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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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