Chapter
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1
|
3118-3121
|
22) Consider the interval (– ∞, – 2), i e , when – ∞ < x < – 2
|
1
|
3119-3122
|
Consider the interval (– ∞, – 2), i e , when – ∞ < x < – 2 In this case, we have x – 1 < 0, x + 2 < 0 and x – 3 < 0
|
1
|
3120-3123
|
e , when – ∞ < x < – 2 In this case, we have x – 1 < 0, x + 2 < 0 and x – 3 < 0 (In particular, observe that for x = –3, f ′(x) = (x – 1) (x + 2) (x – 3) = (– 4) (– 1)
(– 6) < 0)
Therefore,
f ′(x) < 0 when – ∞ < x < – 2
|
1
|
3121-3124
|
, when – ∞ < x < – 2 In this case, we have x – 1 < 0, x + 2 < 0 and x – 3 < 0 (In particular, observe that for x = –3, f ′(x) = (x – 1) (x + 2) (x – 3) = (– 4) (– 1)
(– 6) < 0)
Therefore,
f ′(x) < 0 when – ∞ < x < – 2 Thus, the function f is decreasing in (– ∞, – 2)
|
1
|
3122-3125
|
In this case, we have x – 1 < 0, x + 2 < 0 and x – 3 < 0 (In particular, observe that for x = –3, f ′(x) = (x – 1) (x + 2) (x – 3) = (– 4) (– 1)
(– 6) < 0)
Therefore,
f ′(x) < 0 when – ∞ < x < – 2 Thus, the function f is decreasing in (– ∞, – 2) Consider the interval (– 2, 1), i
|
1
|
3123-3126
|
(In particular, observe that for x = –3, f ′(x) = (x – 1) (x + 2) (x – 3) = (– 4) (– 1)
(– 6) < 0)
Therefore,
f ′(x) < 0 when – ∞ < x < – 2 Thus, the function f is decreasing in (– ∞, – 2) Consider the interval (– 2, 1), i e
|
1
|
3124-3127
|
Thus, the function f is decreasing in (– ∞, – 2) Consider the interval (– 2, 1), i e , when – 2 < x < 1
|
1
|
3125-3128
|
Consider the interval (– 2, 1), i e , when – 2 < x < 1 In this case, we have x – 1 < 0, x + 2 > 0 and x – 3 < 0
(In particular, observe that for x = 0, f ′(x) = (x – 1) (x + 2) (x – 3) = (–1) (2) (–3)
= 6 > 0)
So
f ′(x) > 0 when – 2 < x < 1
|
1
|
3126-3129
|
e , when – 2 < x < 1 In this case, we have x – 1 < 0, x + 2 > 0 and x – 3 < 0
(In particular, observe that for x = 0, f ′(x) = (x – 1) (x + 2) (x – 3) = (–1) (2) (–3)
= 6 > 0)
So
f ′(x) > 0 when – 2 < x < 1 Thus,
f is increasing in (– 2, 1)
|
1
|
3127-3130
|
, when – 2 < x < 1 In this case, we have x – 1 < 0, x + 2 > 0 and x – 3 < 0
(In particular, observe that for x = 0, f ′(x) = (x – 1) (x + 2) (x – 3) = (–1) (2) (–3)
= 6 > 0)
So
f ′(x) > 0 when – 2 < x < 1 Thus,
f is increasing in (– 2, 1) Now consider the interval (1, 3), i
|
1
|
3128-3131
|
In this case, we have x – 1 < 0, x + 2 > 0 and x – 3 < 0
(In particular, observe that for x = 0, f ′(x) = (x – 1) (x + 2) (x – 3) = (–1) (2) (–3)
= 6 > 0)
So
f ′(x) > 0 when – 2 < x < 1 Thus,
f is increasing in (– 2, 1) Now consider the interval (1, 3), i e
|
1
|
3129-3132
|
Thus,
f is increasing in (– 2, 1) Now consider the interval (1, 3), i e , when 1 < x < 3
|
1
|
3130-3133
|
Now consider the interval (1, 3), i e , when 1 < x < 3 In this case, we have
x – 1 > 0, x + 2 > 0 and x – 3 < 0
|
1
|
3131-3134
|
e , when 1 < x < 3 In this case, we have
x – 1 > 0, x + 2 > 0 and x – 3 < 0 So,
f ′(x) < 0 when 1 < x < 3
|
1
|
3132-3135
|
, when 1 < x < 3 In this case, we have
x – 1 > 0, x + 2 > 0 and x – 3 < 0 So,
f ′(x) < 0 when 1 < x < 3 Thus,
f is decreasing in (1, 3)
|
1
|
3133-3136
|
In this case, we have
x – 1 > 0, x + 2 > 0 and x – 3 < 0 So,
f ′(x) < 0 when 1 < x < 3 Thus,
f is decreasing in (1, 3) Finally, consider the interval (3, ∞), i
|
1
|
3134-3137
|
So,
f ′(x) < 0 when 1 < x < 3 Thus,
