625 - 144| = 481$$
|
integer
|
jee-main-2021-online-17th-march-morning-shift
|
1ktbc8v27
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then :
|
[{"identifier": "A", "content": "$${(1 - x)^2}f'(x) - 2{(f(x))^2} = 0$$"}, {"identifier": "B", "content": "$${(1 + x)^2}f'(x) + 2{(f(x))^2} = 0$$"}, {"identifier": "C", "content": "$${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$"}, {"identifier": "D", "content": "$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$"}]
|
["C"]
| null |
$$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$<br><br>$${\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x $$<br><br>or $$f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )$$<br><br>$$ = \cos {\tan ^{ - 1}}\left( {{{2\sqrt x } \over {1 - x}}} \right)$$<br><br>$$f(x) = {{1 - x} \over {1 + x}}$$<br><br>Now, $$f'(x) = {{ - 2} \over {{{(1 + x)}^2}}}$$<br><br>or $$f'(x){(1 - x)^2} = - 2{\left( {{{1 - x} \over {1 + x}}} \right)^2}$$<br><br>or $${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$.
|
mcq
|
jee-main-2021-online-26th-august-morning-shift
|
1ktg2zxf8
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
If $$y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right)$$, then $${{dy} \over {dx}}$$ at $$x = {{5\pi } \over 6}$$ is :
|
[{"identifier": "A", "content": "$$ - {1 \\over 2}$$"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "0"}]
|
["A"]
| null |
We have,
<br/><br/>$$
y(x)=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)
$$
<br/><br/>$$
=\cot ^{-1} \frac{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|+\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|-\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}
$$
<br/><br/>$$
\left[\text { as } \cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}=1\right.
$$
<br/><br/>and $\left.\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\right]$
<br/><br/>$$
\begin{aligned}
& =\cot ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}\right) \forall x \in\left(\frac{\pi}{2}, \pi\right) \\
&
\end{aligned}
$$
<br/><br/>$$
=\cot ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)=\cot ^{-1}\left(\tan \frac{x}{2}\right)
$$
<br/><br/>$$
=\frac{\pi}{2}-\tan ^{-1}\left(\tan \frac{x}{2}\right)
$$
<br/><br/>$$
\begin{aligned}
& \therefore y^{\prime}(x)=\frac{\pi}{2}-\frac{x}{2} \\\\
& \Rightarrow \frac{d y}{d x}=y^{\prime}(x)=-\frac{1}{2}=y^{\prime}\left(\frac{5 \pi}{6}\right)
\end{aligned}
$$
|
mcq
|
jee-main-2021-online-27th-august-evening-shift
|
1l5banuqb
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
<p>If $$y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}$$, then</p>
|
[{"identifier": "A", "content": "$$xy'' + 2y' = 0$$"}, {"identifier": "B", "content": "$${x^2}y'' - 6y + {{3\\pi } \\over 2} = 0$$"}, {"identifier": "C", "content": "$${x^2}y'' - 6y + 3\\pi = 0$$"}, {"identifier": "D", "content": "$$xy'' - 4y' = 0$$"}]
|
["B"]
| null |
<p>Let $${x^3} = \theta \Rightarrow {\theta \over 2} \in \left( {{\pi \over 4},\,{{3\pi } \over 4}} \right)$$</p>
<p>$$\therefore$$ $$y = {\tan ^{ - 1}}(\sec \theta - \tan \theta )$$</p>
<p>$$ = {\tan ^{ - 1}}\left( {{{1 - \sin \theta } \over {\cos \theta }}} \right)$$</p>
<p>$$\therefore$$ $$y = {\pi \over 4} - {\theta \over 2}$$</p>
<p>$$y = {\pi \over 4} - {{{x^3}} \over 2}$$</p>
<p>$$\therefore$$ $$y' = {{ - 3{x^2}} \over 2}$$</p>
<p>$$y'' = - 3x$$</p>
<p>$$\therefore$$ $${x^2}y'' - 6y + {{3\pi } \over 2} = 0$$</p>
|
mcq
|
jee-main-2022-online-24th-june-evening-shift
|
luxwdj6m
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
<p>If $$\log _e y=3 \sin ^{-1} x$$, then $$(1-x^2) y^{\prime \prime}-x y^{\prime}$$ at $$x=\frac{1}{2}$$ is equal to</p>
|
[{"identifier": "A", "content": "$$9 e^{\\pi / 2}$$\n"}, {"identifier": "B", "content": "$$9 e^{\\pi / 6}$$\n"}, {"identifier": "C", "content": "$$3 e^{\\pi / 2}$$\n"}, {"identifier": "D", "content": "$$3 e^{\\pi / 6}$$"}]
|
["A"]
| null |
<p>$$\begin{aligned}
&\log _e y=3 \sin ^{-1} x\\
&\begin{aligned}
& y=e^{3 \sin ^{-1} x} \\
& \frac{d y}{d x}=e^{3 \sin ^{-1} x} \cdot \frac{3}{\sqrt{1-x^2}}
\end{aligned}
\end{aligned}$$</p>
<p>$$\sqrt{1-x^2} \frac{d y}{d x}=3 y$$</p>
<p>Again differentiate</p>
<p>$$\begin{aligned}
& \sqrt{1-x^2} \cdot y^{\prime \prime}-\frac{2 x}{2 \sqrt{1-x^2}} y^{\prime}=3 y^{\prime} \\
& (1-x)^2 y^{\prime \prime}-x y^{\prime}=3 y^{\prime}\left(\sqrt{1-x^2}\right)
\end{aligned}$$</p>
<p>So value of $$3 y^{\prime}\left(\sqrt{1-x^2}\right)$$ at $$x=\frac{1}{2}$$</p>
<p>$$\begin{aligned}
& 3 \cdot \frac{3}{\sqrt{1-x^2}} e^{\sin ^{-1} x}\left(\sqrt{1-x^2}\right) \\
& =9 e^{3 \frac{\pi}{6}}=9 e^{\frac{\pi}{2}}
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-9th-april-evening-shift
|
lvc57bcq
|
maths
|
differentiation
|
differentiation-of-inverse-trigonometric-function
|
<p>Let $$f:(-\infty, \infty)-\{0\} \rightarrow \mathbb{R}$$ be a differentiable function such that $$f^{\prime}(1)=\lim _\limits{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right)$$. Then $$\lim _\limits{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^2-2 \log _e a$$ is equal to</p>
|
[{"identifier": "A", "content": "$$\\frac{5}{2}+\\frac{\\pi}{8}$$\n"}, {"identifier": "B", "content": "$$\\frac{3}{8}+\\frac{\\pi}{4}$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{4}+\\frac{\\pi}{8}$$\n"}, {"identifier": "D", "content": "$$\\frac{3}{2}+\\frac{\\pi}{4}$$"}]
|
["A"]
| null |
<p>Let $$f^{\prime}(1)=k$$</p>
<p>$$\Rightarrow \quad \lim _\limits{x \rightarrow 0} \frac{f(x)}{x^2}=k \quad\left(\frac{0}{0}\right)$$</p>
<p>$$\begin{aligned}
& \lim _\limits{x \rightarrow 0} \frac{f^{\prime}(x)}{2 x}=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)}{2}=k \\
\Rightarrow & f^{\prime \prime}(0)=2 k
\end{aligned}$$</p>
<p>Given information is not complete.</p>
|
mcq
|
jee-main-2024-online-6th-april-morning-shift
|
3BblMHpmENyy47EL
|
maths
|
differentiation
|
differentiation-of-logarithmic-function
|
If $$x = {e^{y + {e^y} + {e^{y + .....\infty }}}}$$ , $$x > 0,$$ then $${{{dy} \over {dx}}}$$ is
|
[{"identifier": "A", "content": "$${{1 + x} \\over x}$$ "}, {"identifier": "B", "content": "$${1 \\over x}$$ "}, {"identifier": "C", "content": "$${{1 - x} \\over x}$$"}, {"identifier": "D", "content": "$${x \\over {1 + x}}$$ "}]
|
["C"]
| null |
$$x = {e^{y + {e^{y + .....\infty }}}}\,\, \Rightarrow x = {e^{y + x}}.$$
<br><br>Taking log.
<br><br>$$\log \,\,x = y + x$$
<br><br>$$ \Rightarrow {1 \over x} = {{dy} \over {dx}} + 1$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {1 \over x} - 1 = {{1 - x} \over x}$$
|
mcq
|
aieee-2004
|
jmcGBPjprXE1bdtL
|
maths
|
differentiation
|
differentiation-of-logarithmic-function
|
If $${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}},$$ then $${{{dy} \over {dx}}}$$ is
|
[{"identifier": "A", "content": "$${y \\over x}$$ "}, {"identifier": "B", "content": "$${{x + y} \\over {xy}}$$ "}, {"identifier": "C", "content": "$$xy$$ "}, {"identifier": "D", "content": "$${x \\over y}$$"}]
|
["A"]
| null |
$${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}$$
<br><br>$$ \Rightarrow m\ln x + n\ln y = \left( {m + n} \right)\ln \left( {x + y} \right)$$
<br><br>Differentiating both sides.
<br><br>$$\therefore$$ $${m \over x} + {n \over y}{{dy} \over {dx}} = {{m + n} \over {x + y}}\left( {1 + {{dy} \over {dx}}} \right)$$
<br><br>$$ \Rightarrow \left( {{m \over x} - {{m + n} \over {x + y}}} \right) = \left( {{{m + n} \over {x + y}} - {n \over y}} \right){{dy} \over {dx}}$$
<br><br>$$ \Rightarrow {{my - nx} \over {x\left( {x + y} \right)}} = \left( {{{my - nx} \over {y\left( {x + y} \right)}}} \right){{dy} \over {dx}}$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {y \over x}$$
|
mcq
|
aieee-2006
|
oMWsMZ2BcQMVesDBdN3ng
|
maths
|
differentiation
|
differentiation-of-logarithmic-function
|
If xlog<sub>e</sub>(log<sub>e</sub>x) $$-$$ x<sup>2</sup> + y<sup>2</sup> = 4(y > 0), then $${{dy} \over {dx}}$$ at x = e is equal to :
|
[{"identifier": "A", "content": "$${{\\left( {1 + 2e} \\right)} \\over {2\\sqrt {4 + {e^2}} }}$$"}, {"identifier": "B", "content": "$${{\\left( {1 + 2e} \\right)} \\over {\\sqrt {4 + {e^2}} }}$$"}, {"identifier": "C", "content": "$${{\\left( {2e - 1} \\right)} \\over {2\\sqrt {4 + {e^2}} }}$$"}, {"identifier": "D", "content": "$${e \\over {\\sqrt {4 + {e^2}} }}$$"}]
|
["C"]
| null |
Differentiating with respect to x,
<br><br>$$x.{1 \over {\ell nx}}.{1 \over x} + \ell n(\ell nx) - 2x + 2y.{{dy} \over {dx}} = 0$$
<br><br>at $$x = e$$ we get
<br><br>$$1 - 2e + 2y{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2y}}$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2\sqrt {4 + {e^2}} }}\,\,$$
<br><br>as $$y(e) = \sqrt {4 + {e^2}} $$
|
mcq
|
jee-main-2019-online-11th-january-morning-slot
|
ltTrASebDziGIvmUP3p9h
|
maths
|
differentiation
|
differentiation-of-logarithmic-function
|
For x > 1, if (2x)<sup>2y</sup> = 4e<sup>2x$$-$$2y</sup>,
<br/><br/>then (1 + log<sub>e</sub> 2x)<sup>2</sup> $${{dy} \over {dx}}$$ is equal to :
|
[{"identifier": "A", "content": "$${{x\\,{{\\log }_e}2x - {{\\log }_e}2} \\over x}$$"}, {"identifier": "B", "content": "log<sub>e</sub> 2x"}, {"identifier": "C", "content": "x log<sub>e</sub> 2x"}, {"identifier": "D", "content": "$${{x\\,{{\\log }_e}2x + {{\\log }_e}2} \\over x}$$"}]
|
["A"]
| null |
(2x)<sup>2y</sup> = 4e<sup>2x-2y</sup>
<br><br>2y$$\ell $$n2x = $$\ell $$n4 + 2x $$-$$ 2y
<br><br>y = $${{x + \ell n2} \over {1 + \ell n2x}}$$
<br><br>y ' = $${{\left( {1 + \ell n2x} \right) - \left( {x + \ell n2} \right){1 \over x}} \over {{{\left( {1 + \ell n2x} \right)}^2}}}$$
<br><br>y '$${\left( {1 + \ell n2x} \right)^2} = \left[ {{{x\ell n2x - \ell n2} \over x}} \right]$$
|
mcq
|
jee-main-2019-online-12th-january-morning-slot
|
1ktbitz1b
|
maths
|
differentiation
|
differentiation-of-logarithmic-function
|
If y = y(x) is an implicit function of x such that log<sub>e</sub>(x + y) = 4xy, then $${{{d^2}y} \over {d{x^2}}}$$ at x = 0 is equal to ___________.
