NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Maths
Chapter : 6
Chapter Name : Application of Derivatives

Q1 Find the rate of change of the area of a circle with respect to its radius r when
(a) r = 3 cm (b) r = 4 cm

Answer. The area of a circle (A) with radius (r) is given by,
A = πr 2
Now, the rate of change of the area with respect to its radius is given by,
dA d
dr
=
dr ( )
πr 2 = 2πv

(a) When r = 3cm
dA
= 2π(3) = 6π
dr
Hence, the area of the circle is changing at the rate of 6π cm 2 /s when its radius is 3 cm.
(b) When r = 4 cm,
dA
dr = 2π(4) = 8π
Hence, the area of the circle is changing at the rate of 6 n cm 2/s when its radius is 4 cm.

Page : 197 , Block Name : Exercise 6.1

Q2 The volume of a cube is increasing at the rate of 8 cm 3 /s. How fast is the surface area increasing when the
length of an edge is 12 cm?

Answer. Let x be the length of a side, V be the volume, and s be the surface area of the cube. Then, V = x 3 and S = 6
x 2 where x is a function of time t.
dV
It is given that dt = 8cm 3 / s
Then , by using the chain rule, we have:

( ) ( )
dV d d dx dx
8 = dt = dt x 3 = dx x 3 ⋅ dt = 3x 2 ⋅ dt
dx 8
⇒ dt =
3x 2
dS d d dx
dt ( ) dx (
Now, = 6x = 2 6x ) ⋅ 2
dt dt

= 12x ⋅
dx
dt
= 12x ⋅
( )
3x
8
2
=
32
x

dS 32 8
= cm 2 / s = cm 2 / s
dt 12 3
8
Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of 3 cm 2 / s.

Page : 197 , Block Name : Exercise 6.1

Q3 The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is
increasing when the radius is 10 cm.

Answer. The area of a circle (A) with radius (r) is given by,
A = πr 2
Now, the rate of change of area (A) with respect to time (t) is given by,

( )
dA d dr dr
dt
= dt ππ 2 ⋅ dt = 2πr dt [ By chain rule ]

It is given that,
dr
dt
= 3cm / s
dA
dt
= 2πr(3) = 6πr
Thus, when r = 10cm,
dA
dt
= 6π(10) = 60πcm 2 / s
Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60ncm 2 / s.

Page : 197 , Block Name : Exercise 6.1

Q4 An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing
when the edge is 10 cm long?

Answer. Let x be the length of a side and V be the volume of the cube. Then,
V = x3
dV dx
dt
= 3x 2 ⋅ dt
It is given that,
dx
dt
= 3cm / s
dV
dt
= 3x 2(3) = 9x 2

Thus, when x = 10cm,
dV
dt
= 9(10) 2 = 900cm 3 / s
Hence, the volume of the cube is increasing at the rate of 900 cm 3/s when the edge is 10 cm long.

Page : 197 , Block Name : Exercise 6.1

Q5 A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the
radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Answer. The area of a circle (A) with radius (r) is given by A = πr 2.
Therefore, the rate of change of area (A) with respect to time(t) is given by ,

( ) ( )
dA d d dr dr
dt = dt πr 2 = dr πr 2 dt = 2πr dt [ By chain rule]
dr
It is given that dt = 5cm / s
Thus, when r = 8cm,
dA
dt
= 2π(8)(5) = 80π
Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80ncm 2 / s

Page : 197 , Block Name : Exercise 6.1

Q6 The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Answer. The circumference of a circle (C) with radius (r) is given by C = 2πr.
Therefore, the rate of change of circumference (C) with respect to time (t) is given by,
dC dC dr
dt
= dr ⋅ dt ( By chain rule)
d dr
= (2πr)
dr dt
dr
= 2π ⋅
dt
dr
It is given that = 0.7cm / s
dt
Hence, the rate of increase of the circumference is 2π(0.7) = 1.4πcm / s.

Page : 198 , Block Name : Exercise 6.1

Q7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4
cm/minute. When x = 8cm and y = 6cm,
nd the rates of change of (a) the perimeter, and (b) the area of the rectangle.

Answer. Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4
cm/minute, we have:
dx
dt
= − 5cm / min

(a) The perimeter (P) of a rectangle is given by,
P = 2(x + y)

dP
(
∴ dt = 2 dt + dt
dx dy
) = 2( − 5 + 4) = − 2cm / min

Hence, the perimeter is decreasing at the rate of 2 cm/min. (b) The area (A) of a rectangle is given by,
A=x×y
dA dx dy
∴ dt = dt ⋅ y + x ⋅ dt = − 5y + 4x
When x = 8 cm and y = 6 cm
dA
dt
= ( − 5 × 6 + 4 × 8)cm 2 / min = 2cm 2 / min
Hence, the area of the rectangle is increasing at the rate of 2cm 2 / min.

Page : 198 , Block Name : Exercise 6.1

Q8 A balloon, which always remains spherical on in ation, is being in ated by pumping in 900 cubic centimetres
of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Answer. The volume of a sphere (V) with radius (r) is given by,
4
V = 3 πr 3
∴Rate of change of volume (V) with respect to time (t) is given by,

dV dV dr
dt
= dr ⋅ dt [By chain rule]

d
( )4
= dr 3 πr 3 ⋅ dt
dr

dr
= 4πr 2 ⋅ dt
dV
dt
= 900cm 3 / s
dr
∴ 900 = 4πr 2 ⋅ dt
dr 900 225
⇒ dt = =
4πr 2 πr 2
Therefore, when radius = 15cm
dr 225 1
dt
= = π
π ( 15 ) 2
Hence, the rate at which the radius of the balloon increases when the radius is 15cm
1
π
cm / s

Page : 198 , Block Name : Exercise 6.1

Q9 A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing
with the radius when the later is 10 cm.

4
Answer. The volume of a sphere (V) with radius (r) is given by V = 3 πr 3. Rate of change of volume (V) with respect
to its radius (r) is given by,
dV
dr = dr
d
( )
4
3 πr
3
4
( )
= 3 π 3r 2 = 4πr 2

Therefore, when radius = 10cm,
dV
dr
= 4π(10) 2 = 400π
Hence, the volume of the balloon is increasing at the rate of 400 cm 3 / s.

Page : 198 , Block Name : Exercise 6.1

Q10 A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from
the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away
from the wall ?

Answer. Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x away
from the wall.
Then, by Pythagoras theorem, we have:
x 2 + y 2 = 25[ Length of the ladder = 5m]
⇒y= √25 − x 2
Then, the rate of change of height (y) with respect to time (t) is given by,
dy −x dx
dt
= ⋅ dt
√25 − x 2

It is given that

dy − 2x
∴ dt =
√25 − x2
Now, when x = 4m, we have:
dy −2×4 8
dt
= = − 3
√ 25 − 4 2

Hence, the height of the ladder on the wall is decreasing at the rate of
8
3
cm / s.

Page : 198 , Block Name : Exercise 6.1

Q11 A particle moves along the curve 6y = x 3 + 2. Find the points on the curve at which the y-coordinate is
changing 8 times as fast as the x-coordinate.

Answer. The equation of the curve is given as:
6y = x 3 + 2
The rate of change of the position of the particle with respect to time (t) is given by,
dy dx
6 dt = 3x 2 dt + 0
dy dx
⇒ 2 dt = x 2 dt
When the y-coordinate of the particle changes 8 times as fast as the

x -coordinate i.e.,
( dy
dt
dx
)
= 8 dt , we have:

( )
2 8 dt
dx
= x 2 dt
dx

dx dx
⇒ 16 dt = x 2 dt

( )
dx
⇒ x 2 − 16 dt = 0

⇒ x 2 = 16
⇒x= ±4
43 + 2 66
x = 4, y = 6
= 6 = 11
( − 4 )3 + 2 62 31
x = − 4, y = = − 6 = − 3
6 .

Hence, the points required on the curve are (4, 11) and
( − 31
− 4, 3
)
Page : 198 , Block Name : Exercise 6.1

1
Q12 The radius of an air bubble is increasing at the rate of 2 cm / s cm/s. At what rate is the volume of the bubble
increasing when the radius is 1 cm?

Answer. The air bubble is in the shape of a sphere. Now, the volume of an air bubble (V) with radius (r) is given by,
4
V = 3 πr 3
The rate of change of volume (V) with respect to time (t) is given by,

dV 4 d 3 dr
dt
=π
3 dr
r ⋅
dt( )
4 dr
= π 3r 2
3 dt ( )
dr
= 4πr 2
dt
dr 1
dt
= 2 cm / s
Therefore, when r = 1cm t

dV
dt
= 4π(1) 2 2 () 1
= 2πcm 3 / s

Hence, the rate at which the volume of the bubble increases is 2ncm 3 / s .

Page : 198 , Block Name : Exercise 6.1

3
Q13 A balloon, which always remains spherical, has a variable diameter 2 (2x + 1). Find the rate of change of its
volume with respect to x.

Answer. The volume of a sphere (V) with radius (r) is given by,
4
V = 3 πr 3

It is given that:
3
Diameter = 2 (2x + 1)
3
⇒ r = 4 (2x + 1)

()
4 3 3 9
∴ V = 3π 4 (2x + 1) 3 = 16 π(2x + 1) 3

Hence, the rate of change of volume with respect to x is as
dV 9 d 9 27
dx
= 16 π dx (2x + 1) 3 = 16 π × 3(2x + 1) 2 × 2 = 8 π(2x + 1) 2.

Page : 198 , Block Name : Exercise 6.1

Q14 Sand is pouring from a pipe at the rate of 12 cm 3 /s. The falling sand forms a cone on the ground in such a way
that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone
increasing when the height is 4 cm?

Answer. The volume of a cone (V) with radius (r) and height (h) is given by,
1
V = 3 πr 2h

It is given that,
1
h = 6 r ⇒ r = 6h
1
∴ V = 3 π(6h) 2h = 12πh 3
The rate of change of volume with respect to time (t) is given by,

( )
dV d dh
dt
= 12π dh h 3 ⋅ dt [By chain rule]

= 12π (3h )
dh
2
dt
dh
= 36πh 2 dt

It is also given that dt
Therefore, when h = 4cm, we have:
dh 12 1
12 = dt = 36π ( 16 ) = 48π
1
Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of 48π cm/s .

Page : 198 , Block Name : Exercise 6.1

Q15 The total cost C(x) in Rupees associated with the production of x units of an item is given by
C(x) = 0.007x 3 − 0.003x 2 + 15x + 4000.
Find the marginal cost when 17 units are produced.

Answer. Marginal cost is the rate of change of total cost with respect to output.

( )
dC
Marginal cost(MC) = dx = 0.007 3x 2 − 0.003(2x) + 15
= 0.021x 2 − 0.006x + 15

( )
when x = 17, MC = 0.021 17 2 − 0.006(17) + 15
= 0.021(289) − 0.006(17) + 15
= 6.069 − 0.102 + 15
= 20.967
Hence, when 17 units are produced, the marginal cost is Rs. 20.967

Page : 198 , Block Name : Exercise 6.1

Q16 The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 13x 2 + 26x + 15
Find the marginal revenue when x = 7.

Answer. Marginal revenue is the rate of change of total revenue with respect to the number of units sold.
dR
Marginal Revenue (MR) = dx = 13(2x) + 26 = 26x + 26
When x = 7, MR = 26(7) + 26 = 182 + 26 = 208 Hence, the required marginal revenue is Rs 208.

Page : 198 , Block Name : Exercise 6.1

Q17 The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π (B) 12π (C) 8π (D) 11π

Answer.
The area of a circle (A) with radius (r) is given by,
A = πr 2
Therefore, the rate of change of the area with respect to its radius r is

( )
dA d
dr
= dr πr 2 = 2πr
∴When r = 6 cm,

dA
dr
= 2π × 6 = 12πcm 2 / s
dA
dr
= 2π × 6 = 12πcm 2 / s
.
Hence, the required rate of change of the area of a circle is 12mcm 2 / s .
The correct answer is B .

Page : 198 , Block Name : Exercise 6.1

Q18 The total revenue in Rupees received from the sale of x units of a product is given by
R(x) = 3x 2 + 36x + 5. The marginal revenue, when x = 15 is
(A) 116 (B) 96 (C) 90 (D) 126

Answer. Marginal revenue is the rate of change of total revenue with respect to the number of units sold.
dR
∴ Marginal Revenue (MR) = dx = 3(2x) + 36 = 6x + 36
∴When x = 15,
MR = 6(15) + 36 = 90 + 36 = 126 Hence, the required marginal revenue is Rs 126. The correct answer is D.

Page : 199 , Block Name : Exercise 6.1

Q1 Show that the function given by f (x) = 3x + 17 is increasing on R.

Answer. Let x 1 and x 2 be any two numbers in R.
Then, we have:

( ) ( )
x 1 < x 2 ⇒ 3x 1 < 3x 2 ⇒ 3x 1 + 17 < 3x 2 + 17 ⇒ f x 1 < f x 2
Hence, f is strictly increasing on R.
Alternate method:
f'(x) = 3 > 0, in every interval of R.
Thus, the function is strictly increasing on R.

Page : 205 , Block Name : Exercise 6.2

Q2 Show that the function given by f (x) = e 2x is increasing on R.

Answer. Let x 1 and x 2 be any two numbers in R.
Then, we have:

( ) ( )
x 1 < x 2 ⇒ 2x 1 < 2x 2 ⇒ e 2x 1 < e 2x 1 ⇒ f x 1 < f x 2

Hence, f is strictly increasing on R .

Page : 205 , Block Name : Exercise 6.2

Q3
Show that the function given by f(x) = sinx is

( )
(a) increasing in 0, 2
π

(c) neither increasing nor decreasing in (0, π)

Answer. The given function is f(x) = sin x.
∴ f ′(x) = cosx

( )
(a) since for each x ∈ 0, 2 , cosx > 0,
π
we have f ′(x) > 0

( )
π
0, 2
Hence, f is strictly increasing in
2

(b) Since for each x ∈
( )
π
2
, π , cosx < 0

f ′(x) < 0

Hence, f is strictly decreasing in 2 , π . ( )
π

(c) From the results obtained in (a) and (b), it is clear that f is neither increasing nor decreasing in (0, π).

Page : 205 , Block Name : Exercise 6.2

Q4 Find the intervals in which the function f given by f(x) = 2x 2 – 3x is
(a) increasing (b) decreasing

Answer.
The given function is f(x) = 2x 2 − 3x .
f ′(x) = 4x − 3
3
∴ f ′(x) = 0 ⇒ x = 4

( ) ( )
3 3 3
Now, the point 4 divides the real line into two disjoint intervals i.e., − ∞, 4 4
,∞ .
and

In interval ( 3
)
− ∞, 4 , f ′(x) = 4x − 3 < 0

Hence, the given function (f) is strictly decreasing in interval
( 3
− ∞, 4
)
In interval ( ) 3
4
, ∞ , f ′(x) = 4x − 3 > 0

Hence, the given function (f) is strictly increasing in interval 4 , ∞ ( )
3

Page : 205 , Block Name : Exercise 6.2

Q5 Find the intervals in which the function f given by f(x) = 2x 3 − 3x 2– 36x + 7 is
(a) increasing (b) decreasing

Answer.

The given function is f(x) = 2x 3 − 3x 2 − 36x + 7 .

(
f ′(x) = 6x 2 − 6x − 36 = 6 x 2 − x − 6 = 6(x + 2)(x − 3) )
∴ f ′(x) = 0 ⇒ x = − 2, 3
The points x = − 2 and x = 3 divide the real line into three disjoint intervals i.e.,
( − ∞, − 2), ( − 2, 3), and (3, ∞) .

In intervals ( − ∞, − 2) and (3, ∞), f ′′(x) is positive while in interval

( − 2, 3), f ′(x) is negative.
Hence, the given function (f) is strictly increasing in intervals
( − ∞, − 2) and (3, ∞), while function (f) is strictly decreasing in interval
( − 2, 3).