f is decreasing in (1, 3) Finally, consider the interval (3, ∞), i e
|
1
|
3135-3138
|
Thus,
f is decreasing in (1, 3) Finally, consider the interval (3, ∞), i e , when x > 3
|
1
|
3136-3139
|
Finally, consider the interval (3, ∞), i e , when x > 3 In this case, we have x – 1 > 0,
x + 2 > 0 and x – 3 > 0
|
1
|
3137-3140
|
e , when x > 3 In this case, we have x – 1 > 0,
x + 2 > 0 and x – 3 > 0 So f ′(x) > 0 when x > 3
|
1
|
3138-3141
|
, when x > 3 In this case, we have x – 1 > 0,
x + 2 > 0 and x – 3 > 0 So f ′(x) > 0 when x > 3 Thus, f is increasing in the interval (3, ∞)
|
1
|
3139-3142
|
In this case, we have x – 1 > 0,
x + 2 > 0 and x – 3 > 0 So f ′(x) > 0 when x > 3 Thus, f is increasing in the interval (3, ∞) Example 34 Show that the function f given by
f (x) = tan–1(sin x + cos x), x > 0
is always an increasing function in 0 4
, π
|
1
|
3140-3143
|
So f ′(x) > 0 when x > 3 Thus, f is increasing in the interval (3, ∞) Example 34 Show that the function f given by
f (x) = tan–1(sin x + cos x), x > 0
is always an increasing function in 0 4
, π
Solution We have
f (x) = tan–1(sin x + cos x), x > 0
Therefore
f ′(x) =
2
1
(cos
sin )
1
(sin
cos )
x
x
x
x
−
+
+
Fig 6
|
1
|
3141-3144
|
Thus, f is increasing in the interval (3, ∞) Example 34 Show that the function f given by
f (x) = tan–1(sin x + cos x), x > 0
is always an increasing function in 0 4
, π
Solution We have
f (x) = tan–1(sin x + cos x), x > 0
Therefore
f ′(x) =
2
1
(cos
sin )
1
(sin
cos )
x
x
x
x
−
+
+
Fig 6 22
Rationalised 2023-24
APPLICATION OF DERIVATIVES
181
= cos
sin
2
xsin 2
x
x
−
+
(on simplification)
Note that 2 + sin 2x > 0 for all x in 0, 4
π
|
1
|
3142-3145
|
Example 34 Show that the function f given by
f (x) = tan–1(sin x + cos x), x > 0
is always an increasing function in 0 4
, π
Solution We have
f (x) = tan–1(sin x + cos x), x > 0
Therefore
f ′(x) =
2
1
(cos
sin )
1
(sin
cos )
x
x
x
x
−
+
+
Fig 6 22
Rationalised 2023-24
APPLICATION OF DERIVATIVES
181
= cos
sin
2
xsin 2
x
x
−
+
(on simplification)
Note that 2 + sin 2x > 0 for all x in 0, 4
π Therefore
f ′(x) > 0 if cos x – sin x > 0
or
f ′(x) > 0 if cos x > sin x or cot x > 1
Now
cot x > 1 if tan x < 1, i
|
1
|
3143-3146
|
Solution We have
f (x) = tan–1(sin x + cos x), x > 0
Therefore
f ′(x) =
2
1
(cos
sin )
1
(sin
cos )
x
x
x
x
−
+
+
Fig 6 22
Rationalised 2023-24
APPLICATION OF DERIVATIVES
181
= cos
sin
2
xsin 2
x
x
−
+
(on simplification)
Note that 2 + sin 2x > 0 for all x in 0, 4
π Therefore
f ′(x) > 0 if cos x – sin x > 0
or
f ′(x) > 0 if cos x > sin x or cot x > 1
Now
cot x > 1 if tan x < 1, i e
|
1
|
3144-3147
|
22
Rationalised 2023-24
APPLICATION OF DERIVATIVES
181
= cos
sin
2
xsin 2
x
x
−
+
(on simplification)
Note that 2 + sin 2x > 0 for all x in 0, 4
π Therefore
f ′(x) > 0 if cos x – sin x > 0
or
f ′(x) > 0 if cos x > sin x or cot x > 1
Now
cot x > 1 if tan x < 1, i e , if 0
4
x
π
<
<
Thus
f ′(x) > 0 in 0 4
, π
Hence f is increasing function in 0, 4
π
|
1
|
3145-3148
|
Therefore
f ′(x) > 0 if cos x – sin x > 0
or
f ′(x) > 0 if cos x > sin x or cot x > 1
Now
cot x > 1 if tan x < 1, i e , if 0
4
x
π
<
<
Thus
f ′(x) > 0 in 0 4
, π
Hence f is increasing function in 0, 4
π
Example 35 A circular disc of radius 3 cm is being heated
|
1
|
3146-3149
|
e , if 0
4
x
π
<
<
Thus
f ′(x) > 0 in 0 4
, π
Hence f is increasing function in 0, 4