|
[]
| null |
40
|
ln(x + y) = 4xy (At x = 0, y = 1)<br><br>x + y = e<sup>4xy</sup><br><br>$$ \Rightarrow 1 + {{dy} \over {dx}} = {e^{4xy}}\left( {4x{{dy} \over {dx}} + 4y} \right)$$<br><br>At x = 0 <br><br>$${{dy} \over {dx}} = 3$$<br><br>$${{{d^2}y} \over {d{x^2}}} = {e^{4xy}}{\left( {4x{{dy} \over {dx}} + 4y} \right)^2} + {e^{4xy}}\left( {4x{{{d^2}y} \over {d{x^2}}} + 4y} \right)$$<br><br>At x = 0, $${{{d^2}y} \over {d{x^2}}} = {e^0}{(4)^2} + {e^0}(24)$$<br><br>$$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = 40$$
|
integer
|
jee-main-2021-online-26th-august-morning-shift
|
1l57o8xuz
|
maths
|
differentiation
|
differentiation-of-logarithmic-function
|
<p>If $${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5},\,|y| < 2$$, then :</p>
|
[{"identifier": "A", "content": "$${x^2}y'' + xy' - 25y = 0$$"}, {"identifier": "B", "content": "$${x^2}y'' - xy' - 25y = 0$$"}, {"identifier": "C", "content": "$${x^2}y'' - xy' + 25y = 0$$"}, {"identifier": "D", "content": "$${x^2}y'' + xy' + 25y = 0$$"}]
|
["D"]
| null |
<p>$${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5}\,\,\,\,\,\,\,\,\,|y| < 2$$</p>
<p>Differentiating on both side</p>
<p>$$ - {1 \over {\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} }} \times {{y'} \over 2} = {5 \over {{x \over 5}}} \times {1 \over 5}$$</p>
<p>$${{ - xy'} \over 2} = 5\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} $$</p>
<p>Square on both side</p>
<p>$${{{x^2}y{'^2}} \over 4} = 25\left( {{{4 - {y^2}} \over 4}} \right)$$</p>
<p>Diff on both side</p>
<p>$$2xy{'^2} + 2y'y''{x^2} = - 25 \times 2yy'$$</p>
<p>$$xy' + y''{x^2} + 25y = 0$$</p>
|
mcq
|
jee-main-2022-online-27th-june-morning-shift
|
1l58gt0p9
|
maths
|
differentiation
|
differentiation-of-logarithmic-function
|
<p>Let f : R $$\to$$ R satisfy $$f(x + y) = {2^x}f(y) + {4^y}f(x)$$, $$\forall$$x, y $$\in$$ R. If f(2) = 3, then $$14.\,{{f'(4)} \over {f'(2)}}$$ is equal to ____________.</p>
|
[]
| null |
248
|
<p>$$\because$$ $$f(x + y) = {2^x}f(y) + {4^y}f(x)$$ ....... (1)</p>
<p>Now, $$f(y + x){2^y}f(x) + {4^x}f(y)$$ ...... (2)</p>
<p>$$\therefore$$ $${2^x}f(y) + {4^y}f(x) = {2^y}f(x) + {4^x}f(y)$$</p>
<p>$$({4^y} - {2^y})f(x) = ({4^x} - {2^x})f(y)$$</p>
<p>$${{f(x)} \over {{4^x} - {2^x}}} = {{f(y)} \over {{4^y} - {2^y}}} = k$$</p>
<p>$$\therefore$$ $$f(x) = k({4^x} - {2^x})$$</p>
<p>$$\because$$ $$f(2) = 3$$ then $$k = {1 \over 4}$$</p>
<p>$$\therefore$$ $$f(x) = {{{4^x} - {2^x}} \over 4}$$</p>
<p>$$\therefore$$ $$f'(x) = {{{4^x}\ln 4 - {2^x}\ln 2} \over 4}$$</p>
<p>$$f'(x) = {{({{2.4}^x} - {2^x})ln2} \over 4}$$</p>
<p>$$\therefore$$ $${{f'(4)} \over {f'(2)}} = {{2.256 - 16} \over {2.16 - 4}}$$</p>
<p>$$\therefore$$ $$14{{f'(4)} \over {f'(2)}} = 248$$</p>
|
integer
|
jee-main-2022-online-26th-june-evening-shift
|
1l6hy9kbq
|
maths
|
differentiation
|
differentiation-of-logarithmic-function
|
<p>The value of $$\log _{e} 2 \frac{d}{d x}\left(\log _{\cos x} \operatorname{cosec} x\right)$$ at $$x=\frac{\pi}{4}$$ is</p>
|
[{"identifier": "A", "content": "$$-2 \\sqrt{2}$$"}, {"identifier": "B", "content": "$$2 \\sqrt{2}$$"}, {"identifier": "C", "content": "$$-4$$"}, {"identifier": "D", "content": "4"}]
|
["D"]
| null |
<p>Let $$f(x) = {\log _{\cos x}}\cos ec\,x$$</p>
<p>$$ = {{\log \cos ec\,x} \over {\log \cos x}}$$</p>
<p>$$ \Rightarrow f'(x) = {{\log \cos x\,.\,\sin x\,.\,\left( { - \cos ec\,x\cot x - \log \cos ec\,x\,.\,{1 \over {\cos x}}\,.\, - \sin x} \right)} \over {{{(\log \cos x)}^2}}}$$</p>
<p>at $$x = {\pi \over 4}$$</p>
<p>$$f'\left( {{\pi \over 4}} \right) = {{ - \log \left( {{1 \over {\sqrt 2 }}} \right) + \log \sqrt 2 } \over {{{\left( {\log {1 \over {\sqrt 2 }}} \right)}^2}}} = {2 \over {\log \sqrt 2 }}$$</p>
<p>$$\therefore$$ $${\log _e}2f'(x)$$ at $$x = {\pi \over 4} = 4$$</p>
|
mcq
|
jee-main-2022-online-26th-july-evening-shift
|
1ldo5zk2p
|
maths
|
differentiation
|
differentiation-of-logarithmic-function
|
<p>If $$y(x)=x^{x},x > 0$$, then $$y''(2)-2y'(2)$$ is equal to</p>
|
[{"identifier": "A", "content": "$$4(\\log_{e}2)^{2}+2$$"}, {"identifier": "B", "content": "$$8\\log_{e}2-2$$"}, {"identifier": "C", "content": "$$4\\log_{e}2+2$$"}, {"identifier": "D", "content": "$$4(\\log_{e}2)^{2}-2$$"}]
|
["D"]
| null |
$\begin{aligned} & y=x^x \\\\ & y^{\prime}=x^x(1+\ln x) \\\\ & y^{\prime \prime}=x^x(1+\ln x)^2+\frac{x^x}{x} \\\\ & f^{\prime \prime}(2)-2 f^{\prime}(2)=\left(4(1+\ln 2)^2+2\right)-(2)(4(1+\ln 2)) \\\\ & =4\left(1+(\ln 2)^2\right)+2-8 \\\\ & =4(\ln 2)^2-2 \\\\ & \end{aligned}$
|
mcq
|
jee-main-2023-online-1st-february-evening-shift
|
1lh23oisg
|
maths
|
differentiation
|
differentiation-of-logarithmic-function
|
<p>If $$2 x^{y}+3 y^{x}=20$$, then $$\frac{d y}{d x}$$ at $$(2,2)$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$-\\left(\\frac{3+\\log _{e} 16}{4+\\log _{e} 8}\\right)$$"}, {"identifier": "B", "content": "$$-\\left(\\frac{2+\\log _{e} 8}{3+\\log _{e} 4}\\right)$$"}, {"identifier": "C", "content": "$$-\\left(\\frac{3+\\log _{e} 8}{2+\\log _{e} 4}\\right)$$"}, {"identifier": "D", "content": "$$-\\left(\\frac{3+\\log _{e} 4}{2+\\log _{e} 8}\\right)$$"}]
|
["B"]
| null |
Given, $2 x^y+3 y^x=20$ ..........(i)
<br/><br/>Let $u=x^y$
<br/><br/>On taking log both sides, we get
<br/><br/>$\log u=y \log x$
<br/><br/>On differentiating both sides with respect to $x$, we get
<br/><br/>$$
\begin{array}{rlrl}
& \frac{1}{u} \frac{d u}{d x} =y \frac{1}{x}+\log x \frac{d y}{d x} \\\\
& \Rightarrow \frac{d u}{d x} =u\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) \\\\
& \Rightarrow \frac{d u}{d x} =x^y\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) .........(ii)
\end{array}
$$
<br/><br/>Also, let $v=y^x$
<br/><br/>On taking log both sides, we get
<br/><br/>$\log v=x \log y$
<br/><br/>On differentiating both sides, we get
<br/><br/>$$
\begin{array}{rlrl}
& \frac{1}{v} \frac{d v}{d x} =x \frac{1}{y} \frac{d y}{d x}+\log y \cdot 1 \\\\
&\Rightarrow \frac{d v}{d x} =v\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) \\\\
&\Rightarrow \frac{d v}{d x} =y^x\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) ..........(iii)
\end{array}
$$
<br/><br/>Now, from Equation (i), $2 u+3 v=20$
<br/><br/>$$
\begin{aligned}
& \Rightarrow 2 \frac{d u}{d x}+3 \frac{d v}{d x}=0 \\\\
& \Rightarrow 2 x^y\left(\frac{y}{x}+\log x \frac{d y}{d x}\right)+3 y^x\left(\frac{x}{y} \frac{d y}{d x}+\log y\right)=0 \text { [Using Eqs. (ii) and (iii)] }
\end{aligned}
$$
<br/><br/>On putting $x=2$ and $y=2$, we get
<br/><br/>$$
\begin{aligned}
& 2(4)\left(1+\log 2 \frac{d y}{d x}\right)+3(4)\left(\frac{d y}{d x}+\log 2\right)=0 \\\\
& \Rightarrow \frac{d y}{d x}(8 \log 2+12)+(8+12 \log 2)=0 \\\\
& \Rightarrow \frac{d y}{d x}=-\left(\frac{2+3 \log 2}{3+2 \log 2}\right) \\\\
& \Rightarrow \frac{d y}{d x}=-\left(\frac{2+\log 8}{3+\log 4}\right)
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-6th-april-morning-shift
|
jaoe38c1lsfkl66w
|
maths
|
differentiation
|
differentiation-of-logarithmic-function
|
<p>$$\text { Let } y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1 < x<1 \text {. Then at } x=\frac{1}{2} \text {, the value of } 225\left(y^{\prime}-y^{\prime \prime}\right) \text { is equal to }$$</p>
|
[{"identifier": "A", "content": "732"}, {"identifier": "B", "content": "736"}, {"identifier": "C", "content": "742"}, {"identifier": "D", "content": "746"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& y=\log _e\left(\frac{1-x^2}{1+x^2}\right) \\
& \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4}
\end{aligned}$$</p>
<p>Again,</p>
<p>$$\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$$</p>
<p>Again</p>
<p>$$y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$$</p>
<p>at $$\mathrm{x}=\frac{1}{2}$$,</p>
<p>$$y^{\prime}-y^{\prime \prime}=\frac{736}{225}$$</p>
<p>Thus $$225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736$$</p>
|
mcq
|
jee-main-2024-online-29th-january-evening-shift
|
ghiwTEg0WvaIpfIxQkmJb
|
maths
|
differentiation
|
differentiation-of-parametric-function
|
If x $$=$$ 3 tan t and y $$=$$ 3 sec t, then the value of $${{{d^2}y} \over {d{x^2}}}$$ at t $$ = {\pi \over 4},$$ is :
|
[{"identifier": "A", "content": "$${1 \\over {3\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over {6\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${3 \\over {2\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over 6}$$"}]
|
["B"]
| null |
x = 3 tan t and y = 3 sec t
<br><br>So that $${{dx} \over {dt}}$$ = 3sec<sup>2</sup>t and $${{dy} \over {dt}}$$ = 3 sec t tan t
<br><br>$${{dy} \over {dx}}$$ = $${{dy/dt} \over {dx/dt}}$$ = sin t
<br><br>$${{{d^2}y} \over {d{x^2}}}$$ = (cos t)$$.{{dt} \over {dx}}$$
<br><br>$${{{d^2}y} \over {d{x^2}}} = \left( {\cos t} \right).{1 \over {3{{\sec }^2}t}}$$
<br><br>$${{{d^2}y} \over {d{x^2}}}$$ = $${1 \over 3}$$(cos<sup>3</sup> t)
<br><br>$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{t = \pi /4}}$$ = $${1 \over 3} \times {\left( {{1 \over {\sqrt 2 }}} \right)^3} = {1 \over {6\sqrt 2 }}$$
|
mcq
|
jee-main-2019-online-9th-january-evening-slot
|
JDlJzbuL7enwokKQJU7k9k2k5k6v2n8
|
maths
|
differentiation
|
differentiation-of-parametric-function
|
If $$x = 2\sin \theta - \sin 2\theta $$ and $$y = 2\cos \theta - \cos 2\theta $$,<br/>
$$\theta \in \left[ {0,2\pi } \right]$$, then $${{{d^2}y} \over {d{x^2}}}$$ at $$\theta $$ = $$\pi $$ is :
|
[{"identifier": "A", "content": "$${3 \\over 8}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "$${3 \\over 4}$$"}, {"identifier": "D", "content": "-$${3 \\over 4}$$"}]
|
["A"]
| null |
$$x = 2\sin \theta - \sin 2\theta $$
<br><br>$$ \Rightarrow $$ $${{dx} \over {d\theta }}$$ = $$2\cos \theta - 2\cos 2\theta $$
<br><br>$$y = 2\cos \theta - \cos 2\theta $$
<br><br>$$ \Rightarrow $$ $${{dy} \over {d\theta }}$$ = –2sin$$\theta $$ + 2sin2$$\theta $$
<br><br>$${{dy} \over {dx}} = {{{{dy} \over {d\theta }}} \over {{{dx} \over {d\theta }}}}$$ = $${{\sin 2\theta - \sin \theta } \over {\cos \theta - \cos 2\theta }}$$
<br/><br/>This expression is simplified by recognizing that $\sin 2 \theta = 2\sin \theta \cos \theta$ and $\cos 2 \theta = 2\cos^2 \theta -1$, leading to the expression:
<br/><br/>$$ \frac{dy}{dx} = \frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \cdot \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}$$
<br/><br/>This is the rate of change of y with respect to x, as a function of θ.
<br/><br/>Next, we differentiate this function with respect to θ to find $\frac{d^2 y}{dx^2}$, yielding :
<br/><br/>$$ \frac{d^2 y}{dx^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x}=-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}$$
<br/><br/>Here, $\frac{d \theta}{d x}$ is the reciprocal of $\frac{dx}{d\theta}$, so the equation becomes :
<br/><br/>$$ \frac{d^2 y}{dx^2}=\frac{-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2}}{2\left(\cos \theta-\cos2 \theta\right)}$$
<br/><br/>Finally, we evaluate this expression at $\theta = \pi$, yielding :
<br/><br/>$$ \frac{d^2 y}{dx^2}(\pi)=\frac{-3}{4(-1-1)}=\frac{3}{8}$$
<br/><br/>Therefore, the correct answer is (A) $\frac{3}{8}$.
<br/><br/><b>Alternate Method :</b>
<br/><br/>First, let's find the derivatives of x and y with respect to θ :
<br/><br/>1) $$ \frac{dx}{d\theta} = 2\cos \theta - 2\cos 2\theta $$
<br/><br/>2) $$ \frac{dy}{d\theta} = -2\sin \theta + 2\sin 2\theta $$
<br/><br/>We know that $$ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} $$
<br/><br/>So, we substitute 1) and 2) into this equation :
<br/><br/>We have
<br/><br/>$$\frac{dy}{dx} = \frac{-2\sin \theta + 2\sin 2\theta}{2\cos \theta - 2\cos 2\theta} = \frac{\sin 2\theta - \sin \theta}{\cos \theta - \cos 2\theta}$$
<br/><br/>For simplification, let's denote the numerator as $$N = \sin 2\theta - \sin \theta$$ and the denominator as $$D = \cos \theta - \cos 2\theta$$.
<br/><br/>We have to compute $$\frac{d}{d\theta}(\frac{dy}{dx})$$ which is $$\frac{d}{d\theta}(\frac{N}{D})$$.
<br/><br/>We can use the quotient rule for differentiation, which states that if we have a function of the form $$\frac{u}{v}$$, then its derivative is given by $$\frac{vu' - uv'}{v^2}$$.
<br/><br/>So here, $$N' = 2\cos 2\theta - \cos \theta$$ and $$D' = -\sin \theta + 2\sin 2\theta$$.
<br/><br/>Applying the quotient rule :
<br/><br/>$$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{D N' - N D'}{D^2}$$
<br/><br/>Substituting the expressions for $$N'$$, $$D'$$, $$N$$, and $$D$$ we get :
<br/><br/>$$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$$
<br/><br/>Now $$
\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \times \frac{d \theta}{d x}
$$
<br/><br/>= $$\frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$$$$
\times \frac{1}{(2 \cos \theta-2 \cos 2 \theta)}
$$
<br/><br/>$$
\begin{aligned}
\left.\therefore \frac{d^2 y}{d x^2}\right|_{\theta=\pi} & =\frac{(-1-1)(2+1)-(0-0)(-0+0)}{2(-1-1)^3} \\\\
& =\frac{-2 \times 3}{-2 \times 8}=\frac{3}{8}
\end{aligned}
$$
|
mcq
|
jee-main-2020-online-9th-january-evening-slot
|
1l6nm5311
|
maths
|
differentiation
|
differentiation-of-parametric-function
|
<p>Let $$x(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t}$$ and <br/><br/>$$y(t)=2 \sqrt{2} \sin t \sqrt{\sin 2 t}, t \in\left(0, \frac{\pi}{2}\right)$$. <br/><br/>Then $$\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}$$ at $$t=\frac{\pi}{4}$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\frac{-2 \\sqrt{2}}{3}$$"}, {"identifier": "B", "content": "$$\\frac{2}{3}$$"}, {"identifier": "C", "content": "$$\\frac{1}{3}$$"}, {"identifier": "D", "content": "$$ \\frac{-2}{3}$$"}]
|
["D"]
| null |
<p>$$x = 2\sqrt 2 \cos t\sqrt {\sin 2t} ,\,y = 2\sqrt 2 \sin t\sqrt {\sin 2t} $$</p>
<p>$$\therefore$$ $${{dx} \over {dt}} = {{2\sqrt 2 \cos 3t} \over {\sqrt {\sin 2t} }},\,{{dy} \over {dt}} = {{2\sqrt 2 \sin 3t} \over {\sqrt {\sin 2t} }}$$</p>
<p>$$\therefore$$ $${{dy} \over {dx}} = \tan 3t,\,\left( {\mathrm{at}\,t = {\pi \over 4},\,{{dy} \over {dx}} = - 1} \right)$$</p>
<p>and $${{{d^2}y} \over {d{x^2}}} = 3{\sec ^2}3t\,.\,{{dt} \over {dx}} = {{3{{\sec }^2}3t\,.\,\sqrt {\sin 2t} } \over {2\sqrt 2 \cos 3t}}$$</p>
<p>$$\left( {\mathrm{At}\,t = {\pi \over 4},\,{{{d^2}y} \over {d{x^2}}} = - 3} \right)$$</p>
<p>$$\therefore$$ $${{1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \over {{{{d^2}y} \over {d{x^2}}}}} = {2 \over { - 3}} = {{ - 2} \over 3}$$</p>
|
mcq
|
jee-main-2022-online-28th-july-evening-shift
|
lmUN4HWdFdPr7roD
|
maths
|
differentiation
|
methods-of-differentiation
|
$${{{d^2}x} \over {d{y^2}}}$$ equals:
|
[{"identifier": "A", "content": "$$ - {\\left( {{{{d^2}y} \\over {d{x^2}}}} \\right)^{ - 1}}{\\left( {{{dy} \\over {dx}}} \\right)^{ - 3}}$$ "}, {"identifier": "B", "content": "$${\\left( {{{{d^2}y} \\over {d{x^2}}}} \\right)^{}}{\\left( {{{dy} \\over {dx}}} \\right)^{ - 2}}$$ "}, {"identifier": "C", "content": "$$ - \\left( {{{{d^2}y} \\over {d{x^2}}}} \\right){\\left( {{{dy} \\over {dx}}} \\right)^{ - 3}}$$ "}, {"identifier": "D", "content": "$${\\left( {{{{d^2}y} \\over {d{x^2}}}} \\right)^{ - 1}}$$ "}]
|
["C"]
| null |
$${{{d^2}x} \over {d{y^2}}} = {d \over {dy}}\left( {{{dx} \over {dy}}} \right)$$
<br><br>$$ = {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}$$
<br><br>$$ = {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}$$
<br><br>$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^2}}}.{{{d^2}y} \over {d{x^2}}}.{1 \over {{{dy} \over {dx}}}}$$
<br><br>$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^3}}}{{{d^2}y} \over {d{x^2}}}$$
|
mcq
|
aieee-2011
|
DdF7R8tTzx4Kv40prc7k9k2k5khz7tx
|
maths
|
differentiation
|
methods-of-differentiation
|
Let ƒ and g be differentiable functions on R
such that fog is the identity function. If for some
a, b $$ \in $$ R, g'(a) = 5 and g(a) = b, then ƒ'(b) is
equal to :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$${2 \\over 5}$$"}, {"identifier": "D", "content": "$${1 \\over 5}$$"}]
|
["D"]
| null |
Given the function composition f(g(x)) is the identity function, it means f(g(x)) = x for all x.