Page : 205 , Block Name : Exercise 6.2

Q6 Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x 2 + 2x − 5 (b) 10 − 6x − 2x 2
(c) − 2x 3 − 9x 2 − 12x + 1(d)6 − 9x − x 2
(e) (x + 1) 3(x − 3) 3

Answer.
(a) We have,
f(x) = x 2 + 2x − 5
∴ f ′(x) = 2x + 2
Now,
f ′(x) = 0
Point x = − 1 divides the real line into two disjoint intervals i.e., ( − ∞, − 1) and ( − 1, ∞) .
′
In interval ( − ∞ , − 1 ) , f ( x ) = 2x + 2 < 0
∴ f is strictly decreasing in interval ( − ∞, − 1)
Thus, f is strictly decreasing for x < − 1
′
In interval ( − 1 , ∞ ) , f ( x ) = 2x + 2 > 0
∴ f is strictly increasing in interval ( − 1, ∞)

(b) We have,
f(x) = 10 − 6x − 2x 2
∴ f ′(x) = − 6 − 4x
Now,
3
f ′(x) = 0 ⇒ x = − 2
3
x = − 2 divides the real line into two disjoint intervals
The point
( − ∞, − 2
3
) (
and − 2, ∞
3
)
In interval
( − ∞, − 2
3
)
i.e., when
3
x < − 2 , f ′(x) = − 6 − 4x < 0

3
∴ f is strictly increasing for x < − 2

In interval ( 3
− 2, ∞ ) i.e., when
3
x > − 2 , f ′(x) = − 6 − 4x < 0
3
∴ f is strictly decreasing for x > − 2

(c) We have,
f(x) = − 2x 3 − 9x 2 − 12x + 1

( )
∴ f ′(x) = − 6x 2 − 18x − 12 = − 6 x 2 + 3x + 2 = − 6(x + 1)(x + 2)
Now,
f ′(x) = 0 ⇒ x = − 1 and x = − 2
Points x = − 1 and x = − 2 divide the real line into three disjoint interval
i.e., ( − ∞, − 2), ( − 2, − 1), and ( − 1, ∞)
In intervals ( − ∞, − 2) and ( − 1, ∞) i.e., when x < − 2 and x > − 1
f ′(x) = − 6(x + 1)(x + 2) < 0
Now, in interval ( − 2, − 1) i.e., when − 2 < x < − 1, f ′(x) = − 6(x + 1)(x + 2) > 0
∴ f is strictly increasing for − 2 < x < − 1

(d) We have,
f(x) = 6 − 9x − x 2
∴ f ′(x) = − 9 − 2x
Now, f ′
9
(x) = 0 gives x = − 2
9
The point x = − 2 divides the real line into two disjoint intervals i.e.,

( 9
− ∞, − 2 ) ( )
and
9
− 2, ∞

In interval
( ) − ∞, − 2
9

i.e., for
9
9
x < − 2 , f ′(x) = − 9 − 2x > 0

∴ f is strictly increasing for x < − 2

In interval ( 9
− 2, ∞ ) i.e., for
9
x > − 2 , f ′(x) = − 9 − 2x < 0
9
∴ f is strictly decreasing for x > − 2

(e) We have,
f(x) = (x + 1) 3(x − 3) 3
f ′(x) = 3(x + 1) 2(x − 3) 3 + 3(x − 3) 2(x + 1) 3
= 3(x + 1) 2(x − 3) 2[x − 3 + x + 1]
= 3(x + 1) 2(x − 3) 2(2x − 2)
= 6(x + 1) 2(x − 3) 2(x − 1)
Now,
f ′(x) = 0 ⇒ x = − 1, 3, 1
The points x = − 1, x = 1, and x = 3 divide the real line into four disjoint intervals
i.e., ( − ∞, − 1), ( − 1, 1), (1, 3), and ( 3 , ∞ )

In intervals ( − ∞, − 1) and ( − 1, 1), f ′(x) = 6(x + 1) 2(x − 3) 2(x − 1) < 0
∴ f is strictly decreasing in intervals ( − ∞ , − 1 ) and ( − 1, 1)
In intervals (1, 3) and ( 3 , ∞ ) , f ′(x) = 6(x + 1) 2(x − 3) 2(x − 1) > 0
∴ f is strictly increasing in intervals (1, 3) and (3, ∞)

Page : 205 , Block Name : Exercise 6.2

Q7
2x
Show that y = log(1 + x) − 2 + x , x > − 1, is an increasing function of x
.
throughout its domain.

Answer.
We have,
2x
y = log(1 + x) − 2 + x
dy 1 ( 2 + x ) ( 2 ) − 2x ( 1 ) 1 4 x2
∴ dx = 1 + x − = 1+x − 2 =
( 2 + x )2 (2+x) ( 2 + x )2
dy
Now, dx = 0
x2
⇒ =0
( 2 + x )2

⇒ x2 = 0
⇒x=0
since x > − 1, point x = 0 divides the domain ( − 1, ∞) in two disjoint intervals
x < 0 and x > 0 .
When − 1 < x < 0, we have:
x < 0 ⇒ x2 > 0
x < 0 ⇒ x2 > 0
x > − 1 ⇒ (2 + x) > 0 ⇒ (2 + x) 2 > 0
x2
∴ y′ = >0
( 2 + x )2
Also, when x > 0 :
x > 0 ⇒ x 2 > 0, (2 + x) 2 > 0
x2
∴ y′ = >0
( 2 + x )2
Hence, function f is increasing throughout this domain.

Page : 205 , Block Name : Exercise 6.2

Q8 Find the values of x for which y = [x(x − 2)] 2 is an increasing function.

Answer.
We have,

y = [x(x − 2)] 2 = x 2 − 2x [ ] 2

( )
dy
∴ dx = y ′ = 2 x 2 − 2x (2x − 2) = 4x(x − 2)(x − 1)
dy
∴ dx = 0 ⇒ x = 0, x = 2, x = 1
The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals i.e.

( − ∞, 0), (0, 1)(1, 2), and (2, ∞)
dy
In intervals ( − ∞, 0) and (1, 2), dx < 0
∴ y is strictly decreasing in intervals ( − ∞, 0) and (1, 2)
dy
However, in intervals (0, 1) and (2, ∞), dx > 0
∴ y is strictly increasing in intervals (0, 1) and (2, ∞)
∴ y is strictly increasing for 0 < x < 1 and x > 2.

Page : 205 , Block Name : Exercise 6.2

4sin θ
Q9 Prove that y = ( 2 + cos θ ) − θ is an increasing function of θ in 0, 2
[ ]
π

Answer. We have:
4sin θ
y = ( 2 + cos θ ) − θ
dy ( 2 + cos θ ) ( 4cos θ ) − 4sin θ ( − sin θ )
∴ dx = −1
( 2 + cos θ ) 2
8cos θ + 4cos 2 θ + 4sin 2 θ
= −1
( 2 + cos θ ) 2
8cos θ + 4
= −1
( 2 + cos θ ) 2
dy
Now, dx = 0
8cos θ + 4
⇒ =1
( 2 + cos θ ) 2
⇒ 8cosθ + 4 = 4 + cos 2θ + 4cosθ
⇒ cos 2θ − 4cosθ = 0
⇒ cosθ(cosθ − 4) = 0
⇒ cosθ = 0 or cosθ = 4
since cosθ ≠ 4, cosθ = 0
π
cosθ = 0 ⇒ θ = 2
Now,

dy (
8cos θ + 4 − 4 + cos 2 θ + 4cos θ ) 4cos θ − cos 2 θ cos θ ( 4 − cos θ )
dx
= = =
( 2 + cos θ ) 2 ( 2 + cos θ ) 2 ( 2 + cos θ ) 2

( )0, 2
π

In interval 2 ), we have cosθ > 0. Also, 4 > cosθ ⇒ 4 − cosθ > 0
∴ cosθ(4 − cosθ) > 0 and also (2 + cosθ) 2 > 0
cos θ ( 4 − cos θ )
⇒ >0
( 2 + cos θ ) 2
dy
⇒ dx > 0

Therefore, y is strictly increasing in interval 0, 2 .
( ) π

π
Also, the given function is continuous at x = 0 and x = 2

Hence, y is increasing in interval 0, 2 . [ ] π

Page : 205 , Block Name : Exercise 6.2

Q10 Prove that the logarithmic function is increasing on (0, ∞).

Answer.
The given function is f(x) = logx .
1
∴ f ′(x) = x
1
f ′(x) = x > 0
It is clear that for x > 0,
Hence, f(x) = logx is strictly increasing in interval (0, ∞)

Page : 205 , Block Name : Exercise 6.2

Q11 Prove that the function f given by f(x) = x 2 – x + 1 is neither strictly increasing nor decreasing on (– 1, 1).

Answer.
The given function is f(x) = x 2 − x + 1
∴ f ′(x) = 2x − 1
1
Now, f ′(x) = 0 ⇒ x = 2
1
The point 2 divides the interval ( − 1, 1) into two disjoint intervals

i.e., ( ) ( )
− 1, 2
1
and
1
2
,1

( )
Now, in interval
1
− 1, 2 , f ′(x) = 2x − 1 < 0

Therefore, f is strictly decreasing in interval
( )
− 1, 2
1

However, in interval
( ) 1
2
, 1 , f ′(x) = 2x − 1 > 0

Therefore, f is strictly decreasing in interval
( )
1
2
,1

Hence, f is neither strictly increasing nor decreasing in interval ( − 1, 1) .

Page : 206 , Block Name : Exercise 6.2

Q12 Which of the following functions are strictly decreasing on 0, 2
( ) π

?
(A) cos x (B) cos 2x (C) cos 3x (D) tan x

Answer.

(A) Let f 1(x) = cosx
′
∴ f 1 (x) = − sinx

( ) π ′
0, 2 , f 1 (x) = − sinx < 0

∴ f 1(x) = cosx is strictly decreasing in interval 0, 2 ( )
π

(B) Let f 2(x) = cos2x
′
∴ f 2 (x) = − 2sin2x
π
Now, 0 < x < 2 ⇒ 0 < 2x < π ⇒ sin2x > 0 ⇒ − 2sin2x < 0

′
∴ f 2 (x) = − 2sin2x < 0 on 0, 2 ( ) π

∴ f 2(x) = cos2x is strictly decreasing in interval 0, 2
( )
π

(C) Let f 3(x) = cos3x
′
∴ f 3 (x) = − 3sin3x
′
Now , f 3 (x) = 0

⇒ sin3x = 0 ⇒ 3x = π, as x ∈ 0, 2 ( ) π

π
⇒x= 3

( )
π π
x = 3 divides the interval 0, 2
into two disjoint intervals

i.e., 0 ( ) ( )
0, 3
π
and
π π
,
3 2

( ) π
Now, in interval 0, 3 , f 3(x) = − 3sin3x < 0 as 0 < x < 3 ⇒ 0 < 3x < π
[ π
]
( ) π π
However, in interval 3 , 2 , f 3(x) = − 3sin3x > 0 as 3 < x < 2 ⇒ π < 3x < 2
[ π π 3π
]
∴ f 3 is strictly increasing in interval ( ) π π
,
3 2

Hence, f 3 is neither increasing nor decreasing in interval 0, 2
( )
π

(D) Let f 4(x) = tanx
′
∴ f 4 (x) = sec 2x

( ) π ′
In interval 0, 2 , f 4 (x) = sec 2x > 0

∴ f 4 is strictly increasing in interval 0, 2
( ) π

Therefore, functions cos x and cos 2x are strictly decreasing in 0, 2 ( )
π

Hence, the correct answers are A and B.

Page : 206 , Block Name : Exercise 6.2

Q13 On which of the following intervals is the function f given by f(x) = x 100 + sin x –1 decreasing ?
π π
(A) (0, 1) (B) 2
,π (C) 0, 2 (D) None of these

Answer. We have,
f(x) = x 100 + sinx − 1
∴ f ′(x) = 100x 99 + cosx
In interval ( 0 , 1 ) , cosx > 0 and 100x 99 > 0
∴ f ′(x) > 0
Thus, function f is strictly increasing in interval (0, 1) .

( )
π
2
, π , cosx < 0 and 100x 99 > 0. Also, 100x 99 > cosx

∴ f ′(x) > 0 in
( )π
2
,π

Thus, function f is strictly increasing in interval 2 , π
( )
π

( ) π
In interval 0, 2 , cosx > 0 and 100x 99 > 0

∴ 100x 99 + cosx > 0

( )
⇒ f ′(x) > 0 on 0, 2
π

∴ f is strictly increasing in interval 0, 2 ( ) π

Hence, function f is strictly decreasing in none of the intervals. The correct answer is D.

Page : 206 , Block Name : Exercise 6.2

Q14 Find the least value of a such that the function f given f(x) = x 2 + ax + 1 is increasing on (1, 2).

Answer. We have,
f(x) = x 2 + ax + 1
∴ f ′(x) = 2x + a

Now, function f will be increasing in (1, 2), if f ′(x) > 0 in (1, 2) .
f ′(x) > 0
⇒ 2x + a > 0
⇒ 2x > − a
−a
⇒x> 2
Therefore, we have to nd the least value of a such that
−a
x > 2 , when x ∈ (1, 2)
−a
⇒ x > 2 (when 1 < x < 2)
Thus, the least value of a for f to be increasing on (1, 2) is given by,
−a
2
=1
−a
2
=1⇒a= −2
Hence, the required value of a is −2.

Page : 206 , Block Name : Exercise 6.2

1
Q15 Let I be any interval disjoint from [–1, 1]. Prove that the function f given by f(x) = x + x 1 is increasing on I.

Answer.
We have,
1
f(x) = x + x
1
∴ f ′(x) = 1 −
x2
Now,
1
f ′(x) = 0 ⇒ =1⇒x= ±1
x2
The points x = 1 and x = −1 divide the real line in three disjoint intervals i.e.,
( − ∞, − 1), ( − 1, 1), and (1, ∞)
In interval ( − 1, 1), it is observed that:
−1 < x < 1
⇒ x2 < 1
1
⇒1< ,x ≠ 0
x2
1
⇒1− < 0, x ≠ 0
x2
1
∴ f ′(x) = 1 − < 0 on ( − 1, 1) ∼ {0}
x2
∴ f is strictly decreasing on ( − 1, 1) − {0}
In intervals ( − ∞, − 1) and (1, ∞), it is observed that:
x < − 1 or 1 < x
⇒ x2 > 1
1
⇒1>
x2
1
⇒1− >0
x2
1
∴ f ′(x) = 1 − > 0 on ( − ∞, − 1) and (1, ∞)
x2

∴ f is strictly increasing on ( − ∞, 1) and (1, ∞)
Hence, function f is strictly increasing in interval I disjoint from ( − 1, 1) .
Hence, the given result is proved.

Page : 206 , Block Name : Exercise 6.2

( )
π
( ) π
Q16 Prove that the function f given by f(x) = log sin x is increasing on 0, 2 and decreasing on 2 , π .

Answer.
We have,
f(x) = logsinx
1
∴ f ′(x) = sin x cosx = cotx

( ) π
In interval 0, 2 , f ′(x) = cotx > 0

∴ f is strictly increasing in 0, 2
( ) π

In interval ( ) π
2
, π , f ′(x) = cotx < 0
.
∴ f is strictly decreasing in
( )
π
2
,π

Page : 206 , Block Name : Exercise 6.2

( ) π
Q17 Prove that the function f given by f (x) = log |cos x| is decreasing on 0, 2 and increasing on ( ) 3π
2
, 2π .

Answer.
We have,
f(x) = logcosx
1
∴ f ′(x) = cos x ( − sinx) = − tanx

( ) π
In interval 0, 2 , tanx > 0 ⇒ − tanx < 0

( )
∴ f ′(x) < 0 on 0, 2
π

∴ f is strictly decreasing on 0, 2 )( ) π

In interval
( ) π
2
, π , tan x < 0 ⇒ − tanx > 0

∴ f ′(x) > 0 on( ) π
2
,π

∴ f is strictly increasing on
( ) π
2
,π .

Page : 206 , Block Name : Exercise 6.2

Q18 Prove that the function given by f (x) = x 3 − 3x 2 + 3x − 100 is increasing in R.

Answer.
We have,
f(x) = x 3 − 3x 2 + 3x − 100
f ′(x) = 3x 2 − 6x + 3

(
= 3 x 2 − 2x + 1 )
= 3(x − 1) 2
For any x ∈ R, (x − 1) 2 > 0
Thus, f ′(x) is always positive in R .
.
Hence, the given function (f) is increasing in R .

Page : 206 , Block Name : Exercise 6.2

Q19 The interval in which y = x 2e − x is increasing is.
(A) ( − ∞, ∞) (B) ( − 2, 0) (C) (2, ∞) (D) (0, 2)

Answer.
We have,
y = x 2e − x
dy
∴ dx = 2xe − x − x 2e − x = xe − x(2 − x)
dy
Now, dx = 0
⇒ x = 0 and x = 2
The points x = 0 and x = 2 divide the real line into three disjoint intervals
i.e., ( − ∞, 0), (0, 2), and (2, ∞)
In intervals ( − ∞, 0) and (2, ∞), f ′(x) < 0 as e − x is always positive.
∴ f is decreasing on ( − ∞, 0) and (2, ∞)
In interval (0, 2), f ′(x) > 0
∴ f is strictly increasing on (0, 2)
Hence, f is strictly increasing in interval (0, 2).
The correct answer is D.

Page : 206 , Block Name : Exercise 6.2

Q1 Find the slope of the tangent to the curve y = 3x 4 − 4x at x = 4.

Answer.
The given curve is y = 3x 4 − 4x .
Then, the slope of the tangent to the given curve at x = 4 is given by,
dy
3
dx ] = 12x − 4] x − 4 = 12(4) 3 − 4 = 12(64) − 4 = 764

Page : 211 , Block Name : Exercise 6.3

x−1
Q2 Find the slope of the tangent to the curve y = x − 2 , x ≠ 2 at x = 10.

Answer.
x−1
y=
x−2
The given curve is x − 2
dy (x − 2)(1) − (x − 1)(1)
∴ =
dx (x − 2) 2
x−2−x+1 −1
= =
(x − 2) 2 (x − 2) 2
Thus, the slope of the tangent at x = 10 is given by,
dy −1 −1 −1
]
dx x = 10
= ] = = 64 −1
( x − 2 ) 2 x = 10 ( 10 − 2 ) 2 .
64
Hence, the slope of the tangent at x = 10 is

Page : 211 , Block Name : Exercise 6.3

Q3 Find the slope of the tangent to curve y = x 3 − x + 1 at the point whose x-coordinate is 2.