π
Example 35 A circular disc of radius 3 cm is being heated Due to expansion, its
radius increases at the rate of 0
|
1
|
3147-3150
|
, if 0
4
x
π
<
<
Thus
f ′(x) > 0 in 0 4
, π
Hence f is increasing function in 0, 4
π
Example 35 A circular disc of radius 3 cm is being heated Due to expansion, its
radius increases at the rate of 0 05 cm/s
|
1
|
3148-3151
|
Example 35 A circular disc of radius 3 cm is being heated Due to expansion, its
radius increases at the rate of 0 05 cm/s Find the rate at which its area is increasing
when radius is 3
|
1
|
3149-3152
|
Due to expansion, its
radius increases at the rate of 0 05 cm/s Find the rate at which its area is increasing
when radius is 3 2 cm
|
1
|
3150-3153
|
05 cm/s Find the rate at which its area is increasing
when radius is 3 2 cm Solution Let r be the radius of the given disc and A be its area
|
1
|
3151-3154
|
Find the rate at which its area is increasing
when radius is 3 2 cm Solution Let r be the radius of the given disc and A be its area Then
A = πr2
or
dA
dt = 2
dr
πr dt
(by Chain Rule)
Now approximate rate of increase of radius = dr =
0
|
1
|
3152-3155
|
2 cm Solution Let r be the radius of the given disc and A be its area Then
A = πr2
or
dA
dt = 2
dr
πr dt
(by Chain Rule)
Now approximate rate of increase of radius = dr =
0 05
dr
t
dt ∆ =
cm/s
|
1
|
3153-3156
|
Solution Let r be the radius of the given disc and A be its area Then
A = πr2
or
dA
dt = 2
dr
πr dt
(by Chain Rule)
Now approximate rate of increase of radius = dr =
0 05
dr
t
dt ∆ =
cm/s Therefore, the approximate rate of increase in area is given by
dA =
A (
)
d
dt ∆t
= 2
rdr
t
dt
π
∆
= 2π (3
|
1
|
3154-3157
|
Then
A = πr2
or
dA
dt = 2
dr
πr dt
(by Chain Rule)
Now approximate rate of increase of radius = dr =
0 05
dr
t
dt ∆ =
cm/s Therefore, the approximate rate of increase in area is given by
dA =
A (
)
d
dt ∆t
= 2
rdr
t
dt
π
∆
= 2π (3 2) (0
|
1
|
3155-3158
|
05
dr
t
dt ∆ =
cm/s Therefore, the approximate rate of increase in area is given by
dA =
A (
)
d
dt ∆t
= 2
rdr
t
dt
π
∆
= 2π (3 2) (0 05) = 0
|
1
|
3156-3159
|
Therefore, the approximate rate of increase in area is given by
dA =
A (
)
d
dt ∆t
= 2
rdr
t
dt
π
∆
= 2π (3 2) (0 05) = 0 320π cm2/s
(r = 3
|
1
|
3157-3160
|
2) (0 05) = 0 320π cm2/s
(r = 3 2 cm)
Example 36 An open topped box is to be constructed by removing equal squares from
each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the
sides
|
1
|
3158-3161
|
05) = 0 320π cm2/s
(r = 3 2 cm)
Example 36 An open topped box is to be constructed by removing equal squares from
each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the
sides Find the volume of the largest such box
|
1
|
3159-3162
|
320π cm2/s
(r = 3 2 cm)
Example 36 An open topped box is to be constructed by removing equal squares from
each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the
sides Find the volume of the largest such box Rationalised 2023-24
MATHEMATICS
182
Solution Let x metre be the length of a side of the removed squares
|
1
|
3160-3163
|
2 cm)
Example 36 An open topped box is to be constructed by removing equal squares from
each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the
sides Find the volume of the largest such box Rationalised 2023-24
MATHEMATICS
182
Solution Let x metre be the length of a side of the removed squares Then, the height