<br><br>$$ \Rightarrow $$ ƒ'(g(x)) g'(x) = 1
<br><br>put x = a
<br><br>$$ \Rightarrow $$ ƒ'(b) g'(a) = 1
<br><br>$$ \Rightarrow $$ ƒ'(b) = $${1 \over 5}$$
|
mcq
|
jee-main-2020-online-9th-january-evening-slot
|
1l54ubqt9
|
maths
|
differentiation
|
methods-of-differentiation
|
<p>Let f and g be twice differentiable even functions on ($$-$$2, 2) such that $$f\left( {{1 \over 4}} \right) = 0$$, $$f\left( {{1 \over 2}} \right) = 0$$, $$f(1) = 1$$ and $$g\left( {{3 \over 4}} \right) = 0$$, $$g(1) = 2$$. Then, the minimum number of solutions of $$f(x)g''(x) + f'(x)g'(x) = 0$$ in $$( - 2,2)$$ is equal to ________.</p>
|
[]
| null |
4
|
Let $h(x)=f(x) \cdot g^{\prime}(x)$
<br/><br/>
As $f(x)$ is even $f\left(\frac{1}{2}\right)=\left(\frac{1}{4}\right)=0$
<br/><br/>
$\Rightarrow f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{4}\right)=0$
<br/><br/>
and $g(x)$ is even $\Rightarrow g^{\prime}(x)$ is odd <br/><br/>
and $g(1)=2$ ensures one root of $g^{\prime}(x)$ is 0 .
<br/><br/>
So, $h(x)=f(x) \cdot g^{\prime}(x)$ has minimum five zeroes
<br/><br/>
$\therefore h^{\prime}(x)=f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)=0$,
<br/><br/>
has minimum 4 zeroes
|
integer
|
jee-main-2022-online-29th-june-evening-shift
|
1ldswyfk5
|
maths
|
differentiation
|
methods-of-differentiation
|
<p>Let $$f:\mathbb{R}\to\mathbb{R}$$ be a differentiable function that satisfies the relation $$f(x+y)=f(x)+f(y)-1,\forall x,y\in\mathbb{R}$$. If $$f'(0)=2$$, then $$|f(-2)|$$ is equal to ___________.</p>
|
[]
| null |
3
|
$f(x+y)=f(x)+f(y)-1$
<br/><br/>
$$
\begin{aligned}
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\\\
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=f^{\prime}(0)=2 \\\\
& f^{\prime}(x)=2 \Rightarrow d y=2 d x \\\\
& y=2 x+C \\\\
& \mathrm{x}=0, \mathrm{y}=1, \mathrm{c}=1 \\\\
& \mathrm{y}=2 \mathrm{x}+1 \\\\
& |f(-2)|=|-4+1|=|-3|=3
\end{aligned}
$$
|
integer
|
jee-main-2023-online-29th-january-morning-shift
|
1lgq0jgd2
|
maths
|
differentiation
|
methods-of-differentiation
|
<p>For the differentiable function $$f: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$$, let $$3 f(x)+2 f\left(\frac{1}{x}\right)=\frac{1}{x}-10$$, then $$\left|f(3)+f^{\prime}\left(\frac{1}{4}\right)\right|$$ is equal to</p>
|
[{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "$$\\frac{29}{5}$$"}, {"identifier": "C", "content": "$$\\frac{33}{5}$$"}, {"identifier": "D", "content": "7"}]
|
["A"]
| null |
<ol>
<li><p>Given the equation: $$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$$</p>
</li>
<li><p>Replace $$x$$ with $$\frac{1}{x}$$ in the original equation:
<br/>$$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$$</p>
</li>
<li><p>Now, we have two equations:</p>
</li>
</ol>
<p>$$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$$
<br/><br/>$$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$$</p>
<ol>
<li>By adding the two equations, we can find $$f(x)$$:</li>
</ol>
<p>$$5f(x) = \frac{3}{x} - 2x - 10$$</p>
<ol>
<li>Now, let's differentiate both sides with respect to $$x$$:</li>
</ol>
<p>$$5f'(x) = -\frac{3}{x^2} - 2$$</p>
<ol>
<li>Now, we can find the values for $$f(3)$$ and $$f'\left(\frac{1}{4}\right)$$:</li>
</ol>
<p>$$f(3) = \frac{1}{5}(1 - 6 - 10) = -3$$
<br/><br/>$$f'\left(\frac{1}{4}\right) = \frac{1}{5}(-48 - 2) = -10$$</p>
<ol>
<li>Finally, calculate the expression we are interested in :</li>
</ol>
<p>$$\left|f(3) + f'\left(\frac{1}{4}\right)\right| = \left|-3 - 10\right| = 13$$</p>
|
mcq
|
jee-main-2023-online-13th-april-morning-shift
|
lsan2rgn
|
maths
|
differentiation
|
methods-of-differentiation
|
If $y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x$, then $96 y^{\prime}\left(\frac{\pi}{6}\right)$ is equal to :
|
[]
| null |
105
|
$\begin{aligned} & y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x \\\\ & y=\frac{(\sqrt{x}+1)(\sqrt{x})\left((\sqrt{x})^3-1\right)}{(\sqrt{x})\left((\sqrt{x})^2+(\sqrt{x})+1\right)}+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x \\\\ & y=(\sqrt{x}+1)(\sqrt{x}-1)+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x\end{aligned}$
<br/><br/>$\begin{aligned} & y^{\prime}=1-\cos ^4 x \cdot(\sin x)+\cos ^2 x(\sin x \\\\ & y^{\prime}\left(\frac{\pi}{6}\right)=1-\frac{9}{16} \times \frac{1}{2}+\frac{3}{4} \times \frac{1}{2} \\\\ & =\frac{32-9+12}{32}=\frac{35}{32} \\\\ & \therefore 96 y^{\prime}\left(\frac{\pi}{6}\right)=105\end{aligned}$
|
integer
|
jee-main-2024-online-1st-february-evening-shift
|
jaoe38c1lsey8r3c
|
maths
|
differentiation
|
methods-of-differentiation
|
<p>Suppose $$f(x)=\frac{\left(2^x+2^{-x}\right) \tan x \sqrt{\tan ^{-1}\left(x^2-x+1\right)}}{\left(7 x^2+3 x+1\right)^3}$$. Then the value of $$f^{\prime}(0)$$ is equal to</p>
|
[{"identifier": "A", "content": "$$\\pi$$\n"}, {"identifier": "B", "content": "$$\\sqrt{\\pi}$$\n"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$\\frac{\\pi}{2}$$"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\left(2^h+2^{-h}\right) \tan h \sqrt{\tan ^{-1}\left(h^2-h+1\right)}-0}{\left(7 h^2+3 h+1\right)^3 h} \\
& =\sqrt{\pi}
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-29th-january-morning-shift
|
1lsg3xnzo
|
maths
|
differentiation
|
methods-of-differentiation
|
<p>Let $$f: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$$ be a function satisfying $$f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$$ for all $$x, y, f(y) \neq 0$$. If $$f^{\prime}(1)=2024$$, then</p>
|
[{"identifier": "A", "content": "$$x f^{\\prime}(x)+2024 f(x)=0$$\n"}, {"identifier": "B", "content": "$$x f^{\\prime}(x)-2023 f(x)=0$$\n"}, {"identifier": "C", "content": "$$x f^{\\prime}(x)-2024 f(x)=0$$\n"}, {"identifier": "D", "content": "$$x f^{\\prime}(x)+f(x)=2024$$"}]
|
["C"]
| null |
<p>$$f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$$</p>
<p>$$\begin{aligned}
& \mathrm{f}^{\prime}(1)=2024 \\
& \mathrm{f}(1)=1
\end{aligned}$$</p>
<p>Partially differentiating w. r. t. x</p>
<p>$$\begin{aligned}
& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) \cdot \frac{1}{\mathrm{y}}=\frac{1}{\mathrm{f}(\mathrm{y})} \mathrm{f}^{\prime}(\mathrm{x}) \\
& \mathrm{y} \rightarrow \mathrm{x} \\
& \mathrm{f}^{\prime}(1) \cdot \frac{1}{\mathrm{x}}=\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \\
& 2024 \mathrm{f}(\mathrm{x})=\mathrm{xf}^{\prime}(\mathrm{x}) \Rightarrow \mathrm{xf}^{\prime}(\mathrm{x})-2024 \mathrm{~f}(\mathrm{x})=0
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-30th-january-evening-shift
|
1lsg8vyuz
|
maths
|
differentiation
|
methods-of-differentiation
|
<p>Let $$g: \mathbf{R} \rightarrow \mathbf{R}$$ be a non constant twice differentiable function such that $$\mathrm{g}^{\prime}\left(\frac{1}{2}\right)=\mathrm{g}^{\prime}\left(\frac{3}{2}\right)$$. If a real valued function $$f$$ is defined as $$f(x)=\frac{1}{2}[g(x)+g(2-x)]$$, then</p>
|
[{"identifier": "A", "content": "$$f^{\\prime \\prime}(x)=0$$ for atleast two $$x$$ in $$(0,2)$$\n"}, {"identifier": "B", "content": "$$f^{\\prime}\\left(\\frac{3}{2}\\right)+f^{\\prime}\\left(\\frac{1}{2}\\right)=1$$\n"}, {"identifier": "C", "content": "$$f^{\\prime \\prime}(x)=0$$ for no $$x$$ in $$(0,1)$$\n"}, {"identifier": "D", "content": "$$f^{\\prime \\prime}(x)=0$$ for exactly one $$x$$ in $$(0,1)$$"}]
|
["A"]
| null |
<p>$$f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0$$</p>
<p>Also $$\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}{2}\right)}{2}=0, \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$$</p>
<p>$$\Rightarrow \mathrm{f}^{\prime}\left(\frac{3}{2}\right)=\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$$</p>
<p>$$\Rightarrow \operatorname{rootsin}\left(\frac{1}{2}, 1\right)$$ and $$\left(1, \frac{3}{2}\right)$$</p>
<p>$$\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})$$ is zero at least twice in $$\left(\frac{1}{2}, \frac{3}{2}\right)$$</p>
|
mcq
|
jee-main-2024-online-30th-january-morning-shift
|
luy6z53a
|
maths
|
differentiation
|
methods-of-differentiation
|
<p>Let $$f(x)=a x^3+b x^2+c x+41$$ be such that $$f(1)=40, f^{\prime}(1)=2$$ and $$f^{\prime \prime}(1)=4$$. Then $$a^2+b^2+c^2$$ is equal to:</p>
|
[{"identifier": "A", "content": "54"}, {"identifier": "B", "content": "51"}, {"identifier": "C", "content": "73"}, {"identifier": "D", "content": "62"}]
|
["B"]
| null |
<p>Given the polynomial function:</p>
<p>$$f(x) = ax^3 + bx^2 + cx + 41$$</p>
<p>We are provided the following conditions from the problem:</p>
<p>1. $$f(1) = 40$$</p>
<p>2. $$f^{\prime}(1) = 2$$</p>
<p>3. $$f^{\prime \prime}(1) = 4$$</p>
<p>First, calculate $f(1)$:</p>
<p>$$f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = 40$$</p>
<p>Simplifying, we get:</p>
<p>$$a + b + c + 41 = 40$$</p>
<p>Therefore:</p>
<p>$$a + b + c = -1$$</p>
<p>Next, calculate the first derivative $f^{\prime}(x)$:</p>
<p>$$f^{\prime}(x) = 3ax^2 + 2bx + c$$</p>
<p>Given $f^{\prime}(1) = 2$:</p>
<p>$$f^{\prime}(1) = 3a(1)^2 + 2b(1) + c = 2$$</p>
<p>Simplifying, we get:</p>
<p>$$3a + 2b + c = 2$$</p>
<p>Next, calculate the second derivative $f^{\prime \prime}(x)$:</p>
<p>$$f^{\prime \prime}(x) = 6ax + 2b$$</p>
<p>Given $f^{\prime \prime}(1) = 4$:</p>
<p>$$f^{\prime \prime}(1) = 6a(1) + 2b = 4$$</p>
<p>Simplifying, we get:</p>
<p>$$6a + 2b = 4$$</p>
<p>Dividing the entire equation by 2:</p>
<p>$$3a + b = 2$$</p>
<p>We now have three equations:</p>
<p>1. $$a + b + c = -1$$</p>
<p>2. $$3a + 2b + c = 2$$</p>
<p>3. $$3a + b = 2$$</p>
<p>To solve for $a$, $b$, and $c$, follow these steps:</p>
<p>First, subtract the third equation from the second equation:</p>
<p>$$(3a + 2b + c) - (3a + b) = 2 - 2$$</p>
<p>Which simplifies to:</p>
<p>$$b + c = 0$$</p>
<p>So,</p>
<p>$$c = -b$$</p>
<p>Substitute $c = -b$ into the first equation:</p>
<p>$$(a + b - b = -1)$$</p>
<p>Simplifying, we get:</p>
<p>$$a = -1$$</p>
<p>Now substitute $a = -1$ into the third equation:</p>
<p>$$3(-1) + b = 2$$</p>
<p>Which simplifies to:</p>
<p>$$-3 + b = 2$$</p>
<p>Therefore:</p>
<p>$$b = 5$$</p>
<p>Next, since $c = -b$:</p>
<p>$$c = -5$$</p>
<p>Finally, we need to find $a^2 + b^2 + c^2$:</p>
<p>$$a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2$$</p>
<p>Simplifying, we get:</p>
<p>$$1 + 25 + 25 = 51$$</p>
<p>Therefore, the answer is:</p>
<p><strong>Option B: 51</strong></p>
|
mcq
|
jee-main-2024-online-9th-april-morning-shift
|
aLwt0WnHgsFA2mmz
|
maths
|
differentiation
|
successive-differentiation
|
If $$f\left( x \right) = {x^n},$$ then the value of
<p>$$f\left( 1 \right) - {{f'\left( 1 \right)} \over {1!}} + {{f''\left( 1 \right)} \over {2!}} - {{f'''\left( 1 \right)} \over {3!}} + ..........{{{{\left( { - 1} \right)}^n}{f^n}\left( 1 \right)} \over {n!}}$$ is</p>
|
[{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$${{2^n}}$$ "}, {"identifier": "C", "content": "$${{2^n} - 1}$$ "}, {"identifier": "D", "content": "$$0$$"}]
|
["D"]
| null |
$$f\left( x \right) = {x^n} \Rightarrow f\left( 1 \right) = 1$$
<br><br>$$f'\left( x \right) = n{x^{n - 1}} \Rightarrow f'\left( 1 \right) = n$$
<br><br>$$f''\left( x \right) = n\left( {n - 1} \right){x^{n - 2}}$$
<br><br>$$ \Rightarrow f''\left( 1 \right) = n\left( {n - 1} \right)$$
<br><br>$$\therefore$$ $${f^n}\left( x \right) = n!$$
<br><br>$$ \Rightarrow {f^n}\left( 1 \right) = n!$$
<br><br>$$ = 1 - {n \over {1!}} + {{n\left( {n - 1} \right)} \over {2!}}{{n\left( {n - 1} \right)\left( {n - 2} \right)} \over {3!}}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + .... + {\left( { - 1} \right)^n}{{n!} \over {n!}}$$
<br><br>$$ = {}^n\,{C_0} - {}^n\,{C_1} + {}^n\,{C_2} - {}^n\,{C_3}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ...... + {\left( { - 1} \right)^n}\,{}^n{C_n} = 0$$
|
mcq
|
aieee-2003
|
oPx9mrcOVu1a3ZqfY3d42
|
maths
|
differentiation
|
successive-differentiation
|
Let f be a polynomial function such that
<br/><br/>f (3x) = f ' (x) . f '' (x), for all x $$ \in $$ <b>R</b>. Then :
|
[{"identifier": "A", "content": "f (2) + f ' (2) = 28"}, {"identifier": "B", "content": "f '' (2) $$-$$ f ' (2) = 0"}, {"identifier": "C", "content": "f '' (2) $$-$$ f (2) = 4"}, {"identifier": "D", "content": "f (2) $$-$$ f ' (2) + f '' (2) = 10"}]
|
["B"]
| null |
<p>Let $$f(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 1}} + \,\,....\,\, + {a_{n - 1}}x + {a_n}$$</p>
<p>$$f'(x) = {a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,.....\,\, + {a_{n - 1}}$$</p>