Answer.
The given curve is y = x 3 − x + 1
dy
∴ dx = 3x 2 − 1
dy
(
The slope of the tangent to a curve at x 0, y 0 is dx ] ( x , y , y ) )
It is given that x 0 = 2 .
It is given that x 0 = 2. Hence, the slope of the tangent at the point where the x-coordinate is 2 is given by,
dy
]
dx x − 2
= 3x 2 − 1] re − 2 = 3(2) 2 − 1 = 12 − 1 = 11.

Page : 211 , Block Name : Exercise 6.3

Q4 Find the slope of the tangent to the curve y = x 3 − 3x + 2 at the point whose x-coordinate is 3.

Answer.
The given curve is y = x 3 − 3x + 2
dy
∴ dx = 3x 2 − 3
dy
The slope of the tangent to a curve at x 0, y 0 is dx ] x , y
( 0 0) ( )
Hence, the slope of the tangent at the point where the x-coordinate is 3 is given by,
dy
]
dx x − 3
= 3x 2 − 3] x − 3 = 3(3) 2 − 3 = 27 − 3 = 24

Page : 211 , Block Name : Exercise 6.3

π
Q5 Find the slope of the normal to the curve x = acos 3θ, y = asin 3θ at θ = 4 .

Answer. It is given that x = acos 3θ and y = asin 3θ

dx
∴ dθ = 3acos 2θ( − sinθ) = − 3acos 2θsinθ
dy
dθ
= 3asin 2θ(cosθ)

dy ( ) dy
dθ
3acos 2 θcos θ sin θ
∴ dx = = = − cos θ = − tanθ
− 3acos 2 θsin θ
( ) dy
dθ
π
Therefore, the slope of the tangent at θ = 4 is given by,
dy π
] x = − tanθ] θ = π = − tan 4 = − 1
dx θ = 4 4
π
Hence, the slope of the normal at θ = 4 is given by,
dy π
]
dx θ = 4
x = − tanθ] θ = π = − tan 4 = − 1
4
π
θ = 4 is given by,
π
Hence, the slope of the normal at θ = 4 is given by, .
1 −1
π = −1 = 1
slope of the tangent at θ = 4

Page : 211 , Block Name : Exercise 6.3

π
Q6 Find the slope of the normal to the curve x = 1 − asinθ, y = bcos 2θ at θ = 2 .

Answer.
It is given that x = 1 − asinθ and y = bcos 2θ
dx dy
∴ dθ = − acosθ and dθ = 2bcosθ( − sinθ) = − 2bsinθcosθ

∴ dx
dy
=
( )= dy
dθ
− 2bsin θcos θ 2b
= a sinθ
− acos θ
( ) dx
dθ

π
Therefore, the slope of the tangent at θ = 2 is given by,
dy 2b 2b π 2b
] π = a sinθ] θ = π = a sin 2 = a
dx 0 = 2 2
π
Hence, the slope of the normal at 2 is given by,
.
1 −1 a
π = = − 2b
slope of the tangent at θ = 4
( ) 2b
a

Page : 211 , Block Name : Exercise 6.3

Q7 Find points at which the tangent to the curve y = x 3 − 3x 2 − 9x + 7 is parallel to the x-axis.

Answer.
The equation of the given curve is y = x 3 − 3x 2 − 9x + 7
dy
∴ dx = 3x 2 − 6x − 9

Now, the tangent is parallel to the x-axis if the slope of the tangent is zero.
∴ 3x 2 − 6x − 9 = 0 ⇒ x 2 − 2x − 3 = 0
⇒ (x − 3)(x + 1) = 0
⇒ x = 3 or x = − 1
when x = 3, y = (3) 3 − 3(3) 2 − 9(3) + 7 = 27 − 27 − 27 + 7 = − 20
When x = − 1, y = ( − 1) 3 − 3( − 1) 2 − 9( − 1) + 7 = − 1 − 3 + 9 + 7 = 12

Hence, the points at which the tangent is parallel to the x-axis are (3, −20) and (−1, 12).

Page : 211 , Block Name : Exercise 6.3

Q8 Find a point on the curve y = (x– 2) 2 at which the tangent is parallel to the chord joining the points (2, 0) and
(4, 4).

Answer. If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then the slope of the tangent = the
slope of the chord.
The slope of the chord is
4−0 4
4−2
= 2 =2
Now, the slope of the tangent to the given curve at a point (x, y) is given by,
dy
dx
= 2(x − 2)
Since the slope of the tangent = slope of the chord, we have:
2(x − 2) = 2
⇒x−2=1⇒x=3
When x = 3, y = (3 − 2) 2 = 1
Hence, the required point is (3, 1) .

Page : 211 , Block Name : Exercise 6.3

Q9 Find the point on the curve y = x 3 − 11x + 5 at which the tangent is y = x – 11.

Answer.
The equation of the given curve is y = x 3 − 11x + 5.
The equation of the tangent to the given curve is given as y = x − 11 (which is of the
form y = mx + c)
Slope of the tangent = 1
Now, the slope of the tangent to the given curve at the point (x, y) is given by,
dy
2
dx = 3x − 11
Then, we have:
3x 2 − 11 = 1
⇒ 3x 2 = 12
⇒ x2 = 4
⇒x= ±2
When x = 2, y = (2) 3 − 11(2) + 5 = 8 − 22 + 5 = − 9
When x = − 2, y = ( − 2) 3 − 11( − 2) + 5 = − 8 + 22 + 5 = 19
Hence, the required points are (2, −9) and (−2, 19).

Page : 212 , Block Name : Exercise 6.3

Q10 Find the equation of all lines having slope – 1 that are tangents to the curve
1
y = x − 1 , x ≠ 1.

Answer.
1
The equation of the given curve is y = x − 1 , x ≠ 1

The slope of the given curve is to the given curve at any point (x, y) is given by,
dy −1
dx
=
( x − 1 )2
If the slope of the tangent is −1, then we have:
−1
= −1
( x − 1 )2

⇒ (x − 1) 2 = 1
⇒x−1= ±1
⇒ x = 2, 0
When x = 0, y = − 1 and when x = 2, y = 1
Thus, there are two tangents to the given curve having slope −1. These are passing through the points (0, −1) and
(2, 1).
∴The equation of the tangent through (0, −1) is given by,
y − ( − 1) = − 1(x − 0)
⇒y+1= −x
⇒y+x+1=0
∴The equation of the tangent through (2, 1) is given by,
y − 1 = − 1(x − 2)
⇒y−1= −x+2
⇒y−1=
⇒y+x−3=0
Hence, the equations of the required lines are y + x + 1 = 0 and y + x − 3 = 0.

Page : 212 , Block Name : Exercise 6.3

1
Q11 Find the equation of all lines having slope 2 which are tangents to the curve y = x − 3 , x ≠ 3.

Answer. The equation of the given curve is
1
y = x−3, x ≠ 3

The slope of the tangent to the given curve at any point (x, y) is given by,
dy −1
dx
=
( x − 3 )2
If the slope of the tangent is 2, then we have:
−1
=2
( x − 3 )2
⇒ 2(x − 3) 2 = − 1
−1
⇒ (x − 3) 2 = 2
This is not possible since the L.H.S. is positive while the R.H.S. is negative.
Hence, there is no tangent to the given curve having slope 2 .

Page : 212 , Block Name : Exercise 6.3

Q12 Find the equations of all lines having slope 0 which are tangent to the curve
1
y=
x 2 − 2x + 3

Answer.
1
The equation of the given curve is y = x 2 − 2x + 3
The slope of the tangent to the given curve at any point (x, y) is given by,
dy − ( 2x − 2 ) −2(x−1)
dx
= =
( x − 2x + 3 )
2 2
( x − 2x + 3 )
2 2

If the slope of the tangent is 0, then we have:
−2(x−1)
=0
( x 2 − 2x + 3 ) 2

⇒ − 2(x − 1) = 0
⇒x=1
1 1
y = 1−2+3 = 2

When x = 1,

1
Whe equation of the tangent through 1, 2
( ) 1

is given by

y − 2 = 0(x − 1)
1
⇒y− 2 =0
1
⇒y= 2
1
Hence, the equation of the required line is y = 2

Page : 212 , Block Name : Exercise 6.3

x2 y2
Q13 Find points on the curve 9 + 16 = 1 at which tangents are
(i) parallel to x-axis (ii) parallel to y-axis

x2 y2
+ 16 = 1
Answer. The equation of the given curve is 9

On differentiating both sides with respect to x, we have :
2x 2y dy
9 + 16 ⋅ dx = 0
dy − 16x
⇒ dx = 9y
− 16x
(i) The tangent is parallel to the x-axis if the slope of the tangent is i.e., 0 9y =0
which is possible if x = 0.
x2 y2
Then 9 + 16 = 1

⇒ y 2 = 16 ⇒ y = ± 4
Hence, the points at which the tangents are parallel to the x -axis are
(0, 4) and (0, − 4)
(ii) The tangent is parallel to the y-axis if the slope of the normal is 0, which gives
−1 9y
= 16x = 0

( ) − 16x
9y

⇒y=0
x2 y2
Then, 9 + 16 = 1

⇒x= ±3
Hence, the points at which the tangents are parallel to the y -axis are
(3, 0) and ( − 3, 0)

Page : 212 , Block Name : Exercise 6.3

Q14 Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x 4 − 6x 3 + 13x 2 − 10x + 5 at (0, 5)
(ii) y = x 4 − 6x 3 + 13x 2 − 10x + 5 at (1, 3)
(iii) y = x 3 at (1, 1)
(iv) y = x 2 at (0, 0)
π
(v) x = cost, y = sint at t = 4

Answer.
(i) The equation of the curve is y = x 4 − 6x 3 + 10x + 5.
On differentiating with respect to x, we get:
dy
dx
= 4x 3 − 18x 2 + 26x − 10
dy
]
dx ( 0 , 5 )
= − 10
Thus, the slope of the tangent at (0, 5) is − 10. The equation of the tangent is given
y − 5 = − 10(x − 0)
⇒ y − 5 = − 10x ⇒ 10x + y = 5
−1 1
The slope of the normal at (0, 5) is Slope of the tangent at ( 0 , 5 ) = 10
Therefore, the equation of the normal at (0, 5) is given as:
1
y − 5 = 10 (x − 0)

⇒ 10y − 50 = x
⇒ x − 10y + 50 = 0

(ii) The equation of the curve is y = x 4 − 6x 3 + 13x 2 − 10x + 5
On differentiating with respect to x, we get:
dy
3
dx = 4x − 18x 2 + 26x − 10
dy
dx ] ( 1 , 3 ) = 4 − 18 + 26 − 10 = 2
Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as:
y − 3 = 2(x − 1)
⇒ y − 3 = 2x − 2
⇒ y = 2x + 1
−1 −1
The slope of the normal at (1, 3) is Slope of the tangent at ( 1 , 3 ) = 2
Therefore, the equation of the normal at (1, 3) is given as:
1
y − 3 = − 2 (x − 1)

⇒ 2y − 6 = − x + 1
⇒ x + 2y − 7 = 0

(iii) The equation of the curve is y = x 3 .
On differentiating with respect to x, we get:
dy
dx
= 3x 2
dy
2
]
dx ( 1 , 1 )
= 3(1) =3
Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as:
y − 1 = 3(x − 1)
⇒ y = 3x − 2
−1 −1 −1
The slope of the normal at ( 1 , 1 ) is Slope of the tangent at ( 1 , 1 )
= 3
Therefore, the equation of the normal at (1, 1) is given as:
−1
y − 1 = 3 (x − 1)

⇒ 3y − 3 = − x + 1
⇒ x + 3y − 4 = 0

(iv) The equation of the curve is y = x 2 .
On differentiating with respect to x, we get:
dy
dx
= 2x
dy
]
dx ( 0 , 0 )
=0
Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as:
y − 0 = 0(x − 0)
⇒y=0
1
The slope of the normal at (0, 0) is Slope of the tangent at (0, 0) = − 0

defined.

( )
Therefore, the equation of the normal at x 0, y 0 = (0, 0) is given by

x = x0 = 0

(v) The equation of the curve is x = cos t, y = sin t.
x = cost and y = sint
dx dy
∴ dt = − sint, dt = cost

dy ( ) ( )dy
dt
dy
dt
cos t
∴ dx = =dx
= − sin t
= − cott
( ) dx
dt

dy
π
]
dx r = 4
= − cott = − 1
π
The slope of the tangent at t = 4 is − 1
π 1 1
t = 4, x = and y =
√2 √2

Thus, the equation of the tangent to the given curve at t = 4 i.e., at
π
[( )]
1

√2 √2
,
1
is

y−
1

√2
= −1 x−
( ) 1

√2
1 1
⇒x+y− − =0
√2 √2
⇒ x + y − √2 = 0
π −1
The slope of the normal at t = 4 π =1
Slope of the tangent at t = 4

Therefore, the equation of the normal to the given curve at t = 4 i.e., at
π
[( )]
1
,
√2 √2
1

is

y−
1

√2 ( )
=1 x−
1

√2 .
⇒x=y

Page : 212 , Block Name : Exercise 6.3

Q15 Find the equation of the tangent line to the curve y = x 2 − 2x + 7 which is
(a) parallel to the line 2x − y + 9 = 0
(b) perpendicular to the line 5y − 15x = 13.

Answer. The equation of the given curve is y = x 2 − 2x + 7 .
On differentiating with respect to x, we get:
dy
dx
= 2x − 2
(a) The equation of the line is 2x − y + 9 = 0.
2x − y + 9 = 0 ∴ y = 2x + 9
This is of the form y = mx + c.
∴Slope of the line = 2
If a tangent is parallel to the line 2x − y + 9 = 0, then the slope of the tangent is equal to the slope of the line.
Therefore, we have:
2 = 2x − 2
⇒ 2x = 4
⇒x=2
Now, x = 2
⇒y=4−4+7=7
Thus, the equation of the tangent passing through (2, 7) is given by,
y − 7 = 2(x − 2)
⇒ y − 2x − 3 = 0
Hence, the equation of the tangent line to the given curve (which is parallel to line 2x − y + 9 = 0) is y − 2x − 3 = 0.
(b) The equation of the line is 5y − 15x = 13 .
13
5y- 15x = 13 y = 3x + 5

This is of the form y = mx + c
Slope of the line = 3
If a tangent is perpendicular to the line 5y - 15x = 13, then the slope of the tangent is

−1 −1
slope of the line
= 3
−1
⇒ 2x − 2 = 3
−1
⇒ 2x = 3 + 2
5
⇒ 2x = 3
5
⇒x= 6
5
Now, x = 6
25 10 25 − 60 + 252 217
⇒ y = 36 − 6 + 7 = 36
= 36

Thus, the equation of the tangent passing through 6 , 36 ( )
5 217
is given by,

217
y − 36 = − 3 x − 6
1
( ) 5

36y − 217 −1
⇒ 36
= 18 (6x − 5)

⇒ 36y − 217 = − 2(6x − 5)
⇒ 36y − 217 = − 12x + 10
⇒ 36y + 12x − 227 = 0
Hence, the equation of the tangent line to the given curve (which is perpendicular to line 5y − 15x = 13) is
36y + 12x − 227 = 0.

Page : 212 , Block Name : Exercise 6.3

Q16 Show that the tangents to the curve y = 7(\x ^{3}\) + 11 at the points where x = 2 and x = – 2 are parallel.

Answer.
The equation of the given curve is y = 7x 3 + 11
dy
∴ dx = 21x 2
dy
The slope of the tangent to a curve at x 0, y 0 is dx ] x , y
( 0 0) ( )
Therefore, the slope of the tangent at the point where x = 2 is given by,
dy
]
dx x = − 2
= 21(2) 2 = 84
It is observed that the slopes of the tangents at the points where x = 2 and x = −2 are equal. Hence, the two
tangents are parallel.

Page : 212 , Block Name : Exercise 6.3

Q17 Find the points on the curve y = x 3 at which the slope of the tangent is equal to the y-coordinate of the point.

Answer. The equation of the given curve is y = x 3 .
dy
∴ dx = 3x 2
The slope of the tangent at the point (x, y) is given by,
dy
] = 3x 2
dx ( x , y )
When the slope of the tangent is equal to the y-coordinate of the point, then y = 3x 2 . Also, we have y = x 3.

3x 2 = x 3
x 2(x − 3) = 0
x = 0, x = 3
When x = 0, then y = 0 and when x = 3, then y = 3(3) 2 = 27.
Hence, the required points are (0, 0) and (3, 27).

Page : 212 , Block Name : Exercise 6.3

Q18 For the curve y = 4x 3 – 2x 5 , nd all the points at which the tangent passes through the origin.