of the box is x, length is 8 – 2x and breadth is 3 – 2x (Fig 6
|
1
|
3161-3164
|
Find the volume of the largest such box Rationalised 2023-24
MATHEMATICS
182
Solution Let x metre be the length of a side of the removed squares Then, the height
of the box is x, length is 8 – 2x and breadth is 3 – 2x (Fig 6 23)
|
1
|
3162-3165
|
Rationalised 2023-24
MATHEMATICS
182
Solution Let x metre be the length of a side of the removed squares Then, the height
of the box is x, length is 8 – 2x and breadth is 3 – 2x (Fig 6 23) If V(x) is the volume
of the box, then
Fig 6
|
1
|
3163-3166
|
Then, the height
of the box is x, length is 8 – 2x and breadth is 3 – 2x (Fig 6 23) If V(x) is the volume
of the box, then
Fig 6 23
V(x) = x(3 – 2x) (8 – 2x)
= 4x3 – 22x2 + 24x
Therefore
2
V ( )
12
44
24
4(
3)(3
2)
V ( )
24
44
x
x
x
x
x
x
x
′
=
−
+
=
−
−
′′
=
−
Now
V′(x) = 0 gives
x =3, 32
|
1
|
3164-3167
|
23) If V(x) is the volume
of the box, then
Fig 6 23
V(x) = x(3 – 2x) (8 – 2x)
= 4x3 – 22x2 + 24x
Therefore
2
V ( )
12
44
24
4(
3)(3
2)
V ( )
24
44
x
x
x
x
x
x
x
′
=
−
+
=
−
−
′′
=
−
Now
V′(x) = 0 gives
x =3, 32 But x ≠ 3 (Why
|
1
|
3165-3168
|
If V(x) is the volume
of the box, then
Fig 6 23
V(x) = x(3 – 2x) (8 – 2x)
= 4x3 – 22x2 + 24x
Therefore
2
V ( )
12
44
24
4(
3)(3
2)
V ( )
24
44
x
x
x
x
x
x
x
′
=
−
+
=
−
−
′′
=
−
Now
V′(x) = 0 gives
x =3, 32 But x ≠ 3 (Why )
Thus, we have
x =32
|
1
|
3166-3169
|
23
V(x) = x(3 – 2x) (8 – 2x)
= 4x3 – 22x2 + 24x
Therefore
2
V ( )
12
44
24
4(
3)(3
2)
V ( )
24
44
x
x
x
x
x
x
x
′
=
−
+
=
−
−
′′
=
−
Now
V′(x) = 0 gives
x =3, 32 But x ≠ 3 (Why )
Thus, we have
x =32 Now
2
2
V
24
44
28
0
3
3
′′
=
−
= −
<
|
1
|
3167-3170
|
But x ≠ 3 (Why )
Thus, we have
x =32 Now
2
2
V
24
44
28
0
3
3
′′
=
−
= −
<
Therefore,
x =32
is the point of maxima, i
|
1
|
3168-3171
|
)
Thus, we have
x =32 Now
2
2
V
24
44
28
0
3
3
′′
=
−
= −
<
Therefore,
x =32
is the point of maxima, i e
|
1
|
3169-3172
|
Now
2
2
V
24
44
28
0
3
3
′′
=
−
= −
<
Therefore,
x =32
is the point of maxima, i e , if we remove a square of side 2
3
metre from each corner of the sheet and make a box from the remaining sheet, then
the volume of the box such obtained will be the largest and it is given by
2
V 3
=
3
2
2
2
2
4
22
24
3
3
3
−
+
=
200 m3
27
Example 37 Manufacturer can sell x items at a price of rupees 5
−100
x
each
|
1
|
3170-3173
|
Therefore,
x =32
is the point of maxima, i e , if we remove a square of side 2
3
metre from each corner of the sheet and make a box from the remaining sheet, then
the volume of the box such obtained will be the largest and it is given by
2
V 3
=
3
2
2
2
2
4
22
24
3
3
3
−
+
=
200 m3
27
Example 37 Manufacturer can sell x items at a price of rupees 5
−100
x
each The
cost price of x items is Rs
5x
+500
|
1
|
3171-3174
|
e , if we remove a square of side 2
3
metre from each corner of the sheet and make a box from the remaining sheet, then
the volume of the box such obtained will be the largest and it is given by
2
V 3
=
3
2
2
2
2
4
22
24
3
3
3
−
+
=
200 m3
27
Example 37 Manufacturer can sell x items at a price of rupees 5
−100
x
each The
cost price of x items is Rs
5x
+500
Find the number of items he should sell to earn
maximum profit
|
1
|
3172-3175
|
, if we remove a square of side 2
3