<p>$$f''(x) = {a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}$$</p>
<p>Now,</p>
<p>$$f(3x) = {3^n}{a_0}{x^n} + {3^{n - 1}}{a_1}{x^{n - 1}} + {3^{n - 2}}{a_2}{x^{n - 2}} + \,\,....\,\, + 3{a_{n - 1}} + {a_n}$$</p>
<p>$$f'(x)\,.\,f''(x) = [{a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,....\,\, + {a_{n - 1}}]$$</p>
<p>$$[{a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}]$$</p>
<p>Comparing highest powers of x, we get</p>
<p>$${3^n}{a_0}{x_n} = a_0^2(n - 1){x^{n - 1 + n - 2}} = a_0^2{n^2}(n - 1){x^{2n - 3}}$$</p>
<p>Therefore, $$2n - 3 = n$$</p>
<p>$$\Rightarrow$$ n = 3 and $${3^n}{a_0} = a_0^2{n^2}(n - 1)$$</p>
<p>$$ \Rightarrow {a_0} = 27 = {3 \over 2}$$</p>
<p>Therefore, $$f(x) = {3 \over 2}{x^3} + {a_1}{x^2} + {a_2}x + {a_3}$$</p>
<p>$$f'(x) = {9 \over 2}{x^2} + 2{a_1}x + {a_2}$$</p>
<p>$$f''(x) = 9x + 2{a_1}$$</p>
<p>$$f(3x) = {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}$$</p>
<p>Now, $$f(3x) = f'(x)\,.\,f''(x)$$</p>
<p>$$ \Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}$$</p>
<p>$$ \Rightarrow \left( {{9 \over 2}{x^2} + 2{a_1}x + {a_2}} \right)(9x + 2{a_1})$$</p>
<p>$$ \Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3} = {{81} \over 2}{x^3} + [9{a_1} + 18{a_1}]{x^2} + [4a_1^2 + 9{a_2}]x + 2{a_1}{a_2}$$</p>
<p>Comparing the coefficients, we get</p>
<p>$$9{a_1} = 27{a_1}$$</p>
<p>$$ \Rightarrow {a_1} = 0,\,3{a_2} = 4a_1^2 + 9{a_2} = 9{a_2}$$</p>
<p>$$ \Rightarrow {a_2} = 0$$</p>
<p>Therefore, $$f(x) = {3 \over 2}{x^3}$$</p>
<p>$$f'(x) = {9 \over 2}{x^2}$$</p>
<p>$$f''(x) = 9x$$</p>
<p>Hence, $$f''(2) - f'(x) = 18 - 18 = 0$$</p>
|
mcq
|
jee-main-2017-online-9th-april-morning-slot
|
7QKM8xqcNeiQvnGv4PFtQ
|
maths
|
differentiation
|
successive-differentiation
|
Let f : R $$ \to $$ R be a function such that f(x) = x<sup>3</sup> + x<sup>2</sup>f'(1) + xf''(2) + f'''(3), x $$ \in $$ R. Then f(2) equals -
|
[{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "$$-$$ 2"}, {"identifier": "C", "content": "$$-$$ 4"}, {"identifier": "D", "content": "8"}]
|
["B"]
| null |
f(x) = x<sup>3</sup> + x<sup>2</sup>f '(1) + xf ''(2) + f '''(3)
<br><br>$$ \Rightarrow $$ f '(x) = 3x<sup>2</sup> + 2xf '(1) + f ''(x) . . . . . (1)
<br><br>$$ \Rightarrow $$ f ''(x) = 6x + 2f '(1) . . . . . . (2)
<br><br>$$ \Rightarrow $$ f '''(x) = 6 . . . . . .(3)
<br><br>put x = 1 in equation (1) :
<br><br>f '(1) = 3 + 2f '(1) + f ''(2) . . . . .(4)
<br><br>put x = 2 in equation (2) :
<br><br>f ''(2) = 12 + 2f '(1) . . . . .(5)
<br><br>from equation (4) & (5) :
<br><br>$$-$$3 $$-$$ f '(1) = 12 + 2f'(1)
<br><br>$$ \Rightarrow $$ 3f '(1) = $$-$$ 15
<br><br>$$ \Rightarrow $$ f '(1) = $$-$$ 5 $$ \Rightarrow $$ f ''(2) = 2 . . . . .(2)
<br><br>put x = 3 in equation (3) :
<br><br>f ''' (3) = 6
<br><br>$$ \therefore $$ f(x) = x<sup>3</sup> $$-$$ 5x<sup>2</sup> + 2x + 6
<br><br>f(2) = 8 $$-$$ 20 + 4 + 6 = $$-$$ 2
|
mcq
|
jee-main-2019-online-10th-january-morning-slot
|
xrX2b9GpaeqoQ7utJtjgy2xukf0qq2t5
|
maths
|
differentiation
|
successive-differentiation
|
If y<sup>2</sup> + log<sub>e</sub> (cos<sup>2</sup>x) = y, <br/>$$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, then :
|
[{"identifier": "A", "content": "|y''(0)| = 2"}, {"identifier": "B", "content": "|y'(0)| + |y''(0)| = 3"}, {"identifier": "C", "content": "y''(0) = 0"}, {"identifier": "D", "content": "|y'(0)| + |y\"(0)| = 1"}]
|
["A"]
| null |
Given y<sup>2</sup> + log<sub>e</sub> (cos<sup>2</sup>x) = y .....(1)
<br><br>Put x = 0, we get
<br><br>y<sup>2</sup> + log<sub>e</sub> (1) = y
<br><br>$$ \Rightarrow $$ y<sup>2</sup> = y
<br><br>$$ \Rightarrow $$ y = 0, 1
<br><br>Differentiating (1) we get
<br><br>2yy' + $${1 \over {\cos x}}\left( { - \sin x} \right)$$ = y'
<br><br>$$ \Rightarrow $$ 2yy' - 2tanx = y' ....(2)
<br><br>From (2) when x = 0, y = 0 then y'(0) = 0
<br><br>From (2) when x = 0, y = 1 then
<br><br>2y' = y'
<br><br>$$ \Rightarrow $$ y'(0) = 0
<br><br>Again differentiating (2) we get
<br><br>2(y')<sup>2</sup>
+ 2yy'' – 2sec<sup>2</sup>x = y''
<br><br>from (2) when x = 0, y = 0, y’(0) = 0 then
<br><br>y”(0) = -2
<br><br>Also from (2) when x = 0, y = 1, y’(0) = 0 then
<br><br>y”(0) = 2
<br><br>$$ \therefore $$ |y''(0)| = 2
|
mcq
|
jee-main-2020-online-3rd-september-morning-slot
|
1l56u56af
|
maths
|
differentiation
|
successive-differentiation
|
<p>If $$y(x) = {\left( {{x^x}} \right)^x},\,x > 0$$, then $${{{d^2}x} \over {d{y^2}}} + 20$$ at x = 1 is equal to ____________.</p>
|
[]
| null |
16
|
<p>$$\because$$ $$y(x) = {\left( {{x^x}} \right)^x}$$</p>
<p>$$\therefore$$ $$y = {x^{{x^2}}}$$</p>
<p>$$\therefore$$ $${{dy} \over {dx}} = {x^2}\,.\,{x^{{x^2} - 1}} + {x^{{x^2}}}\ln x\,.\,2x$$</p>
<p>$$\therefore$$ $${{dx} \over {dy}} = {1 \over {{x^{{x^2} + 1}}(1 + 2\ln x)}}$$ ..... (i)</p>
<p>Now, $${{{d^2}x} \over {dx^2}} = {d \over {dx}}\left( {{{\left( {{x^{{x^2} + 1}}(1 + 2\ln x)} \right)}^{ - 1}}} \right)\,.\,{{dx} \over {dy}}$$</p>
<p>$$ = {{ - x{{\left( {{x^{{x^2} + 1}}(1 + 2\ln x)} \right)}^{ - 2}}\,.\,{x^{{x^2}}}(1 + 2\ln x)({x^2} + 2{x^2}\ln x + 3)} \over {{x^{{x^2}}}(1 + 2\ln x)}}$$</p>
<p>$$ = {{ - {x^{{x^2}}}(1 + 2\ln x)({x^3} + 3 + 2{x^2}\ln x)} \over {{{\left( {{x^{{x^2}}}(1 + 2\ln x)} \right)}^3}}}$$</p>
<p>$${{{d^2}x} \over {d{y^2}(at\,x = 1)}} = - 4$$</p>
<p>$$\therefore$$ $${{{d^2}x} \over {d{y^2}(at\,x = 1)}} + 20 = 16$$</p>
|
integer
|
jee-main-2022-online-27th-june-evening-shift
|
1l6klla1m
|
maths
|
differentiation
|
successive-differentiation
|
<p>For the curve $$C:\left(x^{2}+y^{2}-3\right)+\left(x^{2}-y^{2}-1\right)^{5}=0$$, the value of $$3 y^{\prime}-y^{3} y^{\prime \prime}$$, at the point $$(\alpha, \alpha)$$, $$\alpha>0$$, on C, is equal to ____________.</p>
|
[]
| null |
16
|
<p>$$\because$$ $$C:({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$$ for point ($$\alpha$$, $$\alpha$$)</p>
<p>$${\alpha ^2} + {\alpha ^2} - 3 + {({\alpha ^2} - {\alpha ^2} - 1)^5} = 0$$</p>
<p>$$\therefore$$ $$\alpha = \sqrt 2 $$</p>
<p>On differentiating $$({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$$ we get</p>
<p>$$x + yy' + 5{({x^2} - {y^2} - 1)^4}(x - yy') = 0$$ ...... (i)</p>
<p>When $$x = y = \sqrt 2 $$ then $$y' = {3 \over 2}$$</p>
<p>Again on differentiating eq. (i) we get :</p>
<p>$$1 + {(y')^2} + yy'' + 20({x^2} - {y^2} - 1)(2x - 2yy')(x - y'y) + 5{({x^2} - {y^2} - 1)^4}(1 - y{'^2} - yy'') = 0$$</p>
<p>For $$x = y = \sqrt 2 $$ and $$y' = {3 \over 2}$$ we get $$y'' = - {{23} \over {4\sqrt 2 }}$$</p>
<p>$$\therefore$$ $$3y' - {y^3}y'' = 3\,.\,{3 \over 2} - {\left( {\sqrt 2 } \right)^3}\,.\,\left( { - {{23} \over {4\sqrt 2 }}} \right) = 16$$</p>
|
integer
|
jee-main-2022-online-27th-july-evening-shift
|
1ldon088o
|
maths
|
differentiation
|
successive-differentiation
|
<p>Let $$f(x) = 2x + {\tan ^{ - 1}}x$$ and $$g(x) = {\log _e}(\sqrt {1 + {x^2}} + x),x \in [0,3]$$. Then</p>
|
[{"identifier": "A", "content": "there exists $$\\widehat x \\in [0,3]$$ such that $$f'(\\widehat x) < g'(\\widehat x)$$"}, {"identifier": "B", "content": "there exist $$0 < {x_1} < {x_2} < 3$$ such that $$f(x) < g(x),\\forall x \\in ({x_1},{x_2})$$"}, {"identifier": "C", "content": "$$\\min f'(x) = 1 + \\max g'(x)$$"}, {"identifier": "D", "content": "$$\\max f(x) > \\max g(x)$$"}]
|
["D"]
| null |
$$
\begin{aligned}
& f^{\prime}(x)=2+\frac{1}{1+x^2}, g^{\prime}(x)=\frac{1}{\sqrt{x^2+1}} \\\\
& f^{\prime \prime}(x)=-\frac{2 x}{\left(1+x^2\right)^2}<0 \\\\
& g^{\prime \prime}(x)=-\frac{1}{2}\left(x^2+1\right)^{-3 / 2} \cdot 2 x<0 \\\\
& \left.f^{\prime}(x)\right|_{\min }=f^{\prime}(3)=2+\frac{1}{10}=\frac{21}{10} \\\\
& \left.g^{\prime}(x)\right|_{\max }=g^{\prime}(0)=1 \\\\
& \left.f^{\prime}(x)\right|_{\max }=f(3)=2+\tan ^{-1} 3 \\\\
& \left.g(x)\right|_{\max }=g(3)=\ln (3+\sqrt{10})<\ln <7<2
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-1st-february-morning-shift
|
1ldoofa8p
|
maths
|
differentiation
|
successive-differentiation
|
<p>If $$f(x)=x^{2}+g^{\prime}(1) x+g^{\prime \prime}(2)$$ and $$g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)$$, then the value of $$f(4)-g(4)$$ is equal to ____________.</p>
|
[]
| null |
14
|
Let $g^{\prime}(1)=a$ and $g^{\prime \prime}(2)=b$
<br/><br/>$\Rightarrow f(x)=x^{2}+a x+b$
<br/><br/>Now, $f(1)=1+a+b ; f^{\prime}(x)=2 x+a ; f^{\prime \prime}(x)=2$
<br/><br/>$g(x)=(1+a+b) x^{2}+x(2 x+a)+2$
<br/><br/>$\Rightarrow g(x)=(a+b+3) x^{2}+a x+2$
<br/><br/>$\Rightarrow g^{\prime}(x)=2 x(a+b+3)+a \Rightarrow g^{\prime}(1)=2(a+b+3)$$+a=a$
<br/><br/>$\Rightarrow a+b+3=0$ .......(1)
<br/><br/>$g^{\prime \prime}(x)=2(a+b+3)=b$
<br/><br/>$\Rightarrow 2 a+b+6=0$ ........(2)
<br/><br/>Solving (i) and (ii), we get
<br/><br/>$a=-3$ and $b=0$
<br/><br/>$f(x)=x^{2}-3 x$ and $g(x)=-3 x+2$
<br/><br/>$f(4)=4$ and $g(4)=-12+2=-10$
<br/><br/>$\Rightarrow f(4)-g(4)=16-2=14$
|
integer
|
jee-main-2023-online-1st-february-morning-shift
|
1ldsfuib3
|
maths
|
differentiation
|
successive-differentiation
|
<p>Let $$f$$ and $$g$$ be the twice differentiable functions on $$\mathbb{R}$$ such that</p>
<p>$$f''(x)=g''(x)+6x$$</p>
<p>$$f'(1)=4g'(1)-3=9$$</p>
<p>$$f(2)=3g(2)=12$$.</p>
<p>Then which of the following is NOT true?</p>
|
[{"identifier": "A", "content": "$$g(-2)-f(-2)=20$$"}, {"identifier": "B", "content": "There exists $$x_0\\in(1,3/2)$$ such that $$f(x_0)=g(x_0)$$"}, {"identifier": "C", "content": "$$|f'(x)-g'(x)| < 6\\Rightarrow -1 < x < 1$$"}, {"identifier": "D", "content": "If $$-1 < x < 2$$, then $$|f(x)-g(x)| < 8$$"}]
|
["D"]
| null |
<p>$$f''(x) = g''(x) + 6x$$</p>
<p>$$ \Rightarrow f'(x) = g'(x) + 3{x^2} + C$$</p>
<p>$$f'(1) = g'(1) + 3 + C$$</p>
<p>$$ \Rightarrow g = 3 + 3 + C \Rightarrow C = 3$$</p>
<p>$$ \Rightarrow f'(x) = g'(x) + 3{x^2} + 3$$</p>
<p>$$ \Rightarrow f(x) = g(x) + {x^2} + 3x + C'$$</p>
<p>$$x = 2$$</p>
<p>$$f(2) = g(2) + 14 + C'$$</p>
<p>$$12 = 4 + 14 + C'$$</p>
<p>$$ \Rightarrow C' = - 6$$</p>
<p>$$ \Rightarrow f(x) = g(2) + {x^3} + 3x - 6$$</p>
<p>$$f( - 2) = g( - 2) - 8 - 6 - 6$$</p>
<p>$$g( - 2) - f( - 2) = 20$$</p>
<p>$$f'(x) - g'(x) = 3{x^2} + 3$$</p>
<p>$$x \in ( - 1,1)$$</p>
<p>$$3{x^2} + 3 \in (0,6)$$</p>
<p>$$ \Rightarrow f'(x) - g'(x) \in (0,6)$$<?p>
<p>$$f(x) - g(x) = {x^3} + 3x - 6$$</p>
<p>At $$x = - 1$$</p>
<p>$$|f( - 1) - g( - 1)| = 10$$</p>
<p>$$\therefore$$ Option (4) is false.</p>
|
mcq
|
jee-main-2023-online-29th-january-evening-shift
|
1ldv1t76j
|
maths
|
differentiation
|
successive-differentiation
|
<p>Let $$y(x) = (1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8})(1 + {x^{16}})$$. Then $$y' - y''$$ at $$x = - 1$$ is equal to</p>
|
[{"identifier": "A", "content": "496"}, {"identifier": "B", "content": "976"}, {"identifier": "C", "content": "464"}, {"identifier": "D", "content": "944"}]
|
["A"]
| null |
$$
\begin{aligned}
& y=\frac{1-x^{32}}{1-x}=1+x+x^2+x^3+\ldots+x^{31} \\\\
& y^{\prime}=1+2 x+3 x^2+\ldots+31 x^{30} \\\\
& y^{\prime}(-1)=1-2+3-4+\ldots+31=16 \\\\
& y^{\prime \prime}(x)=2+6 x+12 x^2+\ldots+31.30 x^{29} \\\\
& y^{\prime \prime}(-1)=2-6+12 \ldots 31.30=480 \\\\
& y^{\prime \prime}(-1)-y^{\prime}(-1)=-496
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-25th-january-morning-shift
|
1ldwx6r58
|
maths
|
differentiation
|
successive-differentiation
|
<p>If $$f(x) = {x^3} - {x^2}f'(1) + xf''(2) - f'''(3),x \in \mathbb{R}$$, then</p>
|
[{"identifier": "A", "content": "$$2f(0) - f(1) + f(3) = f(2)$$"}, {"identifier": "B", "content": "$$f(1) + f(2) + f(3) = f(0)$$"}, {"identifier": "C", "content": "$$f(3) - f(2) = f(1)$$"}, {"identifier": "D", "content": "$$3f(1) + f(2) = f(3)$$"}]
|
["A"]
| null |
$$
f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R
$$<br/><br/>
Let $\mathrm{f}^{\prime}(1)=\mathrm{a}, \mathrm{f}^{\prime \prime}(2)=\mathrm{b}, \mathrm{f}^{\prime \prime \prime}(3)=\mathrm{c}$<br/><br/>
$$
\begin{aligned}
& f(x)=x^3-a x^2+b x-c \\\\
& f^{\prime}(x)=3 x^2-2 a x+b \\\\
& f^{\prime \prime}(x)=6 x-2 a \\\\
& f^{\prime \prime \prime}(x)=6 \\\\
& c=6, a=3, b=6 \\\\
& f(x)=x^3-3 x^2+6 x-6 \\\\
& f(1)=-2, f(2)=2, f(3)=12, f(0)=-6 \\\\
& 2 f(0)-f(1)+f(3)=2=f(2)
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-24th-january-evening-shift
|
lsble0m7
|
maths
|
differentiation
|
successive-differentiation
|
Let $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in \mathbf{R}$. Then $f^{\prime}(10)$ is equal to ____________.