Answer. The equation of the given curve is y = 4x 3 − 2x 5.
dy
∴ dx = 12x 2 − 10x 4
Therefore, the slope of the tangent at a point (x, y) is 12x 2 − 10x 4 .
The equation of the tangent at (x, y) is given by,

(
Y − y = 12x 2 − 10x 4 (X − x) )
When the tangent passes through the origin (0, 0), then X = Y = 0.
Therefore, equation (1) reduces to:

(
− y = 12x 2 − 10x 4 ( − x) )
y = 12x 3 − 10x 5
Also, we have y = 4x 3 − 2x 5
∴ 12x 3 − 10x 5 = 4x 3 − 2x 5
⇒ 8x 5 − 8x 3 = 0
⇒ x5 − x3 = 0

(
⇒ x3 x2 − 1 = 0 )
⇒ x = 0, ± 1
When x = 0, y = 4(0) 3 − 2(0) 5 = 0
When x = 1, y = 4(1) 3 − 2(1) 5 = 2
When x = − 1, y = 4( − 1) 3 − 2( − 1) 5 = − 2
Hence, the required points are (0, 0), (1, 2), and (−1, −2).

Page : 212 , Block Name : Exercise 6.3

Q19 Find the points on the curve x 2 + y 2 − 2x − 3 = 0 at which the tangents are parallel to the x-axis.

Answer. The equation of the given curve is x 2 + y 2 − 2x − 3 = 0.
On differentiating with respect to x, we have:
dy
2x + 2y dx − 2 = 0
dy
⇒ y dx = 1 − x
dy 1−x
⇒ dx = y
Now, the tangents are parallel to the x-axis if the slope of the tangent is 0.
1−x
∴ y
=0⇒1−x=0⇒x=1

But rx 2 + y 2 − 2x − 3 = 0 for x = 1

⇒ y2 = 4 y= ±2

Hence, the points at which the tangents are parallel to the x-axis are (1, 2) and (1, −2).

Page : 212 , Block Name : Exercise 6.3

(
Q20 Find the equation of the normal at the point am 2, am 3 for the curve ay 2 = x 3. )
Answer. The equation of the given curve is ay 2 = x 3.
On differentiating with respect to x, we have:
dy
2ay dx = 3x 2
dy 3x 2
⇒ dx = 2ay
dy
The slope of a tangent to the curve at (x0, y0) is dx ] x , x
( 0 0)
The slope of the tangent to the given curve at am 2, am 3 is ( )
dy ( ) 3a m 3m
3 am 2
2
2 4

dx ( am , a )
] 2 =
2 = = 2 2 3
2a ( am )
2a m 3

Slope of the normal at am 2, am 3 ( )
−1 −2
= = 3m
slope of the tangent at ( am , am )
2 3

Hence, the equation of the normal at am 2, am 3 is given by, ( )
( )
−2
y − am 3 = 3m x − am 2

⇒ 3my − 3am 4 = − 2x + 2am 2

⇒ 2x + 3my − am 2 2 + 3m 2 = 0 ( )
Page : 212 , Block Name : Exercise 6.3

Q21 Find the equation of the normals to the curve y = x 3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Answer. The equation of the given curve is y = x 3 + 2x + 6
The slope of the tangent to the given curve at any point (x, y) is given by,
dy
dx
= 3x 2 + 2
∴ Slope of the normal to the given curve at any point (x, y)
−1
=
Slope of the tangent at the point (x, y)
−1
= 2
3x + 2
The equation of the given line is x + 14y + 4 = 0.
1 4
x + 14y + 4 = 0 y = − 14 x − 14 ( which is of the form y = mx + c)
−1
Slope of the given line = 14

f the normal is parallel to the line, then we must have the slope of the normal being
qual to the slope of the line.

−1 −1
∴ = 14
3x 2 + 2

⇒ 3x 2 + 2 = 14
⇒ 3x 2 + 2 = 12
⇒ x2 = 4
⇒x= ±2
When x = 2, y = 8 + 4 + 6 = 18.
When x = −2, y = − 8 − 4 + 6 = −6.
−1
Therefore, there are two normals to the given curve with slope 14 and passing through the points (2, 18) and (−2,
−6).
Thus, the equation of the normal through (2, 18) is given by,
−1
y − 18 = 14 (x − 2)

⇒ 14y − 252 = − x + 2
⇒ x + 14y − 254 = 0
And, the equation of the normal through (−2, −6) is given by,
−1
y − ( − 6) = 14 [x − ( − 2)]
−1
⇒ y + 6 = 14 (x + 2)

⇒ 14y + 84 = − x − 2
⇒ x + 14y + 86 = 0
Hence, the equations of the normals to the given curve (which are parallel to the given line) are
x + 14y − 254 = 0 and x + 14y + 86 = 0.

Page : 213 , Block Name : Exercise 6.3

Q22 Find the equations of the tangent and normal to the parabola y 2 = 4ax at the point at 2, 2at . ( )
Answer. The equation of the given parabola is y 2 = 4ax.
On differentiating y 2 = 4ax with respect to x, we have:
dy
2y dx = 4a
dy 2a
⇒ dx = y

( )
dy 2a 1
The slope of the tangent at at 2, 2at is dx ] a 2 , 2a = 2at = t
( )
Then, the equation of the tangent at at 2, 2at is given by, ( )
( )
1
y − 2at = t x − at 2

⇒ ty − 2at 2 = x − at 2
⇒ ty = x + at 2
Now, the slope of the normal at at 2, 2at is given by, ( )
−1
= −t
Slope of the tangent at ( at 2 , 2at )
Thus, the equation of the normal at at 2, 2at is given as: ( )

y − 2at = − t x − at 2 ( )
⇒ y − 2at = − tx + at 3
⇒ y = − tx + 2at + at 3

Page : 213 , Block Name : Exercise 6.3

Q23 Prove that the curve x = y 2 and xy = k cut at right angles* if 8k 2 = 1.

Answer. The equations of the given curves are given as x = y 2 and xy = k.
Putting x = y 2 in xy = k, we get:
1
y3 = k ⇒ y = k 3
2
∴ x = k3

Thus, the point of intersection of the given curves is k 3 , k 3 .
( )
2 1

Differentiating x = y 2 with respect to x, we have:
dy dy 1
1 = 2y dx ⇒ dx = 2y

( )
2
Therefore, the slope of the tangent to the curve x = y 2 at k 3 , k 3 is dx ] k 3 , k 3 ) =
1 dy

( 2 1
1

2k 3
1

On differentiating xy = k with respect to x, we have:
dy dy −y
x dx + y = 0 ⇒ dx = x

Slope of the tangent to the curve xy = k at k 3 , k 3 is given by,

1
( )
2 1

( (
dy 2 1
−y 2 1
k3 −1
dx
] k3 ,k3 )= x ] k3 ,k3 )= − 2 = 1
k3 k3
We know that two curves intersect at right angles if the tangents to the curves at the

( ) 2 1
point of intersection i.e., at k 3 , k 3 are perpendicular to each other.

This implies that we should have the product of the tangents as − 1.
Thus, the given two curves cut at right angles if the product of the slopes of their respective tangents at

( ) 2 1
k 3 , k 3 is −1.

i.e.,
( )( )
2
1

2k 3
1
−1

k3
1 = −1

⇒ 2k 3 = 1

⇒ 2k 3
( ) 2 3
= (1) 3

⇒ 8k 2 = 1
Hence, the given two curves cut at right angles if 8k 2 = 1

Page : 213 , Block Name : Exercise 6.3

x2 y2
Q24 Find the equations of the tangent and normal to the hyperbola
a 2 − b2 ( )
= 1 at the point x 0, y 0 .

x2 y2
Answer. Differentiating 2 − = 1 with respect to x, we have:
a b2
2x 2y dy
− 2 dx =0
a2 b
2y dy 2x
⇒ =
b 2 dx a2
dy b 2x
⇒ dx =
a 2y
dy b 2x 0
( )
Therefore, the slope of the tangent at x 0, y 0 ]
is dx ( x , y , )
= 2 .
a y0

Then, the equation of the tangent at (x 0, y 0 )is given by,
b x 2

a y (
x − x0 )
0
y − y0 = 2
0
2 2
⇒ a 2yy 0 − a 2y 0 = b 2xx 0 − b 2x 0
2
⇒ b 2yy 0 − a 2y 0 − b 2xx 0 − b 2x 0 = 0

⇒
xx 0

a2

xx 0
−
yy 0

b2

yy 0
−
( ) x 20

a2
−
y 20

b2
=0

⇒ − −1=0
a2 b2
xx 0 yy 0
⇒ − =1
a2 b2
Now, the slope of the normal at x 0, y 0) is given by
−1 − a 2y 0
=
b 2x 0
Slope of the tangent at ( x0 , y0 )
Hence, the equation of the normal at x 0, y 0 ( ) is given by
− a 2y
(x − x0 )
0
y − y0 =
b 2x 0

y − y0 − ( x − x0 )
⇒ =
a 2y 0 b 2x 0

y − y0 ( x − x0 )
⇒ + =0
a 2y 0 b 2x 0

Page : 213 , Block Name : Exercise 6.3

Q25 Find the equation of the tangent to the curve y = √3x − 2 which is parallel to the line 4x − 2y + 5 = 0.

Answer.

The equation of the given curve is y = √3x − 2 .
The slope of the tangent to the given curve at any point (x, y) is given by,
dy 3
dx
=
2√3x − 2

The equation of the given line is 4x − 2y + 5 = 0.
5
4x − 2y + 5 = 0 y = 2x + 2 ( which is of the form y = mx + c)
∴Slope of the line = 2
Now, the tangent to the given curve is parallel to the line 4x − 2y − 5 = 0
if the slope of the tangent is equal to the slope of the line.
3
=2
2√3x − 2

3
⇒ √3x − 2 = 4
9
⇒ 3x − 2 = 16
9 41
⇒ 3x = 16 + 2 = 16
41
⇒ x = 48

When x = 48 , y =
41
√( ) 41
3 48 −2=
√
41
16
−2=
√
41 − 32
16
=
√
9
16
3
= 4

∴Equation of the tangent passing through the point 48 , 4 is given by,
( ) 41 3

3
y − 4 = 2 x − 48 ( ) 41

⇒
4y − 3
4
=2 ( ) 48x − 41
48

48x − 41
⇒ 4y − 3 = 6

⇒ 24y − 18 = 48x − 41
⇒ 48x − 24y = 23
Hence, the equation of the required tangent is 48x - 24y = 23.

Page : 213 , Block Name : Exercise 6.3

Q26 The slope of the normal to the curve y = 2x 2 + 3sin x at x = 0 is
1 1
(A) 3 (B) 3 (C) − 3 (D) − 3

Answer.
The equation of the given curve is y = 2x 2 + 3sinx
Slope of the tangent to the given curve at x = 0 is given by,
dy
]
dx x − 0
= 4x + 3cosx] x − 0 = 0 + 3cos0 = 3
Hence, the slope of the normal to the given curve at x = 0 is
−1 −1
Slope of the tangent at x = 0 = 3
.
The correct answer is D .

Page : 213 , Block Name : Exercise 6.3

Q27 The line y = x + 1 is a tangent to the curve y 2 = 4x at the point
(A) (1, 2) (B) (2, 1) (C) (1, − 2) (D) ( − 1, 2)

Answer.
The equation of the given curve is y 2 = 4x
Differentiating with respect to x, we have:
dy dy 2
2y dx = 4 ⇒ dx = y
Therefore, the slope of the tangent to the given curve at any point (x, y) is given by,
dy 2
dx
= y
The given line is y = x + 1 (which is of the form y = mx + c)
∴ Slope of the line = 1
The line y = x + 1 is a tangent to the given curve if the slope of the line is equal to the slope of the tangent. Also,
the line must intersect the curve.
Thus, we must have:
2
y = 1

⇒y=2
Now , y = x + 1 ⇒ x = y − 1 ⇒ x = 2 − 1 = 1
Hence, the line y = x + 1 is a tangent to the given curve at the point (1, 2). The correct answer is A.

Page : 213 , Block Name : Exercise 6.3

Q1 Using differentials, nd the approximate value of each of the following up to 3 places of decimal.
(i) √25.3
(ii) √49.5
(iii) √0.6
1
(iv) (0.009) 3
1
(v) (0.999) 10
1
(vi) (15) 4
1
(vii) (26) 3
1
(viii) (255) 4
1
(ix) (82) 4
1
(x) (401) 2
1
(xi) (0.0037) 2
1
(xii) (26.57) 3
1
(xiii) (81.5) 4
3
(xiv) (3.968) 2
1
(xv) (32.15) 5

Answer.

(i) √25.3

Consider y = √x . Let x = 25 and Δx = 0.3 .
Then,
Δy = √x + Δx − √x = √25.3 − √25 = √25.3 − 5

⇒ √25.3 = Δy + 5
Now, dy is approximately equal to ∆y and is given by,

dy =
() dy
dx
Δx =
1
2√ x
(0.3)

1
= (0.3) = 0.03
2√25

Hence, the approximate value of √25.3 is 0.03 + 5 = 5.03
(ii)
√49.5
Consider y = √x. Let x = 49 and Δx = 0.5.
Then,
Δy = √x + Δx − √x = √49.5 − √49 = √49.5 − 7

⇒ √49.5 = 7 + Δy
Now, dy is approximately equal to Δy and is given by,

dy = () dy
dx
Δx =
2√ x
(0.5)
1

1 1
= (0.5) = 14 (0.5) = 0.035
2√49

Hence, the approximate value of f√49.5 is 7 + 0.035 = 7.035.

(iii) √0.6

Consider y = √x. Let x = 1 and Δx = − 0.4
Then,
Δy = √x + Δx − √x = √0.6 − 1

⇒ √0.6 = 1 + Δy
Now, dy is approximately equal to ∆y and is given by,

dy = ( ) dy
dx
Δx =
2√x
1
(Δx)

1
=
( − 0.4) = − 0.2
2
Hence, the approximate value of √0.6 is 1 + (−0.2) = 1 − 0.2 = 0.8.

1
(iv) (0.009) 3
1
Consider y = x 3 . Let x = 0.008 and Δx = 0.001
Then,
1 1 1 1 1
Δy = (x + Δx) 3 − (x) 3 = (0.009) 3 − (0.008) 3 = (0.009) 3 − 0.2
1
⇒ (0.009) 3 = 0.2 + Δy

Now, dy is approximately equal to ∆y and is given by,

dy =
( )
dy
dx
Δx =
1

3(x) 3
2 (Δx)

1 0.001
= (0.001) = = 0.008
3 × 0.04 0.12
1
Hence, the approximate value of (0.009) 3 is 0.2 + 0.008 = 0.208.

1
(v)(0.999) 10
1
Consider y = (x) 10 . Let x = 1 and ∆x = −0.001.
Then,
1 1 1
Δy = (x + Δx) 10 − (x) 10 = (0.999) 10 − 1
1
⇒ (0.999) 10 = 1 + Δy
Now, dy is approximately equal to Δy and is given by,

dy = ()
dy
dx
Δx =
1

10 ( x ) 10
9 (Δx)\

[asy = (x) 10

1
1

]
= 10 ( − 0.001) = − 0.0001
1
Hence, the approximate value of (0.999) 10 is 1 + (−0.0001) = 0.9999.

1
(vi) (15) 4
1
Consider y = x 4 ⋅ Let x = 16 and Δx = − 1
Then,
1 1 1 1 1
Δy = (x + Δx) 4 − x 4 = (15) 4 − (16) 4 = (15) 4 − 2
1
⇒ (15) 4 = 2 + Δy
Now, dy is approximately equal to Δy and is given by,

dy =
( )
dy
dx
Δx =
1

4(x) 4
3 (Δx)

1 −1 −1
= 3 ( − 1) = = = − 0.03125
4×8 32
4(16) 4
1
Hence, the approximate value of (15) 4 is 2 + (−0.03125) = 1.96875.

1
(vii) (26) 3
1
Consider y = (x) 3 . Let x = 27 and Δx = − 1
Then,
1 1 1 1 1
Δy = (x + Δx) 3 − (x) 3 = (26) 3 − (27) 3 = (26) 3 − 3
1
⇒ (26) 3 = 3 + Δy
Now, dy is approximately equal to ∆y and is given by,

dy = ( )
dy
dx
Δx =
1

3(x) 3
2 (Δx)

1 −1 ¯
= 2 ( − 1) = = − 0.0370
27
3(27) 3
1
Hence, the approximate value of (26) 3 is 3 + (−0.0370) = 2.9629.

1
(viii) (255) 4
1
Consider y = (x) 4 . Let x = 256 and Δx = − 1
Then,
1 1 1 1 1
Δy = (x + Δx) 4 − (x) 4 = (255) 4 − (256) 4 = (255) 4 − 4
1
⇒ (255) 4 = 4 + Δy
Now, dy is approximately equal to ∆y and is given by,

dy =
( )
dy
dx
Δx =
1

4(x) 4
3 (Δx)

1 −1
= 3 ( − 1) = = − 0.0039
4 × 43
4(256) 4
1
Hence, the approximate value of (255) 4 is 4 + (−0.0039) = 3.9961.

1
( ix )(82) 4
1
Consider y = x 4 . Let x = 81 and Δx = 1
Then,
1 1 1 1 1
Δy = (x + Δx) 4 − (x) 4 = (82) 4 − (81) 4 = (82) 4 − 3
1
⇒ (82) 4 = Δy + 3
Now, dy is approximately equal to Δy and is given by,

dy =
( )
dy
dx
Δx =
1

4(x) 4
3 (Δx)

1 1 1
= 3 (1) = 3
= = 0.009
4(3) 108
4(81) 4
1
Hence, the approximate value of (82) 4 is 3 + 0.009 = 3.009.