metre from each corner of the sheet and make a box from the remaining sheet, then
the volume of the box such obtained will be the largest and it is given by
2
V 3
=
3
2
2
2
2
4
22
24
3
3
3
−
+
=
200 m3
27
Example 37 Manufacturer can sell x items at a price of rupees 5
−100
x
each The
cost price of x items is Rs
5x
+500
Find the number of items he should sell to earn
maximum profit Rationalised 2023-24
APPLICATION OF DERIVATIVES
183
Solution Let S(x) be the selling price of x items and let C(x) be the cost price of x
items
|
1
|
3173-3176
|
The
cost price of x items is Rs
5x
+500
Find the number of items he should sell to earn
maximum profit Rationalised 2023-24
APPLICATION OF DERIVATIVES
183
Solution Let S(x) be the selling price of x items and let C(x) be the cost price of x
items Then, we have
S(x) =
2
5
5
100
100
x
x
x
x
−
=
−
and
C(x) =
500
5
x +
Thus, the profit function P(x) is given by
P(x) =
2
S( )
C( )
5
500
100
5
x
x
x
x
x
−
=
−
−
−
i
|
1
|
3174-3177
|
Find the number of items he should sell to earn
maximum profit Rationalised 2023-24
APPLICATION OF DERIVATIVES
183
Solution Let S(x) be the selling price of x items and let C(x) be the cost price of x
items Then, we have
S(x) =
2
5
5
100
100
x
x
x
x
−
=
−
and
C(x) =
500
5
x +
Thus, the profit function P(x) is given by
P(x) =
2
S( )
C( )
5
500
100
5
x
x
x
x
x
−
=
−
−
−
i e
|
1
|
3175-3178
|
Rationalised 2023-24
APPLICATION OF DERIVATIVES
183
Solution Let S(x) be the selling price of x items and let C(x) be the cost price of x
items Then, we have
S(x) =
2
5
5
100
100
x
x
x
x
−
=
−
and
C(x) =
500
5
x +
Thus, the profit function P(x) is given by
P(x) =
2
S( )
C( )
5
500
100
5
x
x
x
x
x
−
=
−
−
−
i e P(x) =
2
24
500
5
100
x −x
−
or
P′(x) = 24
5
x50
−
Now P′(x) = 0 gives x = 240
|
1
|
3176-3179
|
Then, we have
S(x) =
2
5
5
100
100
x
x
x
x
−
=
−
and
C(x) =
500
5
x +
Thus, the profit function P(x) is given by
P(x) =
2
S( )
C( )
5
500
100
5
x
x
x
x
x
−
=
−
−
−
i e P(x) =
2
24
500
5
100
x −x
−
or
P′(x) = 24
5
x50
−
Now P′(x) = 0 gives x = 240 Also
1
P ( )
50
x
−
′′
=
|
1
|
3177-3180
|
e P(x) =
2
24
500
5
100
x −x
−
or
P′(x) = 24
5
x50
−
Now P′(x) = 0 gives x = 240 Also
1
P ( )
50
x
−
′′
= So
1
P (240)
0
−50
′′
=
<
Thus, x = 240 is a point of maxima
|
1
|
3178-3181
|
P(x) =
2
24
500
5
100
x −x
−
or
P′(x) = 24
5
x50
−
Now P′(x) = 0 gives x = 240 Also
1
P ( )
50
x
−
′′
= So
1
P (240)
0
−50
′′
=
<
Thus, x = 240 is a point of maxima Hence, the manufacturer can earn maximum
profit, if he sells 240 items
|
1
|
3179-3182
|
Also
1
P ( )
50
x
−
′′
= So
1
P (240)
0
−50
′′
=
<
Thus, x = 240 is a point of maxima Hence, the manufacturer can earn maximum
profit, if he sells 240 items Miscellaneous Exercise on Chapter 6
1
|
1
|
3180-3183
|
So
1
P (240)
0
−50
′′
=
<
Thus, x = 240 is a point of maxima Hence, the manufacturer can earn maximum
profit, if he sells 240 items Miscellaneous Exercise on Chapter 6
1 Show that the function given by
log
( )
x
f x
x
=
has maximum at x = e
|
1
|
3181-3184
|
Hence, the manufacturer can earn maximum
profit, if he sells 240 items Miscellaneous Exercise on Chapter 6
1 Show that the function given by
log
( )
x
f x
x
=
has maximum at x = e 2
|
1
|
3182-3185
|
Miscellaneous Exercise on Chapter 6
1 Show that the function given by
log
( )
x
f x
x
=
has maximum at x = e 2 The two equal sides of an isosceles triangle with fixed base b are decreasing at