|
[]
| null |
202
|
<p>$$\begin{aligned}
& f(x)=x^3+x^2 \cdot f^{\prime}(1)+x \cdot f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) \\
& f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2) \\
& f^{\prime \prime}(x)=6 x+2 f^{\prime}(1) \\
& f^{\prime \prime \prime}(x)=6 \\
& f^{\prime}(1)=-5, f^{\prime \prime}(2)=2, f^{\prime \prime \prime}(3)=6 \\
& f(x)=x^3+x^2 \cdot(-5)+x \cdot(2)+6 \\
& f^{\prime}(x)=3 x^2-10 x+2 \\
& f^{\prime}(10)=300-100+2=202
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-27th-january-morning-shift
|
1lsg94q54
|
maths
|
differentiation
|
successive-differentiation
|
<p>If $$f(x)=\left|\begin{array}{ccc}
2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\
3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\
2 \cos ^4 x & 3+2 \sin ^4 x & \sin ^2 2 x
\end{array}\right|,$$ then $$\frac{1}{5} f^{\prime}(0)=$$ is equal to :</p>
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "6"}]
|
["C"]
| null |
<p>$$\begin{aligned}
& \left|\begin{array}{ccc}
2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\
3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\
2 \cos ^4 x & 3+2 \sin ^2 4 x & \sin ^2 2 x
\end{array}\right| \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\
& \left|\begin{array}{ccc}
2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\
3 & 0 & -3 \\
0 & 3 & -3
\end{array}\right| \\
& \mathrm{f}(\mathrm{x})=45 \\
& \mathrm{f}^{\prime}(\mathrm{x})=0 \\
&
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-30th-january-morning-shift
|
lv2er9ry
|
maths
|
differentiation
|
successive-differentiation
|
<p>Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a thrice differentiable function such that $$f(0)=0, f(1)=1, f(2)=-1, f(3)=2$$ and $$f(4)=-2$$. Then, the minimum number of zeros of $$\left(3 f^{\prime} f^{\prime \prime}+f f^{\prime \prime \prime}\right)(x)$$ is __________.</p>
|
[]
| null |
5
|
<p>$$\because f: R \rightarrow R \text { and } f(0)=0, f(1)=1, f(2)=-1 \text {, }$$</p>
<p>$$f(3)=2$$ and $$f(4)=-2$$ then</p>
<p>$$f(x)$$ has atleast 4 real roots.</p>
<p>Then $$f(x)$$ has atleast 3 real roots and $$f^{\prime}(x)$$ has atleast 2 real roots.</p>
<p>Now we know that</p>
<p>$$\begin{aligned}
\frac{d}{d x}\left(f^3 \cdot f^{\prime \prime}\right) & =3 f^2 \cdot f^{\prime} \cdot f^{\prime \prime}+f^3 \cdot f^{\prime \prime \prime} \\
& =f^2\left(3 f^{\prime} \cdot f^{\prime}+f \cdot f^{\prime \prime}\right)
\end{aligned}$$</p>
<p>Here $$f^3 \cdot f'$$ has atleast 6 roots.</p>
<p>Then its differentiation has atleast 5 distinct roots.</p>
|
integer
|
jee-main-2024-online-4th-april-evening-shift
|
lv9s204y
|
maths
|
differentiation
|
successive-differentiation
|
<p>If $$y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}$$, then at $$\theta=\frac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\frac{1}{2}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$"}, {"identifier": "D", "content": "2"}]
|
["D"]
| null |
<p>$$\begin{aligned}
& y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2} \\
& =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{4 \cos ^3 \theta+8 \cos ^2 \theta+2 \cos \theta-2} \\
& =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\
& =\frac{1}{2(1+\cos \theta)}=\frac{1}{4 \cos ^2 \theta / 2}=\frac{\sec ^2 \theta / 2}{4} \\
& y^{\prime}(\theta)=\frac{1}{4}\left(2 \sec \frac{\theta}{2} \cdot \sec \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \cdot \frac{1}{2}\right) \\
& =\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2}
\end{aligned}$$</p>
<p>$$y^{\prime \prime}(\theta)=\frac{1}{4}\left(\tan \frac{\theta}{2}\right)\left(\sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2}\right) +\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \sec ^2 \frac{\theta}{2} \cdot \frac{1}{2}$$</p>
<p>$$\begin{aligned}
& \text { at } \theta=\frac{\pi}{2}, y(\theta)=\frac{1}{2}, y^{\prime}(\theta)=\frac{1}{2}, y^{\prime \prime}(\theta)=1 \\
& \therefore \quad y+y^{\prime}+y^{\prime \prime}=2
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-5th-april-evening-shift
|
lvc57b43
|
maths
|
differentiation
|
successive-differentiation
|
<p>$$\text { If } f(x)=\left\{\begin{array}{ll}
x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\
0 & , x=0
\end{array}\right. \text {, then }$$</p>
|
[{"identifier": "A", "content": "$$f^{\\prime \\prime}(0)=0$$\n"}, {"identifier": "B", "content": "$$f^{\\prime \\prime}(0)=1$$\n"}, {"identifier": "C", "content": "$$f^{\\prime \\prime}\\left(\\frac{2}{\\pi}\\right)=\\frac{24-\\pi^2}{2 \\pi}$$\n"}, {"identifier": "D", "content": "$$f^{\\prime \\prime}\\left(\\frac{2}{\\pi}\\right)=\\frac{12-\\pi^2}{2 \\pi}$$"}]
|
["C"]
| null |
<p>Given the function:</p>
<p>$ f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{array}\right. $</p>
<p>we need to find its second derivative at specific points.</p>
<p>First, let’s compute the first derivative $ f^{\prime}(x) $:</p>
<p>$ f^{\prime}(x) = 3x^2 \sin \left( \frac{1}{x} \right) - x \cos \left( \frac{1}{x} \right) $</p>
<p>Next, the second derivative $ f^{\prime \prime}(x) $ is:</p>
<p>$ f^{\prime \prime}(x) = 6x \sin \left(\frac{1}{x}\right) - 3x \cos \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) - \frac{1}{x} \sin \left(\frac{1}{x}\right) $</p>
<p>Therefore, evaluating the second derivative at $ x = \frac{2}{\pi} $:</p>
<p>$ f^{\prime \prime} \left(\frac{2}{\pi}\right) = 6 \left( \frac{2}{\pi} \right) \sin \left(\frac{\pi}{2}\right) - 3 \left( \frac{2}{\pi} \right) \cos \left(\frac{\pi}{2}\right) - \cos \left( \frac{\pi}{2} \right) - \frac{\pi}{2} \sin \left( \frac{\pi}{2} \right) $</p>
<p>Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(\frac{\pi}{2}) = 0$, this simplifies to:</p>
<p>$ f^{\prime \prime} \left( \frac{2}{\pi} \right) = \frac{12}{\pi} - \frac{\pi}{2} = \frac{24 - \pi^2}{2\pi} $</p>
<p>Finally, note that $ f^{\prime}(0) $ is not defined, as it involves terms like $\frac{1}{x}$ when $ x = 0 $.</p>
|
mcq
|
jee-main-2024-online-6th-april-morning-shift
|
1krw1fh1n
|
maths
|
ellipse
|
chord-of-ellipse
|
Let an ellipse $$E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, $${a^2} > {b^2}$$, passes through $$\left( {\sqrt {{3 \over 2}} ,1} \right)$$ and has eccentricity $${1 \over {\sqrt 3 }}$$. If a circle, centered at focus F($$\alpha$$, 0), $$\alpha$$ > 0, of E and radius $${2 \over {\sqrt 3 }}$$, intersects E at two points P and Q, then PQ<sup>2</sup> is equal to :
|
[{"identifier": "A", "content": "$${8 \\over 3}$$"}, {"identifier": "B", "content": "$${4 \\over 3}$$"}, {"identifier": "C", "content": "$${{16} \\over 3}$$"}, {"identifier": "D", "content": "3"}]
|
["C"]
| null |
$${3 \over {2{a^2}}} + {1 \over {{b^2}}} = 1$$ and $$1 - {{{b^2}} \over {{a^2}}} = {1 \over 3}$$<br><br>$$ \Rightarrow {a^2} = 3{b^2} = 3$$ <br><br>$$ \Rightarrow {{{x^2}} \over 3} + {{{y^2}} \over 2} = 1$$ ...... (i)<br><br>Its focus is (1, 0)<br><br>Now, equation of circle is <br><br>$${(x - 1)^2} + {y^2} = {4 \over 3}$$ ..... (ii)<br><br>Solving (i) and (ii) we get<br><br>$$y = \pm {2 \over {\sqrt 3 }},x = 1$$<br><br>$$ \Rightarrow P{Q^2} = {\left( {{4 \over {\sqrt 3 }}} \right)^2} = {{16} \over 3}$$
|
mcq
|
jee-main-2021-online-25th-july-morning-shift
|
1l59l0lop
|
maths
|
ellipse
|
chord-of-ellipse
|
<p>The line y = x + 1 meets the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$$ at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)<sup>2</sup> is equal to :</p>
|
[{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "8"}]
|
["A"]
| null |
<p>Let point (a, a + 1) as the point of intersection of line and ellipse.</p>
<p>So, $${{{a^2}} \over 4} + {{{{(a + 1)}^2}} \over 2} = 1 \Rightarrow {a^2} + 2({a^2} + 2a + 1) = 4$$</p>
<p>$$ \Rightarrow 3{a^2} + 4a - 2 = 0$$</p>
<p>If roots of this equation are $$\alpha$$ and $$\beta$$.</p>
<p>So, $$P(\alpha ,\,\alpha + 1)$$ and $$Q(\beta ,\,\beta + 1)$$</p>
<p>$$PQ = 4{r^2} = {(\alpha - \beta )^2} + {(\alpha - \beta )^2}$$</p>
<p>$$ \Rightarrow 9{r^2} = {9 \over 4}(2{(\alpha - \beta )^2})$$</p>
<p>$$ = {9 \over 2}\left[ {{{(\alpha + \beta )}^2} - 4\alpha \beta } \right]$$</p>
<p>$$ = {9 \over 2}\left[ {{{\left( { - {4 \over 3}} \right)}^2} + {8 \over 3}} \right]$$</p>
<p>$$ = {1 \over 2}[16 + 24] = 20$$</p>
|
mcq
|
jee-main-2022-online-25th-june-evening-shift
|
1lguvb7is
|
maths
|
ellipse
|
chord-of-ellipse
|
<p>Consider ellipses $$\mathrm{E}_{k}: k x^{2}+k^{2} y^{2}=1, k=1,2, \ldots, 20$$. Let $$\mathrm{C}_{k}$$ be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse $$\mathrm{E}_{k}$$. If $$r_{k}$$ is the radius of the circle $$\mathrm{C}_{k}$$, then the value of $$\sum_\limits{k=1}^{20} \frac{1}{r_{k}^{2}}$$ is :</p>
|
[{"identifier": "A", "content": "2870"}, {"identifier": "B", "content": "3210"}, {"identifier": "C", "content": "3320"}, {"identifier": "D", "content": "3080"}]
|
["D"]
| null |
We have, $E_K=K x^2+K^2 y^2=1, K=1,2, \ldots 20$
<br><br>$\Rightarrow \frac{x^2}{\frac{1}{K}}+\frac{y^2}{\frac{1}{K^2}}=1$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln5ufvyu/8420caac-c5f0-403f-a175-e82f7eb8d93d/8ea00c50-5f75-11ee-8999-67742721d03c/file-6y3zli1ln5ufvyv.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ln5ufvyu/8420caac-c5f0-403f-a175-e82f7eb8d93d/8ea00c50-5f75-11ee-8999-67742721d03c/file-6y3zli1ln5ufvyv.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 11th April Morning Shift Mathematics - Ellipse Question 12 English Explanation">
<br><br>Equation of $A B$ is $\frac{x}{\frac{1}{\sqrt{K}}}+\frac{y}{\frac{1}{K}}=1$
<br><br> or $ \sqrt{K} x+K y=1$
<br><br>$r_K$ is the radius of circle $C_K$,
<br><br>So, $r_K=$ perpendicular distance from $(0,0)$ to $\mathrm{AB}$
<br><br>$$
\begin{aligned}
r_K & =\frac{|0+0-1|}{\sqrt{(\sqrt{K})^2+K^2}} \\\\
& =\frac{1}{\sqrt{K+K^2}} \\\\
\frac{1}{r_K^2} & =K+K^2
\end{aligned}
$$
<br><br>$$
\begin{aligned}
\sum_{K=1}^{20} \frac{1}{r_K^2} & =\sum_{K=1}^{20} K+\sum_{K=1}^{20} K^2 \\\\
& =\frac{20 \times 21}{2}+\frac{20 \times 21 \times 41}{6} \\\\
& =210+2870=3080
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-11th-april-morning-shift
|
lsbkn1ul
|
maths
|
ellipse
|
chord-of-ellipse
|
The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\left(1, \frac{2}{5}\right)$, is equal to :
|
[{"identifier": "A", "content": "$\\frac{\\sqrt{1691}}{5}$"}, {"identifier": "B", "content": "$\\frac{\\sqrt{2009}}{5}$"}, {"identifier": "C", "content": "$\\frac{\\sqrt{1541}}{5}$"}, {"identifier": "D", "content": "$\\frac{\\sqrt{1741}}{5}$"}]
|
["A"]
| null |
<p>Equation of chord with given middle point.</p>
<p>$$\begin{aligned}
& T=S_1 \\
& \frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100} \\
& \frac{8 x+5 y}{200}=\frac{8+2}{200} \\
& y=\frac{10-8 x}{5} \quad \text{.... (i)}
\end{aligned}$$</p>
<p>$$\frac{x^2}{25}+\frac{(10-8 x)^2}{400}=1$$ (put in original equation)</p>
<p>$$\begin{aligned}
& \frac{16 x^2+100+64 x^2-160 x}{400}=1 \\
& 4 x^2-8 x-15=0 \\
& x=\frac{8 \pm \sqrt{304}}{8} \\
& x_1=\frac{8+\sqrt{304}}{8} ; x_2=\frac{8-\sqrt{304}}{8}
\end{aligned}$$</p>
<p>Similarly, $$y=\frac{10-18 \pm \sqrt{304}}{5}=\frac{2 \pm \sqrt{304}}{5}$$</p>
<p>$$\begin{aligned}
& \mathrm{y}_1=\frac{2-\sqrt{304}}{5} ; \mathrm{y}_2=\frac{2+\sqrt{304}}{5} \\
& \text { Distance }=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2} \\
& =\sqrt{\frac{4 \times 304}{64}+\frac{4 \times 304}{25}}=\frac{\sqrt{1691}}{5} \\
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-27th-january-morning-shift
|
EvZIdx6Rf9PPNIN0
|
maths
|
ellipse
|
common-tangent
|
<b>STATEMENT-1 :</b> An equation of a common tangent to the parabola $${y^2} = 16\sqrt 3 x$$ and the ellipse $$2{x^2} + {y^2} = 4$$ is $$y = 2x + 2\sqrt 3 $$
<p><b>STATEMENT-2 :</b>If line $$y = mx + {{4\sqrt 3 } \over m},\left( {m \ne 0} \right)$$ is a common tangent to the parabola $${y^2} = 16\sqrt {3x} $$and the ellipse $$2{x^2} + {y^2} = 4$$, then $$m$$ satisfies $${m^4} + 2{m^2} = 24$$</p>
|
[{"identifier": "A", "content": "Statement-1 is false, Statement-2 is true."}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1."}, {"identifier": "C", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."}, {"identifier": "D", "content": "Statement-1 is true, Statement-2 is false."}]
|
["B"]
| null |
Given equation of ellipse is $$2{x^2} + {y^2} = 4$$
<br><br>$$ \Rightarrow {{2{x^2}} \over 4} + {{{y^2}} \over 4} = 1 \Rightarrow {{{x_2}} \over 2} + {{{y^2}} \over 4} = 1$$
<br><br>Equation of tangent to the ellipse $${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$$ is
<br><br>$$y = mx \pm \sqrt {2{m^2} + 4} \,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>( as equation of tangent to the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
<br><br>is $$y=mx+c$$ where $$c = \pm \sqrt {{a^2}{m^2} + {b^2}} $$ )
<br><br>Now, Equation of tangent to the parabola
<br><br>$${y^2} = 16\sqrt 3 x$$ is $$y = mx + {{4\sqrt 3 } \over m}\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br> ( as equation of tangent to the parabola
<br><br>$${y^2} = 4ax$$ is $$y = mx + {a \over m}$$ )
<br><br>On comparing $$(1)$$ and $$(2),$$ we get
<br><br>$${{4\sqrt 3 } \over m} = \pm \sqrt {2{m^2} + 4} $$
<br><br>Squaring on both the sides, we get
<br><br>$$16\left( 3 \right) = \left( {2{m^2} + 4} \right){m^2}$$
<br><br>$$ \Rightarrow 48 = {m^2}\left( {2{m^2} + 4} \right) \Rightarrow 2{m^4} + 4{m^2} - 48 = 0$$
<br><br>$$ \Rightarrow {m^4} + 2{m^2} - 24 = 0 \Rightarrow \left( {{m^2} + 6} \right)\left( {{m^2} - 4} \right) = 0$$
<br><br>$$ \Rightarrow {m^2} = 4$$ ( as $${m^2} \ne - 6$$ ) $$ \Rightarrow m = \pm 2$$
<br><br>$$ \Rightarrow $$ Equation of common tangents are $$y = \pm 2x \pm 2\sqrt 3 $$
<br><br>Thus, statement - $$1$$ is true.