1
(x)(401) 2
1
Consider y = x 2 . Let x = 400 and ∆x = 1.
Then,
Δy = √x + Δx − √x = √401 − √400 = √401 − 20

⇒ √401 = 20 + Δy
Now, dy is approximately equal to Δy and is given by,

dy = ( )
dy
dx
Δx =
1
2√x
(Δx)

1 1
= = 0.025 (1) =
2 × 20 40
Hence, the approximate value of √401 is 20 + 0.025 = 20.025.

1
(xi)(0.0037) 2
1
Consider y = x 2 . Let x = 0.0036 and ∆x = 0.0001.
Then,
1 1 1 1 1
Δy = (x + Δx) 2 − (x) 2 = (0.0037) 2 − (0.0036) 2 = (0.0037) 2 − 0.06
1
⇒ (0.0037) 2 = 0.06 + Δy
Now, dy is approximately equal to ∆y and is given by,

dy =
( )
dy
dx
Δx =
1
2√x
(Δx)

1
= (0.0001)
2 × 0.06
0.0001
= = 0.00083
0.12
1
Thus, the approximate value of (0.0037) 2 is 0.06 + 0.00083 = 0.06083.

1
(xii)(26.57) 3
1
Consider y = x 3 . Let x = 27 and ∆x = −0.43.
Then,
1 1 1 1 1
Δy = (x + Δx) 3 − x 3 = (26.57) 3 − (27) 3 = (26.57) 3 − 3
1
⇒ (26.57) 3 = 3 + Δy
Now, dy is approximately equal to ∆y and is given by,

dy =
( )
dy
dx
Δx =
3(x) 3
1
2 (Δx)

1
= ( − 0.43)
3(9)
− 0.43
= = − 0.015
27
1
Hence, the approximate value of (26.57) 3 is 3 + (−0.015) = 2.984
1

(xiii) ( 81.5 ) 4
1
Consider y = x 4 . Let x = 81 and ∆x = 0.5.
Then,
1 1 1 1 1
Δy = (x + Δx) 4 − (x) 4 = (81.5) 4 − (81) 4 = (81.5) 4 − 3
1
⇒ (81.5) 4 = 3 + Δy
Now, dy is approximately equal to ∆y and is given by,

dy =
( ) dy
dx
Δx =
1

4(x) 4
3 (Δx)

1 0.5
= (0.5) = = 0.0046
4(3) 3 108
1

Hence, the approximate value of ( 81.5 ) 4 is 3 + 0.0046 = 3.0046.

3
(xiv)(3.968) 2
3
Consider y = x 2 . Let x = 4 and ∆x = − 0.032.
Then,
3 3 3 3 3
Δy = (x + Δx) 2 − x 2 = (3.968) 2 − (4) 2 = (3.968) 2 − 8
3
⇒ (3.968) 2 = 8 + Δy
Now, dy is approximately equal to ∆y and is given by,

( ) dy 3 1
dy = Δx = (x) 2 (Δx)
dx 2
3
= (2)( − 0.032)
2
= − 0.096
3
Hence, the approximate value of (3.968) 2 is 8 + (−0.096) = 7.904.

1
(xv) (32.15) 5
1
Consider y = x 5 . Let x = 32 and Δx = 0.15 .
Then,
1 1 1 1 1
Δy = (x + Δx) 5 − x 5 = (32.15) 5 − (32) 5 = (32.15) 5 − 2
1
⇒ (32.15) 5 = 2 + Δy
Now, dy is approximately equal to ∆y and is given by,

dy =
( ) dy
dx
Δx =
1

5(x) 5
4 ⋅ (Δx)

1
= (0.15)
5 × (2) 4
0.15
= = 0.00187
80
1
Hence, the approximate value of (32.15) 5 is 2 + 0.00187 = 2.00187.

Page : 216 , Block Name : Exercise 6.4

Q2 Find the approximate value of f(2.01), where f(x) = 4x 2 + 5x + 2.

Answer.
Let x = 2 and Δx = 0.01. Then, we have:
f(2.01) = f(x + Δx) = 4(x + Δx) 2 + 5(x + Δx) + 2
Now, Δy = f(x + Δx) − f(x)
f(x + Δx) = f(x) + Δy
( as dx = Δx)
f(x) + f ′(x) ⋅ Δx

( )
⇒ f(2.01) ≈ 4x 2 + 5x + 2 + (8x + 5)Δx

= [4(2) + 5(2) + 2 ] + [8(2) + 5](0.01)
2

= (16 + 10 + 2) + (16 + 5)(0.01)
= 28 + (21)(0.01)
= 28 + 0.21
= 28.21
Hence, the approximate value of f (2.01) is 28.21.

Page : 216 , Block Name : Exercise 6.4

Q3 Find the approximate value of f (5.001), where f(x) = x 3 − 7x 2 + 15.

Answer.
Let x = 5 and Δx = 0.001. Then, we have:
f(5.001) = f(x + Δx) = (x + Δx) 3 − 7(x + Δx) 2 + 15
Now , Δy = f(x + Δx) − f(x)
∴ f(x + Δx) = f(x) + Δy
≈ f(x) + f ′(x) ⋅ Δx (as dx = Δx)

( ) (
⇒ f(5.001) ≈ x 3 − 7x 2 + 15 + 3x 2 − 14x Δx )
[ ] [ ]
= (5) 3 − 7(5) 2 + 15 + 3(5) 2 − 14(5) (0.001) [x = 5, Δx = 0.001]
= (125 − 175 + 15) + (75 − 70)(0.001)
= − 35 + (5)(0.001)
= − 35 + 0.005
= − 34.995
Hence, the approximate value of f (5.001) is −34.995.

Page : 216 , Block Name : Exercise 6.4

Q4 Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.

Answer. The volume of a cube (V) of side x is given by V = x 3.

∴ dV =
( )
dV
dx
Δx

( )
= 3x 2 Δx

( )
= 3x 2 (0.01x) [ as 1% of x is 0.01x]
3
= 0.03x
Hence, the approximate change in the volume of the cube is 0.03x 3m 3

Page : 216 , Block Name : Exercise 6.4

Q5 Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.

Answer. The surface area of a cube (S) of side x is given by S = 6x 2

∴
dS
dx
=
( )
dS
dx
Δx

= (12x)Δx
= (12x)(0.01x) [ as 1% of x is 0.01x]
= 0.12x 2
Hence, the approximate change in the surface area of the cube is 0.12x 2m 2

Page : 216 , Block Name : Exercise 6.4

Q6 If the radius of a sphere is measured as 7 m with an error of 0.02 m, then nd the approximate error in
calculating its volume.

Answer. Let r be the radius of the sphere and ∆r be the error in measuring the radius.
Then,
r = 7 m and ∆r = 0.02 m Now, the volume V of the sphere is given by,
4
V = πr 3
3
dV
∴ = 4πr 2
dr

∴ dV =
( )
dV
dr
Δr

( )
= 4πr 2 Δr

= 4π(7) 2(0.02)m 3 = 3.92πm 3
Hence, the approximate error in calculating the volume is 3.92 π m 3.

Page : 216 , Block Name : Exercise 6.4

Q7 If the radius of a sphere is measured as 9 m with an error of 0.03 m, then nd the approximate error in
calculating its surface area.

Answer. Let r be the radius of the sphere and ∆r be the error in measuring the radius.
Then,
r = 9 m and ∆r = 0.03 m
Now, the surface area of the sphere (S) is given by,
S = 4πr 2
dS
∴ dr = 8πr

∴ dS =
()
dS
dr
Δr

= (8πr)Δr
= 8π(9)(0.03)m 2
= 2.16πm 2
Hence, the approximate error in calculating the surface area is 2.16π m 2.

Page : 216 , Block Name : Exercise 6.4

Q8 If f(x) = 3x 2 + 15x + 5, then the approximate value of f(3.02)
(A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66

Answer.
Let x = 3 and Δx = 0.02. Then, we have:
f(3.02) = f(x + Δx) = 3(x + Δx) 2 + 15(x + Δx) + 5
Now, Δy = f(x + Δx) − f(x)
⇒ f(x + Δx) = f(x) + Δy
≈ f(x) + f ′(x)Δx (As dx = Δx)

( )
⇒ f(3.02) ≈ 3x 2 + 15x + 5 + (6x + 15)Δx

[ ]
= 3(3) 2 + 15(3) + 5 + [6(3) + 15](0.02) [ As x = 3, Δx = 0.02]
= (27 + 45 + 5) + (18 + 15)(0.02)
= 77 + (33)(0.02)
= 77 + 0.66
= 77.66
Hence, the approximate value of f(3.02) is 77.66.
The correct answer is D.

Page : 216 , Block Name : Exercise 6.4

Q9 The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
(A) 0.06x 3m 3 (B) 0.6x 3m 3 (C) 0.09x 3m 3 (D) 0.9x 3m 3

Answer. The volume of a cube (V) of side x is given by V = x 3.

∴ dV = ( )
dV
dx
Δx

( )
= 3x 2 Δx

( )
= 3x 2 (0.03x) [ As 3% of x is 0.03x]

= 0.09x 3m 3
Hence, the approximate change in the volume of the cube is 0.09 x 3m 3 .
The correct answer is C.

Page : 216 , Block Name : Exercise 6.4

Q1 Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = (2x − 1) 2 + 3 (ii) f(x) = 9x 2 + 12x + 2
(iii) f(x) = − (x − 1) 2 + 10 (iv) g(x) = x 3 + 1

Answer.
(i) The given function is f(x) = (2x − 1) 2 + 3
It can be observed that (2x − 1) 2 ≥ 0 for every x . R .
Therefore, f(x) = (2x − 1) 2 + 3 ≥ 3 for every x
The minimum value of f is attained when 2x − 1 = 0
2x − 1 = 0

Minimum value of f = f 2 () (
1 1
= 2⋅ 2 −1 ) 2
+3

Hence, function f does not have a maximum value.

(ii) The given function is f(x) = 9x 2 + 12x + 2 = (3x + 2) 2 − 2
It can be observed that (3x + 2) 2 ≥ 0 for every x .
Therefore, f(x) = (3x + 2) 2 − 2 ≥ − 2 for every x
The minimum value of f is attained when 3x + 2 = 0 .
−2
3x + 2 = 0 x= 3

Minimum value of f = f − 3 ( ) (( ) )
2
= 3
−2
3 +2
2
−2= −2

Hence, function f does not have a maximum value.

(iii) The given function is f(x) = − (x − 1) 2 + 10
It can be observed that (x − 1) 2 ≥ 0 for every x
Therefore, f(x) = − (x − 1) 2 + 10 ≤ 10 for every x R.
The maximum value of f is attained when (x − 1) = 0 .
(x − 1) = 0 x = 0
Maximum value of f = f(1) = − (1 − 1) 2 + 10 = 10
Hence, function f does not have a minimum value.

(iv) The given function is g(x) = x 3 + 1
Hence, function g neither has a maximum value nor a minimum value.

Page : 231 , Block Name : Exercise 6.5

Q2 Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = | x + 2 | − 1 (ii) g(x) = − | x + 1 | + 3
(iii) h(x) = sin(2x) + 5 (iv) f(x) = | sin4x + 3 |
(v) h(x) = x + 1, x ∈ ( − 1, 1)

Answer.
(i) f(x) = | x + 2 | − 1
we know that | x + 2 | ≥ 0 for every x .
Therefore, f(x) = | x + 2 | − 1 ≥ − 1 for every x . R.
The minimum value of f is attained when | x + 2 | = 0
|x + 2| = 0
⇒x= −2
Minimum value of f = f( − 2) = | − 2 + 2 | − 1 = − 1
Hence, function f does not have a maximum value.

(ii) g(x) = − | x + 1 | + 3
We know that − | x + 1 | ≤ 0 for every x R.
Therefore, g(x) = − | x + 1 | + 3 ≤ 3 for every x R.
The maximum value of g is attained when | x + 1 | = 0
|x + 1| = 0
⇒x= −1
Maximum value of g = g( − 1) = − | − 1 + 1 | + 3 = 3
Hence, function g does not have a minimum value.

(iii) h(x) = sin2x + 5
We know that − 1 ≤ sin2x ≤ 1
− 1 + 5 ≤ sin2x + 5 ≤ 1 + 5
4 ≤ sin2x + 5 ≤ 6
Hence, the maximum and minimum values of h are 6 and 4 respectively.

(iv) f(x) = | sin4x + 3 |
we know that − 1 ≤ sin4x ≤ 1
2 ≤ sin4x + 3 ≤ 4
2 ≤ sin4x + 3 | ≤ 4
Hence, the maximum and minimum values of f are 4 and 2 respectively.

(v) h(x) = x + 1, x ( − 1, 1)
x0
Here, if a point x 0 is closest to − 1, then we find 2 + 1 < x 0 + 1
x1 + 1
Also, if x 1 is closest to 1, then x 1 + 1 < 2
+1
+ 1 for all x 0 (-1, 1).
Hence, function h(x) has neither maximum nor minimum value in (-1, 1).

Page : 232 , Block Name : Exercise 6.5

Q3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and
the local minimum values, as the case may be:
(i) f(x) = x 2
π
(iii) h(x) = sinx + cosx, 0 < x < 2
(iv) f(x) = sinx − cosx, 0 < x < 2π
x 2
(v) f(x) = x 3 − 6x 2 + 9x + 15 (vi) g(x) = 2 + x , x > 0
1
(vii) g(x) = (viii) f(x) = x√1 − x, 0 < x < 1
x2 + 2

Answer.
(i) f(x) = x 2
∴ f ′(x) = 2x
Now,
f ′(x) = 0 ⇒ x = 0
Thus, x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.
We have f ∗ (0) = 2, whichispositive.
Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is f(0) =
0.

(ii) g(x) = x 3 − 3x
∴ g ′(x) = 3x 2 − 3
Now,
g ′(x) = 0 ⇒ 3x 2 = 3 ⇒ x = ± 1
g ′(x) = 6x
g ′(1) = 6 > 0
g ′( − 1) = − 6 < 0
By second derivative test, x =1 is a point of local minima and local minimum value of g at x = 1 is g(1) =

1 3 − 3 = 1 − 3 = − 2.
x = − 1 is a point of local maxima and local maximum value of g a
x = − 1 is g(1) = ( − 1) 3 − 3( − 1) = − 1 + 3 = 2

π
(iii) h(x) = sinx + cosx, 0 < x < 2

∴ h ′(x) = cosx − sinx

h ′(x) = 0 ⇒ sinx = cosx ⇒ tanx = 1 ⇒ x = 4 ∈ 0, 2
π
( ) π

h ′′(x) = − sinx − cosx = − (sinx + cosx)

() (
h ′′ 4
π
= −
√2
1
+
√2
1
) = −
√2
2
= − √2 < 0

π
x = 4 is a point of local maxima and the local

Therefore, by second derivative test,

x = 4 is
π
()π
4
π
= sin 4 + cos 4 =
π 1

√2
+
√2
1
= √2

(iv) f(x) = sinx − cosx, 0 < x < 2π
∴ f ′(x) = cosx + sinx
3π 7π
f ′(x) = 0 ⇒ cosx = − sinx ⇒ tanx = − 1 ⇒ x = 4 , 4 ∈ (0, 2π)
f ′′(x) = − sinx + cosx

()3π
f ′′ 4
3π
= − sin 4 + cos 4 = −
3π

√2
1
−
√2
1
= − √2 > 0

()7π
f ′′ 4
7π
= − sin 4 + cos 4 =
7π

√2
1
+
√2
1
= √2 > 0
3π
Therefore, by second derivative test, x = 4 a point of local maxima and the local
3π
maximum value of f at 4
is

()
f 4
3π 3π
= sin 4 − cos 4 =
3π

√2
1
+
1

√2
= √2.
7π
x = 4 is a point of local minima and

the local minimum value of f = 4 fis f 4
7π
() 7π 7π 7π
= sin 4 − cos 4 = −
1

√2
−
1

√2
= − √2

(v)f(x) = x 3 − 6x 2 + 9x + 15
∴ f ′(x) = 3x 2 − 12x + 9

(
f ′(x) = 0 ⇒ 3 x 2 − 4x + 3 = 0 )
⇒ 3(x − 1)(x − 3) = 0
⇒ x = 1, 3
Now, f ′′
(x) = 6x − 12 = 6(x − 2)
f ′′(1) = 6(1 − 2) = − 6 < 0
f ′(3) = 6(3 − 2) = 6 > 0

Therefore, by second derivative test, x = 1 is a point of local maxima and the local maximum value of f at x = 1 is
f(1) = 1 − 6 + 9 + 15 = 19. However, x = 3 is a point of local minima and the local minimum value of f at x = 3 is f(3) =
27 − 54 + 27 + 15 = 15.

x 2
(vi)g(x) = 2 + x , x > 0
1 2
∴ g ′(x) = 2 −
x2
Now,
2 1
g ′(x) = 0 gives = 2 ⇒ x2 = 4 ⇒ x = ± 2
x2
since x > 0, we take x = 2
4
g ′′(x) =
x3
4 1
g ′′(2) = = 2 >0
23
2 2
Therefore, by second derivative test, x =2 is g(2) = 2 + 2 = 1 + 1 = 2

1
g(x) =
x2 + 2
− ( 2x )
∴ g ′(x) =
(vii) (x +2 )
2 2

− 2x
g ′(x) = 0 ⇒ =0⇒x=0
( x2 + 2 ) 2

Now, for values close to x =0 and to the left of 0, g ′(x) > 0. Also, for values close to x = 0 and to the right of 0,
g ′(x) < 0.
1 1
Therefore, by rst derivative test, x = 0 is a point of local maxima and the local maximum value of g(0) is 0 + 2 = 2 .

f(x) = x√1 − x, x > 0
(viii) 1 x
∴ f ′(x) = √1 − x + x ⋅ ( − 1) = √1 − x −
2√ 1 − x 2√ 1 − x
2(1 − x) − x 2 − 3x
= =
2√1 − x 2√1 − x
2 − 3x 2
f ′(x) = 0 ⇒ = 0 ⇒ 2 − 3x = 0 ⇒ x =
2√1 − x 3

( )
[ ]
−1
√1 − x( − 3) − (2 − 3x) 2√1 − x
1
f ′′(x) =
2 1−x

=
√1 − x( − 3) + (2 − 3x) 2√1 − x ( ) 1

2(1 − x)
− 6(1 − x) + (2 − 3x)
=
2(1 − x)
3x − 4
= 3
4(1 − x) 2

() 3
2
−4

f ( )=
3
2 2−4 −1
′′
3
= = <0 3 3 3

( ) () () 4 1− 3
2 2
4
1
3
2
2
1
3
2

2 2
Therefore, by second derivative test, x = 3 is a point of local maxima and the local maximum value of f at x = 3 is .