the rate of 3 cm per second
|
1
|
3183-3186
|
Show that the function given by
log
( )
x
f x
x
=
has maximum at x = e 2 The two equal sides of an isosceles triangle with fixed base b are decreasing at
the rate of 3 cm per second How fast is the area decreasing when the two equal
sides are equal to the base
|
1
|
3184-3187
|
2 The two equal sides of an isosceles triangle with fixed base b are decreasing at
the rate of 3 cm per second How fast is the area decreasing when the two equal
sides are equal to the base 3
|
1
|
3185-3188
|
The two equal sides of an isosceles triangle with fixed base b are decreasing at
the rate of 3 cm per second How fast is the area decreasing when the two equal
sides are equal to the base 3 Find the intervals in which the function f given by
4sin
2
cos
( )
2
cos
x
x
x
x
f x
x
−
−
=
+
is (i) increasing (ii) decreasing
|
1
|
3186-3189
|
How fast is the area decreasing when the two equal
sides are equal to the base 3 Find the intervals in which the function f given by
4sin
2
cos
( )
2
cos
x
x
x
x
f x
x
−
−
=
+
is (i) increasing (ii) decreasing 4
|
1
|
3187-3190
|
3 Find the intervals in which the function f given by
4sin
2
cos
( )
2
cos
x
x
x
x
f x
x
−
−
=
+
is (i) increasing (ii) decreasing 4 Find the intervals in which the function f given by
3
13
( )
,
0
f x
x
x
x
=
+
≠
is
(i) increasing
(ii) decreasing
|
1
|
3188-3191
|
Find the intervals in which the function f given by
4sin
2
cos
( )
2
cos
x
x
x
x
f x
x
−
−
=
+
is (i) increasing (ii) decreasing 4 Find the intervals in which the function f given by
3
13
( )
,
0
f x
x
x
x
=
+
≠
is
(i) increasing
(ii) decreasing Rationalised 2023-24
MATHEMATICS
184
5
|
1
|
3189-3192
|
4 Find the intervals in which the function f given by
3
13
( )
,
0
f x
x
x
x
=
+
≠
is
(i) increasing
(ii) decreasing Rationalised 2023-24
MATHEMATICS
184
5 Find the maximum area of an isosceles triangle inscribed in the ellipse
2
2
2
2
1
x
y
a
b
+
=
with its vertex at one end of the major axis
|
1
|
3190-3193
|
Find the intervals in which the function f given by
3
13
( )
,
0
f x
x
x
x
=
+
≠
is
(i) increasing
(ii) decreasing Rationalised 2023-24
MATHEMATICS
184
5 Find the maximum area of an isosceles triangle inscribed in the ellipse
2
2
2
2
1
x
y
a
b
+
=
with its vertex at one end of the major axis 6
|
1
|
3191-3194
|
Rationalised 2023-24
MATHEMATICS
184
5 Find the maximum area of an isosceles triangle inscribed in the ellipse
2
2
2
2
1
x
y
a
b
+
=
with its vertex at one end of the major axis 6 A tank with rectangular base and rectangular sides, open at the top is to be
constructed so that its depth is 2 m and volume is 8 m3
|
1
|
3192-3195
|
Find the maximum area of an isosceles triangle inscribed in the ellipse
2
2
2
2
1
x
y
a
b
+
=
with its vertex at one end of the major axis 6 A tank with rectangular base and rectangular sides, open at the top is to be
constructed so that its depth is 2 m and volume is 8 m3 If building of tank costs
Rs 70 per sq metres for the base and Rs 45 per square metre for sides
|
1
|
3193-3196
|
6 A tank with rectangular base and rectangular sides, open at the top is to be
constructed so that its depth is 2 m and volume is 8 m3 If building of tank costs
Rs 70 per sq metres for the base and Rs 45 per square metre for sides What is