<br><br>Statement - $$2$$ is obviously true.
|
mcq
|
aieee-2012
|
1uvsN9WuUVTtVaGYvw18hoxe66ijvwq7lwv
|
maths
|
ellipse
|
common-tangent
|
If the tangent to the parabola y<sup>2</sup> = x at a point
($$\alpha $$, $$\beta $$), ($$\beta $$ > 0) is also a tangent to the ellipse,
x<sup>2</sup> + 2y<sup>2</sup> = 1, then $$\alpha $$ is equal to :
|
[{"identifier": "A", "content": "$$\\sqrt 2 + 1$$"}, {"identifier": "B", "content": "$$\\sqrt 2 - 1$$"}, {"identifier": "C", "content": "$$2\\sqrt 2 + 1$$"}, {"identifier": "D", "content": "$$2\\sqrt 2 - 1$$"}]
|
["A"]
| null |
Point P($$\alpha $$, $$\beta $$) is on the parabola y<sup>2</sup> = x
<br><br>$$ \therefore $$ $${\beta ^2} = \alpha $$ ...........(1)
<br><br>Equation of tangent to the parabola y<sup>2</sup> = x
<br><br>at ($$\alpha $$, $$\beta $$) is T = 0
<br><br>$$\beta y = {{x + \alpha } \over 2}$$
<br><br>$$ \Rightarrow $$ $$2\beta y = x + \alpha $$
<br><br>$$ \Rightarrow $$ $$y = {1 \over {2\beta }}x + {\alpha \over {2\beta }}$$
<br><br>For this straight line, m = $${1 \over {2\beta }}$$ and c = $${\alpha \over {2\beta }}$$
<br><br>This tangent is also a tangent to ellipse x<sup>2</sup> + 2y<sup>2</sup> = 1.
<br><br>$$ \Rightarrow $$ $${{{x^2}} \over 1} + {{{y^2}} \over {{1 \over 2}}} = 1$$
<br><br>$$ \therefore $$ $${a^2} = 1$$ and $${b^2} = {1 \over 2}$$.
<br><br>We know the condition of tangency on ellipse is
<br><br>$${c^2} = {a^2}{m^2} + {b^2}$$
<br><br>$$ \Rightarrow $$ $${\left( {{\alpha \over {2\beta }}} \right)^2} = 1{\left( {{1 \over {2\beta }}} \right)^2} + {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $${{{\alpha ^2}} \over {4{\beta ^2}}} = {1 \over {4{\beta ^2}}} + {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $${\alpha ^2} = 1 + 2{\beta ^2}$$
<br><br>$$ \Rightarrow $$ $${\alpha ^2} = 1 + 2\alpha $$
<br><br>$$ \Rightarrow $$ $${\alpha ^2} - 2\alpha - 1 = 0$$
<br><br>$$ \Rightarrow $$ $$\alpha = 1 + \sqrt 2 $$ and $$\alpha = 1 - \sqrt 2 $$
|
mcq
|
jee-main-2019-online-9th-april-evening-slot
|
xtgmJdXktn5EUKYs6j1kluz2mfj
|
maths
|
ellipse
|
common-tangent
|
Let L be a common tangent line to the curves <br/><br/>4x<sup>2</sup> + 9y<sup>2</sup> = 36 and (2x)<sup>2</sup> + (2y)<sup>2</sup> = 31. Then the <br/><br/>square of the slope of the line L is __________.
|
[]
| null |
3
|
Tangent to the curve $${{{x^2}} \over 9} + {{{y^2}} \over {14}} = 1$$ is <br><br>$$y = mx + \sqrt {9{m^2} + 4} $$<br><br>and equation of tangent to the curve $${x^2} + {y^2} = {{31} \over 4}$$ is<br><br>$$y = mx + \sqrt {{{31} \over 4}{{(1 + m)}^2}} $$<br><br>for common tangent $$9{m^2} + 4 = {{31} \over 4} + {{31} \over 4}{m^2}$$<br><br>$$ \Rightarrow {5 \over 4}{m^2} = {{15} \over 4}$$<br><br>$$ \Rightarrow {m^2} = 3$$
|
integer
|
jee-main-2021-online-26th-february-evening-slot
|
1l58fgd83
|
maths
|
ellipse
|
common-tangent
|
<p>If m is the slope of a common tangent to the curves $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$ and $${x^2} + {y^2} = 12$$, then $$12{m^2}$$ is equal to :</p>
|
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "12"}]
|
["B"]
| null |
<p>$${C_1}:{{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$ and $${C_2}:{x^2} + {y^2} = 12$$</p>
<p>Let $$y = mx \pm \,\sqrt {16{m^2} + 9} $$ be any tangent to C<sub>1</sub> and if this is also tangent to C<sub>2</sub> then</p>
<p>$$\left| {{{\sqrt {16{m^2} + 9} } \over {\sqrt {{m^2} + 1} }}} \right| = \sqrt {12} $$</p>
<p>$$ \Rightarrow 16{m^2} + 9 = 12{m^2} + 12$$</p>
<p>$$ \Rightarrow 4{m^2} = 3 \Rightarrow 12{m^2} = 9$$</p>
|
mcq
|
jee-main-2022-online-26th-june-evening-shift
|
1lgvqbyet
|
maths
|
ellipse
|
common-tangent
|
<p>Let a circle of radius 4 be concentric to the ellipse $$15 x^{2}+19 y^{2}=285$$. Then the common tangents are inclined to the minor axis of the ellipse at the angle :</p>
|
[{"identifier": "A", "content": "$$\\frac{\\pi}{4}$$"}, {"identifier": "B", "content": "$$\\frac{\\pi}{3}$$"}, {"identifier": "C", "content": "$$\\frac{\\pi}{6}$$"}, {"identifier": "D", "content": "$$\\frac{\\pi}{12}$$"}]
|
["B"]
| null |
We have, equation of ellipse : $15 x^2+19 y^2=285$
<br/><br/>or $ \frac{x^2}{19}+\frac{y^2}{15}=1$
<br/><br/>Let the coordinate of center of circle be $(0,0)$.
<br/><br/>Equation of circle is $x^2+y^2=16$
<br/><br/>Equation of tangent of ellipse is
<br/><br/>$$
\begin{gathered}
y=m x \pm \sqrt{19 m^2+15} \text { or } \\\\
m x-y \pm \sqrt{19 m^2+15}=0
\end{gathered}
$$
<br/><br/>It is also tangent to the circle $x^2+y^2=16$
<br/><br/>Perpendicular distance from center of circle to tangent $=4$
<br/><br/>$$
\frac{\left|0-0 \pm \sqrt{19 m^2+15}\right|}{\sqrt{m^2+1}}=4
$$
<br/><br/>On squaring both side, we get
<br/><br/>$$
\begin{gathered}
19 m^2+15=16 m^2+16 \\\\
3 m^2=1 \\\\
m^2=\frac{1}{3} \\\\
m= \pm \frac{1}{\sqrt{3}}
\end{gathered}
$$
<br/><br/>$$
\tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\frac{\pi}{6} \text { with } X \text {-axis or }
$$$\frac{\pi}{3}$ with $Y$-axis
<br/><br/>Hence, the common tangents are inclined to the minor axis of the ellipse at an angle of $\frac{\pi}{3}$.
|
mcq
|
jee-main-2023-online-10th-april-evening-shift
|
lhi1qU8ZF8dwtl0X
|
maths
|
ellipse
|
locus
|
The locus of the foot of perpendicular drawn from the centre of the ellipse $${x^2} + 3{y^2} = 6$$ on any tangent to it is :
|
[{"identifier": "A", "content": "$$\\left( {{x^2} + {y^2}} \\right) ^2 = 6{x^2} + 2{y^2}$$ "}, {"identifier": "B", "content": "$$\\left( {{x^2} + {y^2}} \\right) ^2 = 6{x^2} - 2{y^2}$$"}, {"identifier": "C", "content": "$$\\left( {{x^2} - {y^2}} \\right) ^2 = 6{x^2} + 2{y^2}$$ "}, {"identifier": "D", "content": "$$\\left( {{x^2} - {y^2}} \\right) ^2 = 6{x^2} - 2{y^2}$$ "}]
|
["A"]
| null |
Given $$e{q^n}$$ of ellipse can be written as
<br><br>$${{{x^2}} \over 6} + {{{y^2}} \over 2} = 1 \Rightarrow {a^2} = 6,{b^2} = 2$$
<br><br>Now, equation of any variable tangent is
<br><br>$$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} ....\left( i \right)$$
<br><br>where $$m$$ is slope of the tangent
<br><br>So, equation of perpendicular line drawn-
<br><br>from center to tangent is
<br><br>$$y = {{ - x} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>Eliminating $$m,$$ we get
<br><br>$$\left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2}$$
<br><br>$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2}$$
<br><br>$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$$
|
mcq
|
jee-main-2014-offline
|
1ktk74324
|
maths
|
ellipse
|
locus
|
The locus of mid-points of the line segments joining ($$-$$3, $$-$$5) and the points on the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ is :
|
[{"identifier": "A", "content": "$$9{x^2} + 4{y^2} + 18x + 8y + 145 = 0$$"}, {"identifier": "B", "content": "$$36{x^2} + 16{y^2} + 90x + 56y + 145 = 0$$"}, {"identifier": "C", "content": "$$36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$$"}, {"identifier": "D", "content": "$$36{x^2} + 16{y^2} + 72x + 32y + 145 = 0$$"}]
|
["C"]
| null |
General point on $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ is A(2cos$$\theta$$, 3sin$$\theta$$)<br><br>given B($$-$$3, $$-$$5)<br><br>midpoint $$C\left( {{{2\cos \theta - 3} \over 2},{{3\sin \theta - 5} \over 2}} \right)$$<br><br>$$h = {{2\cos \theta - 3} \over 2};k = {{3\sin \theta - 5} \over 2}$$<br><br>$$ \Rightarrow {\left( {{{2h + 3} \over 2}} \right)^2} + {\left( {{{2k + 5} \over 3}} \right)^2} = 1$$<br><br>$$ \Rightarrow 36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$$
|
mcq
|
jee-main-2021-online-31st-august-evening-shift
|
1l58g1v33
|
maths
|
ellipse
|
locus
|
<p>The locus of the mid point of the line segment joining the point (4, 3) and the points on the ellipse $${x^2} + 2{y^2} = 4$$ is an ellipse with eccentricity :</p>
|
[{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over {2\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}]
|
["C"]
| null |
<p>Let $$P(2\cos \theta ,\,\sqrt 2 \sin \theta )$$ be any point on ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$$ and Q(4, 3) and let (h, k) be the mid point of PQ</p>
<p>then $$h = {{2\cos \theta + 4} \over 2},\,k = {{\sqrt 2 \sin \theta + 3} \over 2}$$</p>
<p>$$\therefore$$ $$\cos \theta = h - 2,\,\sin \theta = {{2k - 3} \over {\sqrt 2 }}$$</p>
<p>$$\therefore$$ $${(h - 2)^2} + {\left( {{{2k - 3} \over {\sqrt 2 }}} \right)^2} = 1$$</p>
<p>$$ \Rightarrow {{{{(x - 2)}^2}} \over 1} + {{{{\left( {y - {3 \over 2}} \right)}^2}} \over {{1 \over 2}}} = 1$$</p>
<p>$$\therefore$$ $$e = \sqrt {1 - {1 \over 2}} = {1 \over {\sqrt 2 }}$$</p>
|
mcq
|
jee-main-2022-online-26th-june-evening-shift
|
lsamf0th
|
maths
|
ellipse
|
locus
|
Let $\mathrm{P}$ be a point on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let the line passing through $\mathrm{P}$ and parallel to $y$-axis meet the circle $x^2+y^2=9$ at point $\mathrm{Q}$ such that $\mathrm{P}$ and $\mathrm{Q}$ are on the same side of the $x$-axis. Then, the eccentricity of the locus of the point $R$ on $P Q$ such that $P R: R Q=4: 3$ as $P$ moves on the ellipse, is :
|
[{"identifier": "A", "content": "$\\frac{13}{21}$"}, {"identifier": "B", "content": "$\\frac{\\sqrt{139}}{23}$"}, {"identifier": "C", "content": "$\\frac{\\sqrt{13}}{7}$"}, {"identifier": "D", "content": "$\\frac{11}{19}$"}]
|
["C"]
| null |
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqdjrqo/772aa650-3e8c-46a8-adce-0608922ed01d/09ef1300-cdbd-11ee-a926-9fabe9a328d8/file-6y3zli1lsqdjrqp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqdjrqo/772aa650-3e8c-46a8-adce-0608922ed01d/09ef1300-cdbd-11ee-a926-9fabe9a328d8/file-6y3zli1lsqdjrqp.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Evening Shift Mathematics - Ellipse Question 8 English Explanation 1">
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqdl51c/d5a5c6d5-b0af-43b9-b7e2-7efa69d9eeff/30039c00-cdbd-11ee-a926-9fabe9a328d8/file-6y3zli1lsqdl51d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqdl51c/d5a5c6d5-b0af-43b9-b7e2-7efa69d9eeff/30039c00-cdbd-11ee-a926-9fabe9a328d8/file-6y3zli1lsqdl51d.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Evening Shift Mathematics - Ellipse Question 8 English Explanation 2">
<br>$\begin{aligned} & \mathrm{h}=3 \cos \theta \\\\ & \mathrm{k}=\frac{18}{7} \sin \theta\end{aligned}$
<br><br>$\begin{aligned} & \therefore \text { locus }=\frac{\mathrm{x}^2}{9}+\frac{49 \mathrm{y}^2}{324}=1 \\\\ & \mathrm{e}=\sqrt{1-\frac{324}{49 \times 9}}=\frac{\sqrt{117}}{21}=\frac{\sqrt{13}}{7}\end{aligned}$
|
mcq
|
jee-main-2024-online-1st-february-evening-shift
|
Fdaip3SiqPfH8cxI
|
maths
|
ellipse
|
normal-to-ellipse
|
The eccentricity of an ellipse whose centre is at the origin is $${1 \over 2}$$. If one of its directrices is x = – 4, then the
equation of the normal to it at $$\left( {1,{3 \over 2}} \right)$$ is :
|
[{"identifier": "A", "content": "2y \u2013 x = 2"}, {"identifier": "B", "content": "4x \u2013 2y = 1"}, {"identifier": "C", "content": "4x + 2y = 7"}, {"identifier": "D", "content": "x + 2y = 4"}]
|
["B"]
| null |
Given e = $${1 \over 2}$$ and $${a \over e}$$ = 4
<br><br>$$ \therefore $$ $$a$$ = 2
<br><br>We have b<sup>2</sup> = $$a$$<sup>2</sup> (1 – e<sup>2</sup>) = $$4\left( {1 - {1 \over 4}} \right)$$ = 3
<br><br>$$ \therefore $$ Equation of ellipse is
<br><br>$${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$
<br><br>Now, the equation of normal at $$\left( {1,{3 \over 2}} \right)$$ is
<br><br>$${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}$$
<br><br>$$ \Rightarrow $$ $${{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3$$
<br><br>$$ \Rightarrow $$ 4x – 2y = 1
|
mcq
|
jee-main-2017-offline
|
5Xy9INkIXJfGIM8vGg3rsa0w2w9jx2eobip
|
maths
|
ellipse
|
normal-to-ellipse
|
The tangent and normal to the ellipse 3x<sup>2</sup>
+ 5y<sup>2</sup>
= 32 at the point P(2, 2) meet the x-axis at Q and R,
respectively. Then the area (in sq. units) of the triangle PQR is :
|
[{"identifier": "A", "content": "$${{14} \\over 3}$$"}, {"identifier": "B", "content": "$${{16} \\over 3}$$"}, {"identifier": "C", "content": "$${{68} \\over {15}}$$"}, {"identifier": "D", "content": "$${{34} \\over {15}}$$"}]
|
["C"]
| null |
$$3{x^2} + 5{y^2} = 32$$<br><br>
6x + 10yy' = 0<br><br>
$$ \Rightarrow y' = {{ - 3x} \over {5y}}$$<br><br>
$$ \Rightarrow y{'_{(2,2)}} = - {3 \over 5}$$<br><br>
Tangent $$(y - 2) = - {3 \over 5}(x - 2) \Rightarrow Q\left( {{{16} \over 3},0} \right)$$<br><br>
Normal $$(y - 2) = {5 \over 3}(x - 2) \Rightarrow R\left( {{{4} \over 5},0} \right)$$<br><br>
$$ \therefore $$ Area = $${1 \over 2}(QR) \times 2 = QR = {{68} \over {15}}$$
|
mcq
|
jee-main-2019-online-10th-april-evening-slot
|
ABobMP93kuwLqACJVx3rsa0w2w9jx65dnxr
|
maths
|
ellipse
|
normal-to-ellipse
|
If the normal to the ellipse 3x<sup>2</sup>
+ 4y<sup>2</sup>
= 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent
to the ellipse at P passes through Q(4,4) then PQ is equal to :
|
[{"identifier": "A", "content": "$${{\\sqrt {61} } \\over 2}$$"}, {"identifier": "B", "content": "$${{\\sqrt {221} } \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt {157} } \\over 2}$$"}, {"identifier": "D", "content": "$${{5\\sqrt 5 } \\over 2}$$"}]
|
["D"]
| null |
Equation of ellipse is $${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$<br><br>
Normal at P(2 cos $$\theta $$, $$\sqrt 3 \sin \theta $$) is 2x sin$$\theta $$ - $$\sqrt 3 y\,cos\theta $$ = sin $$\theta $$ cos $$\theta $$ as the normal is parallel to 2x + y = 4<br><br>
$$ \Rightarrow $$ $${2 \over {\sqrt 3 }}\tan \theta = - 2$$<br><br>
$$ \Rightarrow \tan \theta = - \sqrt 3 \,\,......\,(i)$$<br><br>
tangent at P(2 cos $$\theta $$, $$\sqrt 3 \sin \theta $$) is <br><br>
$$\sqrt 3 $$ x cos $$\theta $$ + 2y sin $$\theta $$ = 2 $$\sqrt 3 $$<br><br>
Passes through (4, 4)<br><br>
$$ \Rightarrow $$ 4$$\sqrt 3 $$ cos $$\theta $$ + 8 sin $$\theta $$ = 2$$\sqrt 3 $$ ....... (ii)<br><br>
From (i) and (ii) <br><br>
$$ \Rightarrow P\left( { - 1,{3 \over 2}} \right)\,\& \,Q(4,4)$$<br><br>
$$ \Rightarrow PQ = \sqrt {25 + {{25} \over 4}} = {{5\sqrt 5 } \over 2}$$
|
mcq
|
jee-main-2019-online-12th-april-morning-slot
|
7U4w1GA7rQRFV2KmmT7k9k2k5gjiucr
|
maths
|
ellipse
|
normal-to-ellipse
|
Let the line y = mx and the ellipse 2x<sup>2</sup> + y<sup>2</sup> = 1
intersect at a ponit P in the first quadrant. If the
normal to this ellipse at P meets the co-ordinate axes at $$\left( { - {1 \over {3\sqrt 2 }},0} \right)$$ and (0, $$\beta $$), then $$\beta $$ is equal
to :
|
[{"identifier": "A", "content": "$${{\\sqrt 2 } \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${{2\\sqrt 2 } \\over 3}$$"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 3 }}$$"}]
|
["A"]
| null |
Let P be (x<sub>1</sub>
, y<sub>1</sub>)
<br><br>Equation of normal at P is $${x \over {2{x_1}}} - {y \over {{y_1}}} = {1 \over 2} - 1$$
<br><br>It passes through $$\left( { - {1 \over {3\sqrt 2 }},0} \right)$$
<br><br>$$ \therefore $$ $${{ - 1} \over {6\sqrt 2 {x_1}}} = - {1 \over 2}$$
<br><br>$$ \Rightarrow $$ x<sub>1</sub> = $${1 \over {3\sqrt 2 }}$$
<br><br>Also using 2$$x_1^2$$ + $$y_1^2$$ = 1
<br><br>$$ \Rightarrow $$ $$y_1^2$$ = 1 - $${2 \over {18}}$$
<br><br>$$ \Rightarrow $$ y<sub>1</sub> = $${{2\sqrt 2 } \over 3}$$
<br><br>(0, $$\beta $$) is on the normal.