() √ 2 2 2 2 1 2 2√ 3
f 3 = 3 1− 3 = 3
√ 3
=
3√ 3
= 9 .

Page : 232 , Block Name : Exercise 6.5

Q4 Prove that the following functions do not have maxima or minima:
(i) f(x) = e x (ii) g(x) = logx
(iii) h(x) = x 3 + x 2 + x + 1

Answer. (i) We have,
f(x) = e x
∴ f ′(x) = e x
Now, if f ′(x) = 0, then e x = 0. But the exponential function can never assume 0 for any value of x.
Therefore, there does not exist c R such that f ′(c) = 0 .
Hence, function f does not have maxima or minima.

(ii) We have,
g(x) = logx
1
∴ g ′(x) = x
Since log x is de ned for a positive number x, g ′(x) > 0 for any x.
′
Therefore, there does not exist c R such that g (c) = 0
Hence, function g does not have maxima or minima.

(iii) We have,
h(x) = x 3 + x 2 + x + 1
∴ h ′(x) = 3x 2 + 2x + 1
Now,

− 2 ± 2√2i − 1 ± √2i
h(x) = 0 3x 2 + 2x + 1 = 0 x= 6
= 3
∉R
Therefore, there does not exist c R such that h ′(c) = 0
Hence, function h does not have maxima or minima.

Page : 232 , Block Name : Exercise 6.5

Q5 Find the absolute maximum value and the absolute minimum value of the following functions in the given
intervals:
(i) f(x) = x 3, x ∈ [ − 2, 2] ( (ii) f(x) = sinx + cosx, x ∈ [0, π]

1
(iii) f(x) = 4x − 2 x 2, x ∈
[ ]
− 2, 2
9
(iv) f(x) = (x − 1) 2 + 3, x ∈ [ − 3, 1]

Answer.
(i) The given function is f(x) = x 3 .
∴ f ′(x) = 3x 2
Now,
f ′(x) = 0 ⇒ x = 0
Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [-2, 2].
f(0) = 0
f( − 2) = ( − 2) 3 = − 8
f(2) = (2) 3 = 8
Hence, we can conclude that the absolute maximum value of f on [−2, 2] is 8 occurring at x = 2. Also, the absolute
minimum value of f on [−2, 2] is −8 occurring at x = −2.

(ii) The given function is f(x) = sinx + cosx
∴ f ′(x) = cosx − sinx
Now,
π
f ′(x) = 0 ⇒ sinx = cosx ⇒ tanx = 1 ⇒ x = 4
π
Then, we evaluate the value of f at critical point x = 4 and at the end points of the interval [0, π].

()
f 4
π π π
= sin 4 + cos 4 =
1

√2
+
1

√2
=
2

√2
= √2

f(0) = sin0 + cos0 = 0 + 1 = 1
f(π) = sinπ + cosπ = 0 − 1 = − 1
π
Hence, we can conclude that the absolute maximum value of f on [0, π] is √2 occurring at x = 4 and the absolute
minimum value of f on [0, π] is −1 occurring at x = π.

1
f(x) = 4x − 2 x 2
(iii) The given function is 1
∴ f ′(x) = 4 − 2 (2x) = 4 − x
Now,
f ′(x) = 0 ⇒ x = 4

Then, we evaluate the value of f at critical point x = 4 and at the end points of the interval
[ ]
− 2, 2
9

1
f(4) = 16 − 2 (16) = 16 − 8 = 8
1
f( − 2) = − 8 − 2 (4) = − 8 − 2 = − 10

() () ()
f 2
9 9
=4 2
1
− 2 2
9 2 81
= 18 − 8 = 18 − 10.125 = 7.875

Hence, we can conclude that the absolute maximum value of f on [ ] 9
− 2, 2 is 8 occurring at x = 4 and the absolute

minimum value of f on [ ] 9
− 2, 2 is −10 occurring at x = −2.

(iv) The given function is f(x) = (x − 1) 2 + 3
∴ f ′(x) = 2(x − 1)
Now,
f ′(x) = 0 ⇒ 2(x − 1) = 0 x=1
Then, we evaluate the value of f at critical point x = 1 and at the end points of the interval [−3, 1].
f(1) = (1 − 1) 2 + 3 = 0 + 3 = 3
f( − 3) = ( − 3 − 1) 2 + 3 = 16 + 3 = 19
Hence, we can conclude that the absolute maximum value of f on [−3, 1] is 19 occurring at x = −3 and the minimum
value of f on [−3, 1] is 3 occurring at x = 1.

Page : 232 , Block Name : Exercise 6.5

Q6 Find the maximum pro t that a company can make, if the pro t function is given by
p(x) = 41 − 72x − 18x 2

Answer. The pro t function is given as p(x) = 41 − 72x − 18x 2.
∴ p ′(x) = − 24 − 36x
p ′′(x) = − 36
Now
Now,
− 24 2
p ′(x) = 0 ⇒ x = 36 = − 3

Also,

p ′′
( ) −2
3
= − 36 < 0
2
By second derivative test, x = − 3 is the point of local maxima of p.

∴ Maximum profit = p −
( ) 2
3

( ) ( )
= 41 − 24 −
2
3
− 18 −
2 2
3
= 41 + 16 − 8
= 49
Hence, the maximum pro t that the company can make is 49 units.

Page : 232 , Block Name : Exercise 6.5

Q7 Find both the maximum value and the minimum value of
3x 4 − 8x 3 + 12x 2 − 48x + 25 on the interval [0, 3]

Answer.
Let f(x) = 3x 4 − 8x 3 + 12x 2 − 48x + 25
∴ f ′(x) = 12x 3 − 24x 2 + 24x − 48

(
= 12 x 3 − 2x 2 + 2x − 4)
= 12 [x (x − 2) + 2(x − 2) ]
2

= 12(x − 2) (x + 2 )
2

Now, f ′(x) = 0 gives x = 2 or x 2 + 2 = 0 for which there are no real roots.
Therefore, we consider only x = 2 ∈ [0, 3] .
Now, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3] .
f(2) = 3(16) − 8(8) + 12(4) − 48(2) + 25
= 48 − 64 + 48 − 96 + 25
= − 39
f(0) = 3(0) − 8(0) + 12(0) − 48(0) + 25
= 25
f(3) = 3(81) − 8(27) + 12(9) − 48(3) + 25
= 243 − 216 + 108 − 144 + 25 = 16
Hence, we can conclude that the absolute maximum value of f on [0, 3] is 25 occurring at x = 0 and the absolute
minimum value of f at [0, 3] is − 39 occurring at x = 2.

Page : 232 , Block Name : Exercise 6.5

Q8 At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?

Answer.
Let f(x) = sin2x
∴ f ′(x) = 2cos2x
Now,
f ′(x) = 0 ⇒ cos2x = 0
π 3π 5π 7π
⇒ 2x = 2 , 2 , 2 , 2
π 3π 5π 7π
⇒ x = 4, 4 , 4 , 4
3π 5π 7π
Then, we evaluate the values of f at critical points x = 4 , 4 , 4 and at the end

points of the interval [0, 2π].

()
f 4
π π
()
3π
= sin 2 = 1, f 4
3π
= sin 2 = − 1

()
5π
f 4
5π
()
= sin 2 = 1, f 4
7π 7π
= sin 2 = − 1

f(0) = sin0 = 0, f(2π) = sin2π = 0
π 5π
Hence, we can conclude that the absolute maximum value of f on [0, 2π] is occurring at x = 4 and x = 4 .

Page : 232 , Block Name : Exercise 6.5

Q9 What is the maximum value of the function sin x + cos x?

Answer.
Let f(x) = sinx + cosx
∴ f ′(x) = cosx − sinx
π 5π
f ′(x) = 0 ⇒ sinx = cosx ⇒ tanx = 1 ⇒ x = 4 , 4 …

f ′′(x) = − sinx − cosx = − (sinx + cosx)
Now, f ′′(x) will be negative when (sin x + cos x) is positive i.e., when sin x and cos x are both positive. Also, we know
that sin x and cos x both are positive in the rst
quadrant. Then, f ′′(x) will be negative when
π
x= 4

Thus, we consider

() (
π
f′ 4
π
= − sin 4 + cos 4
π
) ( )
= −
√2
2
= − √2 < 0

π
By second derivative test, f will be the maximum at x = 4 and the maximum value of f

()
π
f 4
π
= sin 4 + cos 4 =
π

√2
1
×
1

√2
=
√2
2
= √2

Page : 232 , Block Name : Exercise 6.5

Q10 Find the maximum value of 2x 3 − 24x + 107 in the interval [1, 3]. Find the maximum value of the same
function in [–3, –1].

Answer.
Let f(x) = 2x 3 − 24x + 107

∴ f ′(x) = 6x 2 − 24 = 6 x 2 − 4 ( )
Now,

( )
f ′(x) = 0 ⇒ 6 x 2 − 4 = 0 ⇒ x 2 = 4 ⇒ x = ± 2

we first consider the interval [1, 3]
Then, we evaluate the value of f at the critical point x = 2 [1, 3] and at the end points of the interval [1, 3].
f(2) = 2(8) − 24(2) + 107 = 16 − 48 + 107 = 75
f(1) = 2(1) − 24(1) + 107 = 2 − 24 + 107 = 85
f(3) = 2(27) − 24(3) + 107 = 54 − 72 + 107 = 89
Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occuring at x = 3.
Next, we consider the interval [−3, −1]. Evaluate the value of f at the critical point x = −2 ∴ [−3, −1] and at the end
points of the interval [1, 3].
f(−3) = 2 (−27) − 24(−3) + 107 = −54 + 72 + 107 = 125
f(−1) = 2(−1) − 24 (−1) + 107 = −2 + 24 + 107 = 129
f(−2) = 2(−8) − 24 (−2) + 107 = −16 + 48 + 107 = 139
Hence, the absolute maximum value of f(x) in the interval [−3, −1] is 139 occurring at x = −2.

Page : 232 , Block Name : Exercise 6.5

Q11 It is given that at x = 1, the function x 4 − 62x 2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find

the value of a.

Answer.
Let f(x) = x 4 − 62x 2 + ax + 9
∴ f ′(x) = 4x 3 − 124x + a
It is given that function f attains its maximum value on the interval [0, 2] at x = 1.
∴ f ′(1) = 0
⇒ 4 − 124 + a = 0
⇒ a = 120
Hence, the value of a is 120 .

Page : 233 , Block Name : Exercise 6.5

Q12 Find the maximum and minimum values of x + sin 2x on [0, 2π].

Answer.
Let f(x) = x + sin2x
∴ f ′(x) = 1 + 2cos2x

1 π
Now, f ′(x) = 0 ⇒ cos2x = − 2 = − cos 3 = cos π − 3
2π
( )
π 2π
= cos 3

2x = 2π ± n∈Z
3′
π
⇒ x = nπ ± 3 , n ∈ Z
π 2π 4π 5π
⇒ x = 3 , 3 , 3 , 3 ∈ [0, 2π]
π 2π 4π 5z
Then, we evaluate the value of f at critical points x = 3 , 3 , 3 , 3 and at the ends point of the interval [0, 2π]

()
f 3
π π 2π
= 3 + sin 3 = 3 + 2
π √3

()
f 3
2π 2π 4π
= 3 + sin 3 = 3 − 2
2π √3

()
f 3
4π 4π 8π
= 3 + sin 3 = 3 + 2
4π √3

()
f 3
5π 5π 10π
= 3 + sin 3 = 3 − 2
5π √3

f(0) = 0 + sin0 = 0
f(2π) = 2π + sin4π = 2π + 0 = 2π
Hence, we can conclude that the absolute maximum value of f(x) in the interval [0, 2π] is 2π occurring at x = 2π and
the absolute minimum value of f(x) in the interval [0, 2π] is 0 occurring at x = 0.

Page : 233 , Block Name : Exercise 6.5

Q13 Find two numbers whose sum is 24 and whose product is as large as possible.

Answer. Let one number be x. Then, the other number is (24 − x).
Let P(x) denote the product of the two numbers. Thus, we have:

P(x) = x(24 − x) = 24x − x 2
∴ P ′(x) = 24 − 2x
P ′′(x) = − 2
Now,
P ′(x) = 0 ⇒ x = 12
Also,
P ′′(12) = − 2 < 0
By second derivative test, x = 12 is the point of local maxima of P. Hence, the product of the numbers is the
maximum when the numbers are 12 and 24 − 12 = 12.

Page : 233 , Block Name : Exercise 6.5

Q14 Find two positive numbers x and y such that x + y = 60 and xy 3 is maximum.

Answer. The two numbers are x and y such that x + y = 60.
Y = 60 - X
Let f(x) = xy 3
⇒ f(x) = x(60 − x) 3
∴ f ′(x) = (60 − x) 3 − 3x(60 − x) 2
= (60 − x) 2[60 − x − 3x]
= (60 − x) 2(60 − 4x)
And, f (x) = − 2(60 − x)(60 − 4x) − 4(60 − x) 2
′

= − 2(60 − x)(60 − 4x + 2(60 − x)]
= − 2(60 − x)(180 − 6x)
= − 12(60 − x)(30 − x)
Now, f ′(x) = 0 ⇒ x = 60 or x = 15
When x = 60, f ∗ (x) = 0
.
When x = 15, f ′(x) = − 12(60 − 15)(30 − 15) = − 12 × 45 × 15 < 0
∴By second derivative test, x = 15 is a point of local maxima of f. Thus, function xy3 is maximum when x = 15 and y
= 60 − 15 = 45. Hence, the required numbers are 15 and 45.

Page : 233 , Block Name : Exercise 6.5

Q15 Find two positive numbers x and y such that their sum is 35 and the product x 2y 5 is a maximum.

Answer.
Let one number be x. Then, the other number is y = (35 − x) .
Let P(x) = x 2y 5 . Then, we have:
P(x) = x 2(35 − x) 5
∴ P ′(x) = 2x(35 − x) 5 − 5x 2(35 − x) 4
= x(35 − x) 4[2(35 − x) − 5x]
= x(35 − x) 4(70 − 7x)
= 7x(35 − x) 4(10 − x)

[
And, P ′′(x) = 7(35 − x) 4(10 − x) + 7x − (35 − x) 4 − 4(35 − x) 3(10 − x) ]
= 7(35 − x) 4(10 − x) − 7x(35 − x) 4 − 28x(35 − x) 3(10 − x)
= 7(35 − x) 3[(35 − x)(10 − x) − x(35 − x) − 4x(10 − x)]

[
= 7(35 − x) 3 350 − 45x + x 2 − 35x + x 2 − 40x + 4x 2 ]
= 7(35 − x) (6x − 120x + 350 )
3 2

Now, P ′(x) = 0 ⇒ x = 0, x = 35, x = 10
When x = 35, f ′(x) = f(x) = 0 and y = 35 − 35 = 0. This will make the product x 2y 5
equal to 0.
When x = 0, y = 35 − 0 = 35 and the product x 2y 2 will be 0 .
x = 0 and x = 35 cannot be the possible values of x .
When x = 10, we have:
P ′′(x) = 7(35 − 10) 3(6 × 100 − 120 × 10 + 350)
= 7(25) 3( − 250) < 0
By second derivative test, P(x) will be the maximum when x = 10 and y = 35 − 10 =
Hence, the required numbers are 10 and 25.