the cost of least expensive tank
|
1
|
3194-3197
|
A tank with rectangular base and rectangular sides, open at the top is to be
constructed so that its depth is 2 m and volume is 8 m3 If building of tank costs
Rs 70 per sq metres for the base and Rs 45 per square metre for sides What is
the cost of least expensive tank 7
|
1
|
3195-3198
|
If building of tank costs
Rs 70 per sq metres for the base and Rs 45 per square metre for sides What is
the cost of least expensive tank 7 The sum of the perimeter of a circle and square is k, where k is some constant
|
1
|
3196-3199
|
What is
the cost of least expensive tank 7 The sum of the perimeter of a circle and square is k, where k is some constant Prove that the sum of their areas is least when the side of square is double the
radius of the circle
|
1
|
3197-3200
|
7 The sum of the perimeter of a circle and square is k, where k is some constant Prove that the sum of their areas is least when the side of square is double the
radius of the circle 8
|
1
|
3198-3201
|
The sum of the perimeter of a circle and square is k, where k is some constant Prove that the sum of their areas is least when the side of square is double the
radius of the circle 8 A window is in the form of a rectangle surmounted by a semicircular opening
|
1
|
3199-3202
|
Prove that the sum of their areas is least when the side of square is double the
radius of the circle 8 A window is in the form of a rectangle surmounted by a semicircular opening The total perimeter of the window is 10 m
|
1
|
3200-3203
|
8 A window is in the form of a rectangle surmounted by a semicircular opening The total perimeter of the window is 10 m Find the dimensions of the window to
admit maximum light through the whole opening
|
1
|
3201-3204
|
A window is in the form of a rectangle surmounted by a semicircular opening The total perimeter of the window is 10 m Find the dimensions of the window to
admit maximum light through the whole opening 9
|
1
|
3202-3205
|
The total perimeter of the window is 10 m Find the dimensions of the window to
admit maximum light through the whole opening 9 A point on the hypotenuse of a triangle is at distance a and b from the sides of
the triangle
|
1
|
3203-3206
|
Find the dimensions of the window to
admit maximum light through the whole opening 9 A point on the hypotenuse of a triangle is at distance a and b from the sides of
the triangle Show that the minimum length of the hypotenuse is
2
2
3
3
3
2
(
)
a
+b
|
1
|
3204-3207
|
9 A point on the hypotenuse of a triangle is at distance a and b from the sides of
the triangle Show that the minimum length of the hypotenuse is
2
2
3
3
3
2
(
)
a
+b 10
|
1
|
3205-3208
|
A point on the hypotenuse of a triangle is at distance a and b from the sides of
the triangle Show that the minimum length of the hypotenuse is
2
2
3
3
3
2
(
)
a
+b 10 Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has
(i) local maxima
(ii) local minima
(iii) point of inflexion
11
|
1
|
3206-3209
|
Show that the minimum length of the hypotenuse is
2
2
3
3
3
2
(
)
a
+b 10 Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has
(i) local maxima
(ii) local minima
(iii) point of inflexion
11 Find the absolute maximum and minimum values of the function f given by
f (x) = cos2 x + sin x, x ∈ [0, π]
12
|
1
|