<br><br>So 0 - $${\beta \over {{{2\sqrt 2 } \over 3}}}$$ = $$ - {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $$\beta $$ = $${{\sqrt 2 } \over 3}$$
|
mcq
|
jee-main-2020-online-8th-january-morning-slot
|
1EvusdwISdLMDs07h4jgy2xukfakgvy9
|
maths
|
ellipse
|
normal-to-ellipse
|
Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is $${1 \over 2}$$. If P(1, $$\beta $$), $$\beta $$ > 0 is a point on this ellipse, then the equation of the normal to it at P is :
|
[{"identifier": "A", "content": "4x \u2013 3y = 2\n"}, {"identifier": "B", "content": "8x \u2013 2y = 5"}, {"identifier": "C", "content": "7x \u2013 4y = 1 "}, {"identifier": "D", "content": "4x \u2013 2y = 1"}]
|
["D"]
| null |
$$e = {1 \over 2}$$ <br><br>$$x = {a \over e} = 4$$<br><br>$$ \Rightarrow $$ a = 2<br><br>$${e^2} = 1 - {{{b^2}} \over {{a^2}}} $$
<br><br>$$\Rightarrow {1 \over 4} = 1 - {{{b^2}} \over 4}$$<br><br>$${{{b^2}} \over 4} = {3 \over 4} \Rightarrow {b^2} = 3$$<br><br>$$ \therefore $$ Ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$<br><br>P(1, $$\beta $$) is on the ellipse<br><br> $${1 \over 4} + {{{\beta ^2}} \over 3} = 1$$<br><br>$${{{\beta ^2}} \over 3} = {3 \over 4} \Rightarrow \beta = {3 \over 2}$$<br><br>$$ \Rightarrow P\left( {1,{3 \over 2}} \right)$$<br><br>Equation of normal $${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}$$<br><br>$$ \Rightarrow $$ $${{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3$$<br><br>$$ \Rightarrow $$ $$4x - 2y = 1$$
|
mcq
|
jee-main-2020-online-4th-september-evening-slot
|
N3TmnUyC4v34yUj1Gpjgy2xukg0cqp9y
|
maths
|
ellipse
|
normal-to-ellipse
|
If the normal at an end of a latus rectum of an
ellipse passes through an extremity of the
minor axis, then the eccentricity e of the ellipse
satisfies :
|
[{"identifier": "A", "content": "e<sup>4</sup> + 2e<sup>2</sup> \u2013 1 = 0"}, {"identifier": "B", "content": "e<sup>4</sup> + e<sup>2</sup> \u2013 1 = 0"}, {"identifier": "C", "content": "e<sup>2</sup> + 2e \u2013 1 = 0"}, {"identifier": "D", "content": "e<sup>2</sup> + e \u2013 1 = 0"}]
|
["B"]
| null |
Equation of normal at $$\left( {ae,{{{b^2}} \over a}} \right)$$
<br><br>$${{{a^2}x} \over {ae}} - {{{b^2}y} \over {{{{b^2}} \over a}}} = {a^2} - {b^2}$$
<br><br>It passes through (0,–b)
<br><br>$$ \therefore $$ $$0 - {{{b^2}\left( { - b} \right)} \over {{{{b^2}} \over a}}} = {a^2} - {b^2}$$
<br><br>$$ \Rightarrow $$ $$a$$b = $${a^2} - {b^2}$$
<br><br>$$ \Rightarrow $$ $$a$$b = $${a^2}{e^2}$$ [as b<sup>2</sup> = $${a^2}\left( {1 - {e^2}} \right)$$]
<br><br>$$ \Rightarrow $$ $$a$$<sup>2</sup>b<sup>2</sup> = $${a^4}{e^4}$$
<br><br>$$ \Rightarrow $$ $${{{{b^2}} \over {{a^2}}}}$$ = e<sup>4</sup>
<br><br>$$ \Rightarrow $$ $${1 - {e^2}}$$ = e<sup>4</sup>
<br><br>$$ \Rightarrow $$ e<sup>4</sup> + e<sup>2</sup> – 1 = 0
|
mcq
|
jee-main-2020-online-6th-september-evening-slot
|
1ldpt2o9m
|
maths
|
ellipse
|
normal-to-ellipse
|
<p>If the maximum distance of normal to the ellipse $$\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1, b < 2$$, from the origin is 1, then the eccentricity of the ellipse is :</p>
|
[{"identifier": "A", "content": "$$\\frac{\\sqrt{3}}{4}$$"}, {"identifier": "B", "content": "$$\\frac{1}{2}$$"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt{2}}$$"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{3}}{2}$$"}]
|
["D"]
| null |
Equation of normal is
<br/><br/>$2 x \sec \theta-b y \operatorname{cosec} \theta=4-b^{2}$
<br/><br/>Distance from $(0,0)=\frac{4-b^{2}}{\sqrt{4 \sec ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta}}$
<br/><br/>Distance is maximum if
<br/><br/>$4 \sec ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta$ is minimum
<br/><br/>$\Rightarrow \tan ^{2} \theta=\frac{\mathrm{b}}{2}$
<br/><br/>$\Rightarrow \frac{4-b^{2}}{\sqrt{4 \cdot \frac{b+2}{2}+b^{2} \cdot \frac{b+2}{b}}}=1$
<br/><br/>$\Rightarrow 4-b^{2}=(b+2) \Rightarrow b^{2}+b-2=0$
<br/><br/>$\Rightarrow(b+2)(b-1)=0$
<br/><br/>$\Rightarrow b=1$
<br/><br/>$\therefore e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$
|
mcq
|
jee-main-2023-online-31st-january-morning-shift
|
1lgpy8jfv
|
maths
|
ellipse
|
normal-to-ellipse
|
<p>Let the tangent and normal at the point $$(3 \sqrt{3}, 1)$$ on the ellipse $$\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$$ meet the $$y$$-axis at the points $$A$$ and $$B$$ respectively. Let the circle $$C$$ be drawn taking $$A B$$ as a diameter and the line $$x=2 \sqrt{5}$$ intersect $$C$$ at the points $$P$$ and $$Q$$. If the tangents at the points $$P$$ and $$Q$$ on the circle intersect at the point $$(\alpha, \beta)$$, then $$\alpha^{2}-\beta^{2}$$ is equal to :</p>
|
[{"identifier": "A", "content": "61"}, {"identifier": "B", "content": "$$\\frac{304}{5}\n$$"}, {"identifier": "C", "content": "60"}, {"identifier": "D", "content": "$$\\frac{314}{5}\n$$"}]
|
["B"]
| null |
$$
\begin{aligned}
& \frac{x^2}{36}+\frac{y^2}{4}=1 \\\\
& T: \frac{3 \sqrt{3} x}{36}+\frac{y}{4}=1 \\\\
& T: \frac{\sqrt{3} x}{12}+\frac{y}{4}=1 \\\\
& N: \frac{x-3 \sqrt{3}}{\frac{3 \sqrt{3}}{36}}=\frac{y-1}{\frac{1}{4}}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \frac{12 x-36 \sqrt{3}}{\sqrt{3}}=4 y-4 \\\\
& 3 x-9 \sqrt{3}=\sqrt{3} y-\sqrt{3} \\\\
& N: 3 x-\sqrt{3} y=8 \sqrt{3} \\\\
& A(0,4) \quad B(0,-8) \\\\
& \text { C: } x^2+(y-4)(y+8)=0 \\\\
& \text { Line } x=2 \sqrt{5} \\\\
& 20+y^2+4 y-32=0 \\\\
& y^2+4 y-12=0 \\\\
& (y+6)(y-2)=0
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\\\
& C: x^2+y^2+4 y-32=0 \\\\
& P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\\\
& T: x x_1+y y_1+2 y+2 y_1-32=0 \\\\
& T_1: 2 \sqrt{5} x-6 y+2 y-12-32=0 \\\\
& \quad 2 \sqrt{5} x-4 y=44 \\\\
& T_1: \sqrt{5} x-2 y=22 .......(i) \\\\
& T_2: 2 \sqrt{5} x+2 y+2 y+4-32=0 \\\\
& 2 \sqrt{5} x+4 y=28 \\\\
& T_2: \sqrt{5} x+2 y=14 .......(ii)
\end{aligned}
$$
<br/><br/>From (i) and (ii), we get
<br/><br/>$$
\begin{aligned}
& \alpha=\frac{18}{\sqrt{5}} \quad \beta=-2 \\\\
& \alpha^2-\beta^2=\frac{304}{5}
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-13th-april-morning-shift
|
rlmIdLEuDAejugVr
|
maths
|
ellipse
|
position-of-point-and-chord-joining-of-two-points
|
The equation of the circle passing through the foci of the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$, and having centre at $$(0,3)$$ is :
|
[{"identifier": "A", "content": "$${x^2} + {y^2} - 6y - 7 = 0$$ "}, {"identifier": "B", "content": "$${x^2} + {y^2} - 6y + 7 = 0$$"}, {"identifier": "C", "content": "$${x^2} + {y^2} - 6y - 5 = 0$$"}, {"identifier": "D", "content": "$${x^2} + {y^2} - 6y + 5 = 0$$"}]
|
["A"]
| null |
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263921/exam_images/s76qynlxlel1kkck45r2.webp" loading="lazy" alt="JEE Main 2013 (Offline) Mathematics - Ellipse Question 76 English Explanation">
<br><br>From the given equation of ellipse, we have
<br><br>$$a = 4,b = 3,e = \sqrt {1 - {9 \over {16}}} $$
<br><br>$$ \Rightarrow e = {{\sqrt 7 } \over 4}$$
<br><br>Now, radius of this circle $$ = {a^2} = 16$$
<br><br>$$ \Rightarrow Focii = \left( { \pm \sqrt 7 ,0} \right)$$
<br><br>Now equation of circle is
<br><br>$${\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = 16$$
<br><br>$${x^2} + y{}^2 - 6y - 7 = 0$$
|
mcq
|
jee-main-2013-offline
|
2oTqNMY0qPaZCT9gEHjgy2xukfjjqpjo
|
maths
|
ellipse
|
position-of-point-and-chord-joining-of-two-points
|
If the co-ordinates of two points A and B <br/>are $$\left( {\sqrt 7 ,0} \right)$$ and $$\left( { - \sqrt 7 ,0} \right)$$ respectively and<br/> P is any
point on the conic, 9x<sup>2</sup> + 16y<sup>2</sup> = 144, then PA + PB is equal to :
|
[{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "6"}]
|
["A"]
| null |
9x<sup>2</sup> + 16y<sup>2</sup> = 144
<br><br>$$ \Rightarrow $$ $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$
<br><br>$$ \therefore $$ a = 4; b = 3;
<br><br>Now e = $$\sqrt {1 - {9 \over {16}}} = {{\sqrt 7 } \over 4}$$
<br><br>A and B are foci
<br><br>PA + PB = 2a = 2 × 4 = 8
|
mcq
|
jee-main-2020-online-5th-september-morning-slot
|
BcheBfrZmexesZ4Um71klt9cr9j
|
maths
|
ellipse
|
position-of-point-and-chord-joining-of-two-points
|
If the curve x<sup>2</sup> + 2y<sup>2</sup> = 2 intersects the line x + y = 1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is :
|
[{"identifier": "A", "content": "$${\\pi \\over 2} - {\\tan ^{ - 1}}\\left( {{1 \\over 4}} \\right)$$"}, {"identifier": "B", "content": "$${\\pi \\over 2} + {\\tan ^{ - 1}}\\left( {{1 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$${\\pi \\over 2} - {\\tan ^{ - 1}}\\left( {{1 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$${\\pi \\over 2} + {\\tan ^{ - 1}}\\left( {{1 \\over 4}} \\right)$$"}]
|
["D"]
| null |
Ellipse : $${x \over 2} + {y \over 1} = 1$$<br><br>Line : $$x + y = 1$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267198/exam_images/lmh3h5mktsgvuq0vvk3v.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Evening Shift Mathematics - Ellipse Question 48 English Explanation"><br><br>Using homogenisation<br><br>$${x^2} + 2{y^2} = 2{(1)^2}$$<br><br>$${x^2} + 2{y^2} = 2{(x + y)^2}$$<br><br>$${x^2} + 2{y^2} = 2{x^2} + 2{y^2} + 4xy$$<br><br>$${x^2} + 4xy = 0$$<br><br>for $$a{x^2} + 2hxy + b{y^2} = 0$$<br><br>$$\tan \theta = \left| {{{2\sqrt {{h^2} - ab} } \over {a + b}}} \right|$$<br><br>$$\tan \theta = \left| {{{2\sqrt {{{(2)}^2} - 0} } \over {1 + 0}}} \right|$$<br><br>$$\tan \theta = - 4$$<br><br>$$\cot \theta = - {1 \over 4}$$<br><br>$$\theta = {\cot ^{ - 1}}\left( { - {1 \over 4}} \right)$$<br><br>$$\theta = \pi - {\cot ^{ - 1}}\left( {{1 \over 4}} \right)$$<br><br>$$\theta = \pi - \left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {{1 \over 4}} \right)} \right)$$<br><br>$$\theta = {\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)$$
|
mcq
|
jee-main-2021-online-25th-february-evening-slot
|
1lgrg5nwu
|
maths
|
ellipse
|
position-of-point-and-chord-joining-of-two-points
|
<p>Let $$\mathrm{P}\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), \mathrm{Q}, \mathrm{R}$$ and $$\mathrm{S}$$ be four points on the ellipse $$9 x^{2}+4 y^{2}=36$$. Let $$\mathrm{PQ}$$ and $$\mathrm{RS}$$ be mutually perpendicular and pass through the origin. If $$\frac{1}{(P Q)^{2}}+\frac{1}{(R S)^{2}}=\frac{p}{q}$$, where $$p$$ and $$q$$ are coprime, then $$p+q$$ is equal to :</p>
|
[{"identifier": "A", "content": "143"}, {"identifier": "B", "content": "147"}, {"identifier": "C", "content": "137"}, {"identifier": "D", "content": "157"}]
|
["D"]
| null |
Given, points $P$ and $R$ are on the ellipse defined by $9x^2+4y^2=36$ which simplifies to $\frac{x^2}{4} + \frac{y^2}{9} = 1$.
This is the standard form of the equation of an ellipse centered at the origin, with semi-major axis $a=3$ along the $y$-axis and semi-minor axis $b=2$ along the $x$-axis.