Page : 233 , Block Name : Exercise 6.5

Q16 Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Answer. Let one number be x. Then, the other number is (16 − x).
Let the sum of the cubes of these numbers be denoted by S(x). Then,
S(x) = x 3 + (16 − x) 3
∴ S ′(x) = 3x 2 − 3(16 − x) 2, S ∗ (x) = 6x + 6(16 − x)
Now, S ′(x) = 0 ⇒ 3x 2 − 3(16 − x) 2 = 0
⇒ x 2 − (16 − x) 2 = 0
⇒ x 2 − 256 − x 2 + 32x = 0
256
⇒ x = 32 = 8
Now, S ′(8) = 6(8) + 6(16 − 8) = 48 + 48 = 96 > 0
By second derivative test, x = 8 is the point of local minima of S .
Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8
and 16 − 8 = 8 .

Page : 233 , Block Name : Exercise 6.5

Q17 A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner
and folding up the aps to form the box. What should be the side of the square to be cut off so that the volume of
the box is the maximum possible

Answer. Let the side of the square to be cut off be x cm. Then, the length and the breadth of the box will be (18 −
2x) cm each and the height of the box is x cm. Therefore, the volume V(x) of the box is given by,
V(x) = x(18 − 2x) 2

∴ V ′(x) = (18 − 2x) 2 − 4x(18 − 2x)
= (18 − 2x)[18 − 2x − 4x]
= (18 − 2x)(18 − 6x)
= 6 × 2(9 − x)(3 − x)
= 12(9 − x)(3 − x)
And, V ′′(x) = 12[ − (9 − x) − (3 − x)]
= − 12(9 − x + 3 − x)
= − 12(12 − 2x)
= − 24(6 − x)
Now, V ′(x) = 0 ⇒ x = 9 or x = 3
If x = 9, then the length and the breadth will become 0.
∴x≠9
⇒x=3
Now, V ′(3) = − 24(6 − 3) = − 72 < 0
By second derivative test, x = 3 is the point of maxima of V. Hence, if we remove a square of side 3 cm from each
corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the
largest possible.

Page : 233 , Block Name : Exercise 6.5

Q18 A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each
corner and folding up the aps. What should be the side of the square to be cut off so that the volume of the box is
maximum ?

Answer. Let the side of the square to be cut off be x cm. Then, the height of the box is x, the length is 45 − 2x, and
the breadth is 24 − 2x. Therefore, the volume V(x) of the box is given by,
V(x) = x(45 − 2x)(24 − 2x)

(
= x 1080 − 90x − 48x + 4x 2

= 4x 3 − 138x 2 + 1080x
∴ V ′(x) = 12x 2 − 276x + 1080

(
= 12 x 2 − 23x + 90 )
= 12(x − 18)(x − 5)
V ′′(x) = 24x − 276 = 12(2x − 23)
Now, (x) = 0
It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet. Thus, x cannot be
equal to 18.
x=5
Now, V ′′(5) = 12(10 − 23) = 12( − 13) = − 156 < 0
∴ By second derivative test, x = 5 is the point of maxima.
Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm.

Page : 233 , Block Name : Exercise 6.5

Q19 Show that of all the rectangles inscribed in a given xed circle, the square has the maximum area.

Answer. Let a rectangle of length l and breadth b be inscribed in the given circle of radius a. Then, the diagonal
passes through the centre and is of length 2a cm.

Now, by applying the Pythagoras theorem, we have:
(2a) 2 = l 2 + b 2
⇒ b 2 = 4a 2 − l 2
⇒b= √4a 2 − l 2
Area of the rectangle, A = I 4a 2 − l 2 √
dA 1 l2
∴
dl
= √ 4a 2 − l 2 + / ( − 2l) = √ 4a 2 − l 2 −
√
2 4a 2 − l 2 √4a 2 − l 2
4a 2 − 2l 2
=
√4a 2 − l 2
√4a − l ( − 4l ) − ( 4a − 2l ) √
( − 2l )
2 2 2 2
d 2A 2 4a 2 − l 2
=
dt 2
( 4a − l ) 2 2

( 4a − l ) ( − 4l ) + l ( 4a − 2l )
2 2 2 2

= 3

( 4a − l ) 2 2 2

2 − 2l ( 6a − l )
3
2 2
− 12a l + 2l
= = 3 3

( 4a − l ) 2 2
( 4a − l )
2 2 2 2

dA
Now, = 0 gives 4a 2 = 2l 2 ⇒ l = √2a
dl
⇒b= √4a 2 − 2a 2 = √2a 2 = √2a
Now, when l = √2a

d 2A (
− 2 ( √2a ) 6a 2 − 2a 2 ) − 8√2a 3
= = = −4<0
dl 2 2√2a 3 2√2a 3

By the second derivative test, when l = √2a, then the area of the rectangle is the maximum.
Since l = b = √2a, the rectangle is a square.
Hence, it has been proved that of all the rectangles inscribed in the given xed circle, the square has the maximum
area.

Page : 233 , Block Name : Exerciser 6.5

Q20 Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to
the diameter of the base.

Answer. Let r and h be the radius and height of the cylinder respectively. Then, the surface area (S) of the cylinder
is given by,

S = 2πr 2 + 2πrh
S − 2πr 2
⇒h=
2πr

=
S
2π r()
1
−r

Let V be the volume of the cylinder. Then,

V = πr 2h = πr 2 2π r
[ () ]
S 1 Sr
− r = 2 − πr 3

dV S d 2V
Then, dr = 2 − 3πr 2, = − 6πr
dr 2
dV S S
Now, dr = 0 ⇒ 2 = 3πr 2 ⇒ r 2 = 6π

When r 2 = 6π , then
S d 2V
dr 2
= − 6π
(√ ) S
6π
<0

S
By second derivative test, the volume is the maximum when r 2 = 6π

S
Now, when r 2 = 6π , then h = 2π
6πr 2
()
1
r
− r = 3r − r = 2r

Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the
diameter.

Page : 233 , Block Name : Exercise 6.5

Q21 Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, nd the
dimensions of the can which has the minimum surface area?

Answer. Let r and h be the radius and height of the cylinder respectively.
Then, volume (V) of the cylinder is given by,
V = πr 2h = 100
100
∴h=
πr 2
Surface area (S) of the cylinder is given by,
200
S = 2πr 2 + 2πrh = 2πr 2 + r
dS 200 d 2S 400
∴ dr = 4πr − , = 4π +
r 2 dr 2 r3
dS 200
dr
= 0 ⇒ 4πr =
r2
200 50
⇒ r 3 = 4π = π
1

⇒r= () 50
π
3

1

Now, it is observed that when r =
()50
π
3
,
d 2S
dr 2
>0
cylinder is ()
50
π
1
3
cm.
By second derivative test, the surface area is the minimum when the radius of the

1 1

When r =
()
50
π
3
,h =
100

( )
2
=
2 × 50
()
50
=2 π
3
cm

(
50 3
π π
2
¯3 1
50 t ( ) 3

Hence, the required dimensions of the can which has the minimum surface area is given
1 1

By radius =
() 50
π
3
cm and height = 2 π
() 50 3
cm.

Page : 233 , Block Name : Exercise 6.5

Q22 A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other
into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is
minimum?

Answer. Let a piece of length l be cut from the given wire to make a square. Then, the other piece of wire to be
made into a circle is of length (28 − l) m.
l
Now, side of square = = 4 .
1
Let r be the radius of the circle. Then, 2πr = 28 − l ⇒ r = 2π (28 − l).
The combined areas of the square and the circle (A) is given by,

A= ( side of the square ṫ + r 2
2

=
l2
16
+π
[ 2π
1
(28 − l)
] 2

l2 1
= + (28 − l) 2
16 4π
dA 2l 2 l 1
∴ dl = 16 + 4π (28 − l)( − 1) = 8 − 2π (28 − l)
d 2A 1 1
= 8 + 2π > 0
dl 2
dA l 1
Now, dl = 0 ⇒ 8 − 2π (28 − l) = 0
πl − 4 ( 28 − l )
⇒ 8π
=0

⇒ (π + 4)l − 112 = 0
112
⇒ l = π+4
112 d 2A
Thus, when l = π + 4 , >0
dl 2
112
∴ By second derivative test, the area (A) is the minimum when l = π + 4
112
Hence, the combined area is the minimum when the length of the wire in making the square is π + 4 cm while the
112 28π
length of the wire in making the circle is 28 − π + 4 = π + 4 cm

Page : 233 , Block Name : Exercise 6.5

8
Q23 Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 27 of the volume of
the sphere.

Answer. Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.

Let V be the volume of the cone.
1
V = 3 πr 2h

Height of the cone is given by,
h = R + AB = R + √R 2 − r 2 [ABC is a right triangle ]
∴V=
1
3 ( √
πr 2 R + R2 − r2 )
1 1
= πr 2R +
πr 2 R 2 − r 2 √
3 3
dV 2 2 1 ( − 2r)
∴ = πrR + πr R 2 − r 2 + πr 2 ⋅ √
dr 3 3 3 2 R2 − r2 √
2 2 1 r3
=
3
πrR +
3 √
πr R 2 − r 2 −
3
π
√R 2 − r 2
2 2πrR 2 − 3πr 3
= πrR +
3
√
3 R2 − r2

2 2πrR 2 − 3πr 3
= πrR +
3
√
3 R2 − r2

( ) ( ) √
( − 2r )

d 2V 2πR
√
3 R 2 − r 2 2πR 2 − 9πr 2 − 2πrR 2 − 3πr 3 ⋅
6 R2 − r2
= +
dr 2 3
(
9 R2 − r2 )
2 (
9 R2 − r2 )(2πR − 9πr ) + 2πr R + 3πr
2 2 2 2 4

= πR + 3
3
27 (R − r ) 2 2 2

dV 2 3πr 3 − 2πrR 2
Now, dr = 0 ⇒ 3 rR =
3 R2 − r2
√
3r 2 − 2R 2
⇒ 2R =
√ R2 − r2 √
⇒ 2R R 2 − r 2 = 3r 2 − 2R 2

( ) (
⇒ 4R 2 R 2 − r 2 = 3r 2 − 2R 2 ) 2

⇒ 4R 4 − 4R 2r 2 = 9r 4 + 4R 4 − 12r 2R 2
⇒ 9r 4 = 8R 2r 2
8
⇒ r2 = 9 R2
8 d 2V
When r 2 = 9 R 2, then <0
dr 2
8
2 2
By second derivative test, the volume of the cone is the maximum when r = 9 R

8 8 1 R 4
When r 2 = 9 R 2, h = R +
√ R2 − 9 R2 = R +
√ 9
R2 = R + 3 = 3 R

Therefore,

1
( )( )
= 3 π 9 R2
8 4
3
R

8
( )
= 27 3 πR 3
4

8
= 27 × ( Volume of the sphere)
8
Hence, the volume of the largest cone that can be inscribed in the sphere is 27 the volume of the sphere.

Page : 233 , Block Name : Exercise 6.5

Q24 Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the
radius of the base.

Answer. Let r and h be the radius and the height (altitude) of the cone respectively. Then, the volume (V) of the
cone is given as:
1 3V
V = 3π πr 2h ⇒ h =
r2
The surface area (S) of the cone is given by,
S = πrl (where l is the slant height)
√
= πr r 2 + h 2

9π 2 √
r 9 2r 6 + V 2

√
= πr r 2 + 2 4 =
π r πr 2

1
=
r √π 2r 6 + 9V 2
6π 2r 5
dS
r⋅
2π 2r 69
− √π 2r 6 + 9V 2
∴ =
dr r2
3π 2r 6 − π 2r 6 − 9V 2
=
√
r 2 π 2r 6 + 9V 2

2π 2r 6 − 9V 2
=
√
r 2 π 2r 6 + 9V 2

2π 2r 6 − 9V 2
=
√
r 2 π 2r 6 + 9V 2
dS 9V 2
Now, dr = 0 ⇒ 2π 2r 6 = 9V 2 ⇒ r 6 =
2π 2
9V 2 d 2S
Thus, it can be easily verified that when r 6 = , >0
2π 2 dr 2
9V 2
By second derivative test, the surface area of the cone is the least when r 6 =
2π 2

( )
1
9V 2 3V 3 2π 2r 6 2 3 √2πr 3
r6 = 2, h = 2 = 9
= 2 ⋅ 3
= √2r
2π πr πr 2 πr

Hence, for a given volume, the right circular cone of the least curved surface has an
Altitude equal to √2 times the radius of the base.

Page : 233 , Block Name : Exercise 6.5

Q25 Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan − 1√2 .

Answer. Let θ be the semi-vertical angle of the cone.

It is clear that θ ∈ 0, 2
[ ] π

Let r, h, and l be the radius, height, and the slant height of the cone respectively. The slant height of the cone is
given as constant.

Now, r = Isinθ and h = Icosθ
The volume (V) of the cone is given by,
1
V = 3 πr 2h

( )
1
= 3 π l 2sin 2θ (lcosθ)
1
= 3 πl 3sin 2θcosθ
dV l 3π
∴
dθ [sin θ( − sinθ) + cosθ(2sinθcosθ) ]
=
3
2

I π 3

3 [
= − sin + 2sinθcos θ ] 3 2

2
d V l π 3

3 [
= 2
− 3sin θcosθ + 2cos θ − 4sin θcosθ ]
2 3 2
dθ
l 3π
=
3 [2cos θ − 7sin θcosθ ]
3 2

dV
Now, dθ = 0

⇒ sin 3θ = 2sinθcos 2θ
⇒ tan 2θ = 2
⇒ tanθ = √2

⇒ θ = tan − 1√2
Now, when θ = tan − 1√2, then tan 2θ = 2 or sin 2θ = 2cos 2θ
Then, we have:

d 2V
dθ 2
l 3π
[
= 3 2cos 3θ − 14cos 3θ = − 4πI 3cos 3θ < 0 for θ ∈ 0, 2] [ ]
π

By second derivative test, the volume (V) is the maximum when θ = tan − 1√2 .

Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is tan − 1√2.

Page : 233 , Block Name : Exercise 6.5

()
1
Q26 Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin − 1 3 .

Answer. Let r be the radius, l be the slant height and h be the height of the cone of given surface area, S.
Also, let α be the semi-vertical angle of the cone.

Then S = πrl + πr 2
S − πr 2
⇒l= πr
…. (1)

Let V be the volume of the cone.
1
Then V = 3 πr 2h
1
⇒ V 2 = 9 π 2r 4h 2
1 2 4 2
=
9
π r l − r2 ( )[Asl = r + h ]
2 2 2

=
1 2 4
9
π r
[( ) ]
S − πr 2 2
πr
− r2

[ ( )
]
2
S − πr 2 − π 2r 4
1
= π 2r 4
9 π 2r 2

1 2 2
=
9 (
r S − 2Sπr 2 )
1
(
⇒ V 2 = Sr 2 S − 2πr 2 ……(2)
9 )
Differentiating (2) with respect to r, we get

( )
dV 1
2V dr = 9 S 2Sr − 8πr 3
dV
For maximum or minimum, put dr = 0

( )
1
⇒ 9 S 2Sr − 8πr 3 = 0

⇒ 2Sr − 8πr 3 = 0 ( As S ≠ 0)
⇒ S = 4πr 2( As r ≠ 0)
S
⇒ r 2 = 4π
Differentiating again with respect to r, we get

2V
d 2V
dr 2
+2
( )
dV 2
dr
=
1
9 (
S 2S − 24πr 2 )
⇒ 2V
d 2V
dr 2
=
1
9 (
S 2S − 24π ×
S
4π )( As
dV
dr
= 0 and r 2 =
S
4π )
1
= S(2S − 6S)
9
4 2
= −
S <0
9
Thus, V is maximum when S = 4πr 2
As S = πrl + πr 2
⇒ 4πr 2 = πrl + πr 2
⇒ 3πr 2 = πrl
⇒ I = 3r
Now, in ΔCOB
OB
sinα =
BC
r
=
l
r
=
3r
1
=
3

⇒ α = Sin − 1
() 1
3

Page : 233 , Block Name : Exercise 6.5

Q27 The point on the curve x 2 = 2y which is nearest to the point (0, 5) is
(A) (2√2, 4) (B) (2√2, 0) (C) (0, 0) (D) (2, 2)

Answer. The given curve is x 2 = 2y.

( )
x2
x, 2
x, the position of the point will be
2

The distance d(x) between the points x, 2
( ) x2
and (0, 5) is given by

d(x) = √ (x − 0) 2 +
( ) √
x2
2 −5
2
=
x4
x 2 + 4 + 25 − 5x 2 =
√
x4
4 − 4x
2
+ 25

′
( x − 8x )
3
( x − 8x )
3

∴ d (x) = =
x4
√x4 − 16x2 + 100
2
√ 4
− 4x 2 + 25

(
⇒ x x2 − 8 = 0 )
⇒ x = 0, ± 2√2
4x 3 − 32x
√ (
x 4 − 16x 2 + 100 3x 2 − 8 − x 3 − 8x ⋅ ) ( ) √ 2 x 4 − 16x 2 + 100
And , d ′(x) =
(x − 16x + 100 ) 4 2

(x − 16x + 100 )(3x − 8 ) − 2(x − 8x )(x − 8x )
4 2 2 3 3

= 3

(x − 16x + 100 ) 4 2 2

(x − 16x + 100 )(3x − 8 ) − 2(x − 8x )
4 2 2 3 2

= 3

(x − 16x + 100 ) 4 2 2

When, x = ± 2√2, d ′(x) > 0
By second derivative test, d(x) is the minimum at x = ± 2√2 .
( 2√ 2 ) 2
x = ± 2√2, y = 2 =4

Hence, the point on the curve x 2 = 2y which is nearest to the point (0, 5) is ( ± 2√2, 4)
The correct answer is A.