3207-3210
|
10 Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has
(i) local maxima
(ii) local minima
(iii) point of inflexion
11 Find the absolute maximum and minimum values of the function f given by
f (x) = cos2 x + sin x, x ∈ [0, π]
12 Show that the altitude of the right circular cone of maximum volume that can be
inscribed in a sphere of radius r is 4
3
r
|
1
|
3208-3211
|
Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has
(i) local maxima
(ii) local minima
(iii) point of inflexion
11 Find the absolute maximum and minimum values of the function f given by
f (x) = cos2 x + sin x, x ∈ [0, π]
12 Show that the altitude of the right circular cone of maximum volume that can be
inscribed in a sphere of radius r is 4
3
r 13
|
1
|
3209-3212
|
Find the absolute maximum and minimum values of the function f given by
f (x) = cos2 x + sin x, x ∈ [0, π]
12 Show that the altitude of the right circular cone of maximum volume that can be
inscribed in a sphere of radius r is 4
3
r 13 Let f be a function defined on [a, b] such that f ′(x) > 0, for all x ∈ (a, b)
|
1
|
3210-3213
|
Show that the altitude of the right circular cone of maximum volume that can be
inscribed in a sphere of radius r is 4
3
r 13 Let f be a function defined on [a, b] such that f ′(x) > 0, for all x ∈ (a, b) Then
prove that f is an increasing function on (a, b)
|
1
|
3211-3214
|
13 Let f be a function defined on [a, b] such that f ′(x) > 0, for all x ∈ (a, b) Then
prove that f is an increasing function on (a, b) 14
|
1
|
3212-3215
|
Let f be a function defined on [a, b] such that f ′(x) > 0, for all x ∈ (a, b) Then
prove that f is an increasing function on (a, b) 14 Show that the height of the cylinder of maximum volume that can be inscribed in
a sphere of radius R is 2R
3
|
1
|
3213-3216
|
Then
prove that f is an increasing function on (a, b) 14 Show that the height of the cylinder of maximum volume that can be inscribed in
a sphere of radius R is 2R
3 Also find the maximum volume
|
1
|
3214-3217
|
14 Show that the height of the cylinder of maximum volume that can be inscribed in
a sphere of radius R is 2R
3 Also find the maximum volume 15
|
1
|
3215-3218
|
Show that the height of the cylinder of maximum volume that can be inscribed in
a sphere of radius R is 2R
3 Also find the maximum volume 15 Show that height of the cylinder of greatest volume which can be inscribed in a
right circular cone of height h and semi vertical angle α is one-third that of the
cone and the greatest volume of cylinder is
3
2
4
27 htan
π
α
|
1
|
3216-3219
|
Also find the maximum volume 15 Show that height of the cylinder of greatest volume which can be inscribed in a
right circular cone of height h and semi vertical angle α is one-third that of the
cone and the greatest volume of cylinder is
3
2
4
27 htan
π
α Rationalised 2023-24
APPLICATION OF DERIVATIVES
185
16
|
1
|
3217-3220
|
15 Show that height of the cylinder of greatest volume which can be inscribed in a
right circular cone of height h and semi vertical angle α is one-third that of the
cone and the greatest volume of cylinder is
3
2
4
27 htan
π
α Rationalised 2023-24
APPLICATION OF DERIVATIVES
185
16 A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314
cubic metre per hour
|
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