<br/><br/>OP is the distance from origin O to point P, which is given by :
<br/><br/>$$OP =r_1 = \sqrt{\left(\frac{2\sqrt{3}}{\sqrt{7}}\right)^2 + \left(\frac{6}{\sqrt{7}}\right)^2} = \sqrt{\frac{12}{7} + \frac{36}{7}} = \sqrt{\frac{48}{7}} = 2\sqrt{\frac{12}{7}}.$$
<br/><br/>1. Let's represent the given point $P\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right)$ in polar coordinates. We can write $P$ as $(r_1 \cos \theta, r_1 \sin \theta)$. Since $P$ lies on the ellipse, it must satisfy the equation of the ellipse. Substituting $x=r_1 \cos \theta$ and $y=r_1 \sin \theta$ into the equation of the ellipse gives us :
<br/><br/> $$\frac{r_1^2 \cos ^2 \theta}{4}+\frac{r_1^2 \sin ^2 \theta}{9}=1$$
<br/><br/> Simplifying this, we obtain :
<br/><br/> $$\frac{\cos ^2 \theta}{4}+\frac{\sin ^2 \theta}{9}=\frac{7}{48} \quad \text{--- (equation 1)}$$
<br/><br/>2. Similarly, if we represent the point $R$ as $(-r_2 \sin \theta, r_2 \cos \theta)$, (the negative sign is due to the fact that line RS is perpendicular to line PQ. Since they are perpendicular, the angle between them is 90 degrees or $\pi/2$ radians. In terms of sin and cos, $\sin(\theta + \pi/2) = \cos(\theta)$ and $\cos(\theta + \pi/2) = -\sin(\theta)$.) it too should satisfy the equation of the ellipse. We have :
<br/><br/> $$\frac{r_2^2 \sin ^2 \theta}{4}+\frac{r_2^2 \cos ^2 \theta}{9}=1$$
<br/><br/> Simplifying this, we obtain :
<br/><br/> $$\frac{\sin ^2 \theta}{4}+\frac{\cos ^2 \theta}{9}=\frac{1}{r_2^2} \quad \text{--- (equation 2)}$$
<br/><br/>3. From equations (1) and (2), we have :
<br/><br/> $$\frac{1}{r_2^2}=\frac{1}{4}+\frac{1}{9}-\frac{7}{48}=\frac{31}{144}$$
<br/><br/>4. Now, note that lines $PQ$ and $RS$ are perpendicular and pass through the origin, so $PQ = 2OP$ and $RS = 2OR$. Thus,
<br/><br/> $$\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4} \left(\frac{1}{r_1^2} + \frac{1}{r_2^2} \right)$$
<br/><br/>5. Substituting the values of $r_1$ and $r_2$, we obtain :
<br/><br/> $$\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4}\left(\frac{7}{48}+\frac{31}{144}\right)=\frac{13}{144} = \frac{p}{q}$$
<br/><br/>6. Hence, $p = 13$ and $q = 144$.
<br/><br/>7. So, the final answer $p+q = 13 + 144 = 157$.
|
mcq
|
jee-main-2023-online-12th-april-morning-shift
|
lv7v4g1l
|
maths
|
ellipse
|
position-of-point-and-chord-joining-of-two-points
|
<p>Let the line $$2 x+3 y-\mathrm{k}=0, \mathrm{k}>0$$, intersect the $$x$$-axis and $$y$$-axis at the points $$\mathrm{A}$$ and $$\mathrm{B}$$, respectively. If the equation of the circle having the line segment $$A B$$ as a diameter is $$x^2+y^2-3 x-2 y=0$$ and the length of the latus rectum of the ellipse $$x^2+9 y^2=k^2$$ is $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime, then $$2 \mathrm{~m}+\mathrm{n}$$ is equal to</p>
|
[{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "13"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "10"}]
|
["C"]
| null |
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwgi30yz/efd8a8c4-b19c-464d-9aab-50da9ed84968/ccf9e7b0-177f-11ef-97dc-2d80937d5077/file-1lwgi30z0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwgi30yz/efd8a8c4-b19c-464d-9aab-50da9ed84968/ccf9e7b0-177f-11ef-97dc-2d80937d5077/file-1lwgi30z0.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Morning Shift Mathematics - Ellipse Question 1 English Explanation"></p>
<p>Equation of circle with $$A B$$ as diameter</p>
<p>$$\begin{aligned}
& \left(x-\frac{k}{2}\right) x+y\left(y-\frac{k}{3}\right)=0 \\
& \Rightarrow x^2+y^2-\frac{k x}{2}-\frac{k y}{3}=0
\end{aligned}$$</p>
<p>Comparing, $$k=6$$</p>
<p>Latus rectum of ellipse</p>
<p>$$\begin{aligned}
& x^2+9 y^2=k^2=6^2 \\
& \Rightarrow \frac{x^2}{6^2}+\frac{y^2}{2^2}=1 \\
& \text { L.R }=\frac{2 b^2}{a}=\frac{2 \times 4}{6}=\frac{4}{3} \\
& m=4 \\
& n=3 \\
& 2 m+n=8+3=11
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-5th-april-morning-shift
|
UpdYGwyWWDYrcvXF
|
maths
|
ellipse
|
question-based-on-basic-definition-and-parametric-representation
|
The eccentricity of an ellipse, with its centre at the origin, is $${1 \over 2}$$. If one of the directrices is $$x=4$$, then the equation of the ellipse is :
|
[{"identifier": "A", "content": "$$4{x^2} + 3{y^2} = 1$$ "}, {"identifier": "B", "content": "$$3{x^2} + 4{y^2} = 12$$"}, {"identifier": "C", "content": "$$4{x^2} + 3{y^2} = 12$$"}, {"identifier": "D", "content": "$$3{x^2} + 4{y^2} = 1$$"}]
|
["B"]
| null |
$$e = {1 \over 2}.\,\,$$ Directrix, $$x = {a \over e} = 4$$
<br><br>$$\therefore$$ $$a = 4 \times {1 \over 2} = 2$$
<br><br>$$\therefore$$ $$b = 2\sqrt {1 - {1 \over 4}} = \sqrt 3 $$
<br><br>Equation of elhipe is
<br><br>$${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1 \Rightarrow 3{x^2} + 4{y^2} = 12$$
|
mcq
|
aieee-2004
|
fqgR73iD6te7y31I
|
maths
|
ellipse
|
question-based-on-basic-definition-and-parametric-representation
|
An ellipse has $$OB$$ as semi minor axis, $$F$$ and $$F$$' its focii and theangle $$FBF$$' is a right angle. Then the eccentricity of the ellipse is :
|
[{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 3 }}$$"}]
|
["A"]
| null |
as $$\angle FBF' = {90^ \circ }$$
<br><br>$$ \Rightarrow F{B^2} + F'{B^2} = FF{'^2}$$
<br><br>$$\therefore$$ $${\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} + \left( {\sqrt {{a^2}{e^2} + {b^2}} } \right) = {\left( {2ae} \right)^2}$$
<br><br>$$ \Rightarrow 2\left( {{a^2}{e^2} + {b^2}} \right) = 4{a^2}{e^2}$$
<br><br>$$ \Rightarrow {e^2} = {{{b^2}} \over {{a^2}}}$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263637/exam_images/evoxpwc1lpvv4diyvenq.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Ellipse Question 84 English Explanation">
<br><br>Also $${e^2} = 1 - {b^2}/{a^2} = 1 - {e^2}$$
<br><br>$$ \Rightarrow 2{e^2} = 1,\,\,e = {1 \over {\sqrt 2 }}$$
|
mcq
|
aieee-2005
|
h1jmOU3BvKJGrV5A
|
maths
|
ellipse
|
question-based-on-basic-definition-and-parametric-representation
|
In the ellipse, the distance between its foci is $$6$$ and minor axis is $$8$$. Then its eccentricity is :
|
[{"identifier": "A", "content": "$${3 \\over 5}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${4 \\over 5}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 5 }}$$"}]
|
["A"]
| null |
$$2ae = 6 \Rightarrow ae = 3;\,\,2b = 8 \Rightarrow b = 4$$
<br><br>$${b^2} = {a^2}\left( {1 - {e^2}} \right);16 = {a^2} - {a^2}{e^2}$$
<br><br>$$ \Rightarrow a{}^2 = 16 + 9 = 25$$
<br><br>$$ \Rightarrow a = 5$$
<br><br>$$\therefore$$ $$e = {3 \over a} = {3 \over 5}$$
|
mcq
|
aieee-2006
|
YPUS6uIIDYQa7Fw1
|
maths
|
ellipse
|
question-based-on-basic-definition-and-parametric-representation
|
A focus of an ellipse is at the origin. The directrix is the line $$x=4$$ and the eccentricity is $${{1 \over 2}}$$. Then the length of the semi-major axis is :
|
[{"identifier": "A", "content": "$${{8 \\over 3}}$$"}, {"identifier": "B", "content": "$${{2 \\over 3}}$$"}, {"identifier": "C", "content": "$${{4 \\over 3}}$$"}, {"identifier": "D", "content": "$${{5 \\over 3}}$$"}]
|
["A"]
| null |
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264328/exam_images/fxzt3slso48o3sotippm.webp" loading="lazy" alt="AIEEE 2008 Mathematics - Ellipse Question 81 English Explanation">
<br><br>Perpendicular distance of directrix from focus
<br><br>$$ = {a \over e} - ae = 4$$
<br><br>$$ \Rightarrow a\left( {2 - {1 \over 2}} \right) = 4$$
<br><br>$$ \Rightarrow a = {8 \over 3}$$
<br><br>$$\therefore$$ Semi major axis $$=8/3$$
|
mcq
|
aieee-2008
|
8BGB4L5m1YJ9F01b
|
maths
|
ellipse
|
question-based-on-basic-definition-and-parametric-representation
|
The ellipse $${x^2} + 4{y^2} = 4$$ is inscribed in a rectangle aligned with the coordinate axex, which in turn is inscribed in another ellipse that passes through the point $$(4,0)$$. Then the equation of the ellipse is :
|
[{"identifier": "A", "content": "$${x^2} + 12{y^2} = 16$$ "}, {"identifier": "B", "content": "$$4{x^2} + 48{y^2} = 48$$ "}, {"identifier": "C", "content": "$$4{x^2} + 64{y^2} = 48$$ "}, {"identifier": "D", "content": "$${x^2} + 16{y^2} = 16$$ "}]
|
["A"]
| null |
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265178/exam_images/ymnjbdlge7ihvtgsvwzr.webp" loading="lazy" alt="AIEEE 2009 Mathematics - Ellipse Question 80 English Explanation">
<br><br>The given ellipse is $${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$$
<br><br>So $$A=(2,0)$$ and $$B = \left( {0,1} \right)$$
<br><br>If $$PQRS$$ is the rectangular in which it is inscribed, then
<br><br>$$P = \left( {2,1} \right).$$
<br><br>Let $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
<br><br>be the ellipse circumscribing the rectangular $$PQRS$$.
<br><br>Then it passes through $$P\,\,(2,1)$$
<br><br>$$\therefore$$ $${4 \over {a{}^2}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( a \right)$$
<br><br>Also, given that, it passes through $$(4,0)$$
<br><br>$$\therefore$$ $${{16} \over {{a^2}}} + 0 = 1 \Rightarrow {a^2} = 16$$
<br><br>$$ \Rightarrow {b^2} = 4/3$$ $$\left[ {\,\,} \right.$$ substituting $${{a^2} = 16\,\,}$$ in $$\left. {e{q^n}\left( a \right)\,\,} \right]$$
<br><br>$$\therefore$$ The required ellipse is $${{{x^2}} \over {16}} + {{{y^2}} \over {4/3}} = 1$$
<br><br>or $${x^2} + 12y{}^2 = 16$$
|
mcq
|
aieee-2009
|
e0Okjyna0slrDuAf
|
maths
|
ellipse
|
question-based-on-basic-definition-and-parametric-representation
|
Equation of the ellipse whose axes of coordinates and which passes through the point $$(-3,1)$$ and has eccentricity $$\sqrt {{2 \over 5}} $$ is :
|
[{"identifier": "A", "content": "$$5{x^2} + 3{y^2} - 48 = 0$$ "}, {"identifier": "B", "content": "$$3{x^2} + 5{y^2} - 15 = 0$$"}, {"identifier": "C", "content": "$$5{x^2} + 3{y^2} - 32 = 0$$"}, {"identifier": "D", "content": "$$3{x^2} + 5{y^2} - 32 = 0$$"}]
|
["D"]
| null |
Let the ellipse be $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
<br><br>It press through $$(-3, 1)$$ so $${9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)$$
<br><br>Also, $${b^2} = {a^2}\left( {1 - 2/5} \right)$$
<br><br>$$ \Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$$
<br><br>So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$
|
mcq
|
aieee-2011
|
nDHwUJAI71QHUK6u
|
maths
|
ellipse
|
question-based-on-basic-definition-and-parametric-representation
|
An ellipse is drawn by taking a diameter of thec circle $${\left( {x - 1} \right)^2} + {y^2} = 1$$ as its semi-minor axis and a diameter of the circle $${x^2} + {\left( {y - 2} \right)^2} = 4$$ is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is :
|
[{"identifier": "A", "content": "$$4{x^2} + {y^2} = 4$$ "}, {"identifier": "B", "content": "$${x^2} + 4{y^2} = 8$$"}, {"identifier": "C", "content": "$$4{x^2} + {y^2} = 8$$"}, {"identifier": "D", "content": "$${x^2} + 4{y^2} = 16$$"}]
|
["D"]
| null |
Equation of circle is $${\left( {x - 1} \right)^2} + {y^2} = 1$$
<br><br>$$ \Rightarrow $$ radius $$=1$$ and diameter $$=2$$
<br><br>$$\therefore$$ Length of semi-minor axis is $$2.$$
<br><br>Equation of circle is $${x^2} + {\left( {y - 2} \right)^2} = 4 = {\left( 2 \right)^2}$$
<br><br>$$ \Rightarrow $$ radius $$=2$$ and diameter $$=4$$
<br><br>$$\therefore$$ Length of semi major axis is $$4$$
<br><br>We know, equation of ellipse is given by
<br><br>$${{{x^2}} \over {\left( {Major\,\,\,axi{s^{\,\,2}}} \right)}} + {{{y^2}} \over {\left( {Minor\,\,\,axi{s^{\,\,2}}} \right)}} = 1$$
<br><br>$$ \Rightarrow {{{x^2}} \over {{{\left( 4 \right)}^2}}} + {{{y^2}} \over {{{\left( 2 \right)}^2}}} = 1$$
<br><br>$$ \Rightarrow {{{x^2}} \over {16}} + {{{y^2}} \over 4} = 1$$
<br><br>$$ \Rightarrow {x^2} + 4{y^2} = 16$$
|
mcq
|
aieee-2012
|
3EgKRF0Qam0Y2DwppaMvV
|
maths
|
ellipse
|
question-based-on-basic-definition-and-parametric-representation
|
Consider an ellipse, whose center is at the origin and its major axis is along the x-axis. If its eccentricity is $${3 \over 5}$$ and the distance between its foci is 6, then the area (in sq. units) of the quadrilatateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is :
|
[{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "80"}, {"identifier": "D", "content": "40"}]
|
["D"]
| null |
e = 3/5 & 2ae = 6 $$ \Rightarrow $$ a = 5
<br><br>$$ \because $$ b<sup>2</sup> = a<sup>2</sup> (1 $$-$$ e<sup>2</sup>)
<br><br>$$ \Rightarrow $$ b<sup>2</sup> = 25(1 $$-$$ 9/25)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265271/exam_images/qlwwjazgwqnl9oqbsmlw.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Online) 8th April Morning Slot Mathematics - Ellipse Question 71 English Explanation">
<br>$$ \Rightarrow $$ b = 4
<br><br>$$ \therefore $$ Area of required quadrilateral
<br><br>= 4(1/2 ab) = 2ab = 40
|
mcq
|
jee-main-2017-online-8th-april-morning-slot
|
pH7BEcGlUt4jdIi97WFap
|
maths
|
ellipse
|
question-based-on-basic-definition-and-parametric-representation
|
The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate
axes and passing through the points (4, −1) and (−2, 2) is :
|
[{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$${2 \\over {\\sqrt 5 }}$$ "}, {"identifier": "C", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt 3 } \\over 4}$$ "}]
|
["C"]
| null |
Centre at (0, 0)
<br><br>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}}$$ = 1
<br><br>at point (4, $$-$$ 1)
<br><br>$${{16} \over {{a^2}}} + {1 \over {{b^2}}}$$ = 1
<br><br>$$ \Rightarrow $$ 16b<sup>2</sup> + a<sup>2</sup> = a<sup>2</sup>b<sup>2</sup> . . . .(i)
<br><br>at point ($$-$$ 2, 2)
<br><br>$${4 \over {{a^2}}} + {4 \over {{b^2}}} = 1$$
<br><br>$$ \Rightarrow $$ 4b<sup>2</sup> + 4a<sup>2</sup> = a<sup>2</sup>b<sup>2</sup> . . . .(ii)
<br><br>$$ \Rightarrow $$ 16b<sup>2</sup> + a<sup>2</sup> = 4a<sup>2</sup> + 4b<sup>2</sup>
<br><br>From equations (i) and (ii)
<br><br>$$ \Rightarrow $$ 3a<sup>2</sup> = 12b<sup>2</sup>
<br><br>$$ \Rightarrow $$ <b>a<sup>2</sup> = 4b<sup>2</sup></b>
<br><br>b<sup>2</sup> = a<sup>2</sup>(1 $$-$$ e<sup>2</sup>)
<br><br>$$ \Rightarrow $$ e<sup>2</sup> = $${3 \over 4}$$
<br><br>$$ \Rightarrow $$ e = $${{\sqrt 3 } \over 2}$$
|
mcq
|
jee-main-2017-online-9th-april-morning-slot
|
Subsets and Splits
Chapter Question Count Chart
Displays the number of questions in each chapter and a graphical representation, revealing which chapters have the most questions for focused study.
SQL Console for archit11/jee_math
Counts the number of occurrences of each paper ID, which could help identify duplicates but lacks deeper analytical insight.