Page : 234 , Block Name : Exercise 6.5

1 − x + x2
Q28 For all real values of x, the minimum value of is
1 + x + x2
1
(A) 0 (B) 1 (C) 3 (D) 3

Answer.

(1 + x + x )( − 1 + 2x) − (1 − x + x )(1 + 2x)
2 2

∴ f ′(x) =
(1 + x + x ) 2 2

− 1 + 2x − x + 2x 2 − x 2 + 2x 3 − 1 − 2x + x + 2x 2 − x 2 − 2x 3
=
(1 + x + x ) 2 2

2 (x − 1 )
2
2
2x − 2
= =
(1 + x + x ) (1 + x + x )
2 2 2 2

∴ f ′(x) = 0 ⇒ x 2 = 1 ⇒ x = ± 1

Now, f ′′(x) =
2 [( ) 2
( ) (
1 + x + x 2 (2x) − x 2 − 1 (2) 1 + x + x 2 (1 + 2x) ) ]
(1 + x + x ) 2 4

4 (1 + x + x )[(1 + x + x )x − ( x − 1 )(1 + 2x) ]
2 2 2

=
(1 + x + x ) 2 4

4 [x + x + x − x − 2x + 1 + 2x ]
2 3 2 3

=
(1 + x + x ) 2 3

4 (1 + 3x − x ) 2

=
(1 + x + x ) 2 3

4(1+3−1) 4(3) 4
And, f ′′(1) = = = 9 >0
( 1 + 1 + 1 )3 ( 3 )3
4(1−3+1)
Also, f ′( − 1) = = 4( − 1) = − 4 < 0
( 1 − 1 + 1 )3
By second derivative test, f is the minimum at x = 1 and the minimum value is given
1−1+1 1
by f(1) = 1 + 1 + 1 = 3

The correct answer is D.

Page : 234 , Block Name : Exercise 6.5

Q29
1
The maximum value of [x(x − 1) + 1] 3 , 0 ≤ x ≤ 1 is
1

(A)
()1
3
3
(B) 2
1
(C) 1 (D) 0

Answer.

1
Let f(x) = [x(x − 1) + 1] 3
2x − 1
∴ f ′(x) = 2
3[x(x−1) +1]3
1
Now, f ′(x) = 0 ⇒ x = 2
1
Then, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 1] {i.e., at x = 0 and x
= 1}.
1
f(0) = [0(0 − 1) + 1] 3 = 1
1
f(1) = [1(1 − 1) + 1] 3 = 1
1 1

() [( ) ] ()
f 2
1
=
1
2
−1
2 +1
3
=
3
4
3

Hence, we can conclude that the maximum value of f in the interval [0, 1] is 1. The correct answer is C.

Page : 234 , Block Name : Exerciser 6.5

Q1 Using differentials, nd the approximate value of each of the following:
1

() 17 1
4
(a) 81 (b)(33) 5

Answer. (a) Consider
1 16 1
y = x 4 ⋅ Let x = 81 and Δx = 81

(a) Consider
1 1
Then, Δy = (x + Δx) 4 − x 4
1 1

= () ()
17
81
4
−
16
81
4

1

=
() 17
81
4
− 3
2

1

∴ () 17
81
4 2
= 3 + Δy

Now, dy is approximately equal to Δy and is given by,

dy =
() dy
dx Δx =
4(x)4
1
3 (Δx)

=
4 ( )
16
81
1
3
4
()1
81
27 1
= 4 × 8 × 81 = 32 × 3 = 96 = 0.010
1 1

1

Hence, the approximate value of 81 () 17 4 2
is 3 = 0.667 + 0.010

= 0.677.

1
(b) Consider y = x 5 . Let x = 32 and Δx = 1

1 1 1 1 1 1
Δy = (x + Δx) 5 − x 5 = (33) 5 − (32) 5 = (33) 5 − 2
1 1
∴ (33) 5 = 2 + Δy

Now, dy is approximately equal to Δy and is given by,

dy =
()
dy
dx
(Δx) =
−1
5 ( x )5
(Δx)

1 1
= − (1) = − 320 = − 0.003
5 ( 2 )6
1 1
Hence, the approximate value of ( 33 ) 5 is 2 + ( − 0.003 )
= 0.5 − 0.003 = 0.497

Page : 242 , Block Name : Miscellaneous Exercise

log x
Q2 Show that the function given by f(x) = x
has maximum at x = e.

Answer.
log x
The given function is f(x) = x
.

′
f (x) =
()
x
1
x
− log x

=
1 − log x
x2 x2

Now, f ′(x) = 0
⇒ 1 − logx = 0
⇒ logx = 1
⇒ logx = loge
⇒x=e

Now, f ′′(x) =
( )
x2 − x
1
− (1 − logx)(2x)

x4
− x − 2x(1 − logx)
=
x4
− 3 + 2logx
=
x3
− 3 + 2loge −3 + 2 −1
Now, f ′′(e) = 3
<0= 3
=
e e e3
Therefore, by second derivative test, f is the maximum at x = e.

Page : 242 , Block Name : Miscellaneous Exercise

Q3 The two equal sides of an isosceles triangle with xed 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 ?

Answer. Let ∆ABC be isosceles where BC is the base of xed length b.
Let the length of the two equal sides of ∆ABC be a.
Draw AD∴BC.

Now, in ∆ADC, by applying the Pythagoras theorem, we have:
b2
AD =
√ a2 − 4

b2

√
1
Area of triangle = (A) = 2 b a 2 − 4

The rate of change of the area with respect to time (t) is given by,
dA 1 2a da ab da
dt
= 2b ⋅ dt
=
b2
√ 4a 2 − b 2 dt
√
2 a2 − 4

It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second.
da
= − 3cm / s
dt
dA − 3ab
∴ =
dt 4a 2 − b 2 √
Then, when a = b, we have:
dA − 3b 2 − 3b 2
= = = − √3b
dt
√4b 2 − b 2 √3b 2
Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of is
√3bcm 2 / s.

Page : 242 , Block Name : Miscellaneous Exercise

Q4 Find the equation of the normal to curve y 2 = 4x at the point (1, 2).

Answer. The equation of the given curve is y 2 = 4x.
Differentiating with respect to x, we have:
dy
2y dx = 4
dy 4 2
⇒ dx = 2y = y
dy 2
∴ dx ] ( 1 , 2 ) = 2 = 1
−1 −1
Now, the slope of the normal at point (1, 2) is dy = 1 = −1
dx ] ( 1 , 2 )
∴Equation of the normal at (1, 2) is y − 2 = −1(x − 1).
∴y−2=−x+1
∴x+y−3=0

Page : 242 , Block Name : Miscellaneous Exercise

Q5 Show that the normal at any point θ to the curve
x = a cosθ + a θ sin θ, y = a sinθ – aθ cosθ
is at a constant distance from the origin.

Answer.
We have x = acosθ + aθsinθ
dx
∴ dθ = − asinθ + asinθ + aθcosθ = aθcosθ
y = asinθ − aθcosθ
dy
∴ dθ = acosθ − acosθ + aθsinθ = aθsinθ
dy dy dθ aθsin θ
∴ dx = dθ ⋅ dx = aθcos θ = tanθ
1
Slope of the normal at any point θ is − tan θ
The equation of the normal at a given point (x, y) is given by,
−1
y − asinθ + aθcosθ = tan θ (x − acosθ − aθsinθ)

⇒ ysinθ − asin 2θ + aθsinθcosθ = − xcosθ + acos 2θ + aθsinθcosθ

(
⇒ xcosθ + ysinθ − a sin 2θ + cosθ + acos 2θ + aθsinθcosθ

⇒ xcosθ + ysinθ − a = 0
Now, the perpendicular distance of the normal from the origin is
| −a| | −a|
= = | − a | , which is independent of θ.
√ cos 2 θ + sin 2 θ √1
Hence, the perpendicular distance of the normal from the origin is constant.

Page : 242 , Block Name : Miscellaneous Exercise

Q6 Find the intervals in which the function f given by
4sin x − 2x − xcos x
f(x) = 2 + cos x

is (i) increasing (ii) decreasing

Answer.
4sinx − 2x − xcosx
f(x) =
2 + cosx
(2 + cosx)(4cosx − 2 − cosx + xsinx) − (4sinx − 2x − xcosx)( − sinx)
∴ f ′(x) =
(2 + cosx) 2
(2 + cosx)(3cosx − 2 + xsinx) + sinx(4sinx − 2x − xcosx)
=
(2 + cosx) 2
6cosx − 4 + 2xsinx + 3cos 2x − 2cosx + xsinxcosx + 4sin 2x − 2xsinx − xsinxcosx
=
(2 + cosx) 2
4cosx − 4 + 3cos 2x + 4sin 2x
=
(2 + cosx) 2
4cosx − 4 + 3cos 2x + 4 − 4cos 2x
=
(2 + cosx) 2
4cosx − cos 2x cosx(4 − cosx)
= =
(2 + cosx) 2 (2 + cosx) 2

Now, f ′(x) = 0
⇒ cosx = 0 or cosx = 4
But, cosx ≠ 4
cosx = 0
π 3π
⇒ x = 2, 2
π 3π
Now, x = 2 and x = 2 divides (0, 2π) into three disjoint intervals i.e.,

( )( ) ( )
π π 3π
0, 2 , 2 , 2 , and
3π
2
, 2π
x 3π

( ) ( ) π 3π 0 < x < 2 and 2 < x < 2π
In intervals 0, 2 and 2
, 2π , f ′(x) > 0

Thus, f(x) is increasing for

In the interval
( )
π 3π
2
, 2 , f ′(x) < 0 π
2 < x < 2 .
3π

Thus, f(x) is decreasing for

Page : 242 , Block Name : Miscellaneous Exercise

1
Q7 Find the intervals in which the function f given by f(x) = x 3 + ,x ≠ 0
x3
(i) increasing (ii) decreasing.

Answer.
1
f(x) = x 3 +
x3
3 3x 6 − 3
∴ f ′(x) = 3x 2 − 3 =
x x4

Then, f ′(x) = 0 ⇒ 3x 6 − 3 = 0 ⇒ x 6 = 1 ⇒ x = ± 1
Now, the points x = 1 and x = − 1 divide the real line into three disjoint intervals
i.e., ( − ∞, − 1), ( − 1, 1), and (1, ∞) .
In intervals ( − ∞ , − 1 ) and ( 1 , ∞ ) i.e., when x < − 1 and x > 1, f ′(x) > 0
Thus, when x < − 1 and x > 1, f is increasing.
In interval ( − 1, 1) i.e., when − 1 < x < 1, f ′(x) < 0
Thus, when − 1 < x < 1, f is decreasing.

Page : 242 , Block Name : Miscellaneous Exercise

x2 y2
Q8 Find the maximum area of an isosceles triangle inscribed in the ellipse 2 + = 1.
a b2

Answer.

x2 y2
The given ellipse is 2 + = 1.
a b2
Let the major axis be along the x −axis. Let ABC be the triangle inscribed in the ellipse where vertex C is at (a, 0).
Since the ellipse is symmetrical with respect to the x−axis and y −axis, we can assume the coordinates of A to be
(−x1, y1) and the coordinates of B to be (−x1, −y1).
2 b
2
Now, we have y 1 = ± a a − x 1 √
Coordinates of A are

(x 1, − a
b
√ a2 − x 1
2
) and the coordinates of B

( b
√a − x
x 1, − a 2 2
1 )
As the point (x , y ) lies on the ellipse, the area of triangle ABC(A) is given by,
1 1

1
A = 2a a (√ 2b
a2 − x 1 +
2
) ( − x1) ( − a √a2 − x21 ) + ( − x1 ) ( − a √a2 − x21 )
b b

2 b 2
√
⇒ A = b a2 − x 1 + x1 a
√a − x 2
1

dA − 2x 1b 2bx 21
∴ dx = b −
1
√
2 a 2 − x 21 + a
√ a 2 − x 21 √
a2 a 2 − x 21

b
=
2 [ − x a + (a − x ) − x ]
1
2 2
1
2
1
√
2
a a − x1

(
b − 2x 21 − x 1a + a 2 )
=
2
√
a a2 − x 1
dA
Now, dx = 0
1

2
⇒ − 2x 1 − x 1a + a 2 = 0

a± √a 2 − 4 ( − 2 ) ( a 2 )
⇒ x1 = 2( −2)

a± √9a2
= −4
a
⇒ x 1 = − a, 2
But, x1 cannot be equal to a.

a2 √3b
√
a b ba
∴ x1 = 2 ⇒ y1 = a a 2 − 4 = 2a √3 = 2

{ }
( − 2x1 )
√ a2 − x 1
2
( − 4x1 − a ) − ( − 2x 21 − x1a + a2 ) 2 a − x 2
d 2A b √2 1
Now, 2 = a 2
dx 1 a2 − x 1

{
(a − x ) ( − 4x − a ) + x ( − 2x − x a + a )
}
2 2 2 2
b 1 1 1 1 1
= 3
a
(a − x ) 2 2
1
2

=
a
{
b 2x 3 − 3a 2x − a 3

2
a − x1
2 2
3

( )
a
}
Also, when x 1 = 2 , then

d 2A
2
dx 1
= a
{ }{ }
b
a3
2 8 − 3 2 − a3

( )3a 2
4
a3

3
2
= a
b
a3
4

( )3a 2
3
− 2 a3 − a3

4
3
2

= − a
b

{} 9
4

( )
3a 2
4
a3

3
2

Thus, the area is the maximum when x 1 = 2
<0

a

Maximum area of the triangle is given by,

A = b a2 −
√
a2
4
+
()√ a b
2 a
a2 −
a2
4

√3
() a b a√3
= ab + ×
2 2 a 2
ab√3 ab√3 3√3
= + = ab
2 4 4

Page : 242 , Block Name : Miscellaneous Exercise

Q9 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 m 3 . 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?

Answer. Let l, b, and h represent the length, breadth, and height of the tank respectively. Then, we have height (h)
= 2 m Volume of the tank = 8 m 3 Volume of the tank = l × b × h
∴8=l×b×2
4
⇒ lb = 4 ⇒ b = 1
Now, area of the base = lb = 4
Area of the 4 walls (A) = 2h (l + b)

∴A=4 l+ l ( ) 4

dA
⇒ dl = 4 1 − 2
( ) 4
l

dA
Now, dl = 0
4
⇒1− 2 =0
l
2
⇒I =4
⇒I= ±2
However, the length cannot be negative.
Therefore, we have l = 4.
4 4
∴b= l = 2 =2

d 2A 32
Now, =
dl 2 l3
d 2A 32
When l = 2, = 8 =4>0
dl 2
Thus, by second derivative test, the area is the minimum when l = 2. We have l = b = h = 2.
∴Cost of building the base = Rs 70 × (lb) = Rs 70 (4) = Rs 280
Cost of building the walls = Rs 2h (l + b) × 45 = Rs 90 (2) (2 + 2) = Rs 8 (90) = Rs 720
Required total cost = Rs (280 + 720) = Rs 1000
Hence, the total cost of the tank will be Rs 1000.

Page : 242 , Block Name : Miscellaneous Exercise

Q10 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.

Answer. Let r be the radius of the circle and a be the side of the square.
Then, we have:

2πr + 4a = k (where k is constant)
k − 2πr
⇒a= 4
The sum of the areas of the circle and the square (A) is given by,
( k − 2πr ) 2
A = πr 2 + a 2 = πr 2 + 16
dA 2 ( k − 2πr ) ( − 2π ) π ( k − 2πr )
∴ dr = 2πr + 16
= 2πr − 4
dA
Now, dr = 0
π ( k − 2πr )
⇒ 2πr = 4
8r = k − 2πr
⇒ (8 + 2π)r = k
k k
⇒ r = 8 + 2π = 2 ( 4 + π )
d 2A π2
Now, 2 = 2π + 2 > 0
dr
k d 2A
∴ When r = 2 ( 4π , >0
dr 2
k
The sum of the areas is least when r = 2 ( 4π )

k
k − 2π [ ] k
2 ( 4π )

r = 2 ( 4π ) , a = 4
k ( 4π ) π − k 4k k
= 44 ( π )
= 4 ( π ) 4 = π = 2r
Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the
circle.

Page : 242 , Block Name : Miscellaneous Exercise

Q11 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.

Answer. Let x and y be the length and breadth of the rectangular window.
x
Radius of the semicircular opening = 2

It is given that the perimeter of the window is 10 m.

πx
∴ x + 2y + 2 = 10

( )
⇒x 1+ 2
π
+ 2y = 10

⇒ 2y = 10 − x 1 + 2 ( ) π

⇒y=5−x 2 + 4
( ) 1π π

∴Area of the window (A) is given by,

A = xy + () π x 2
2 2

=x 5−x
[ ( )] 1π
2
+
1
4
π

( )
= 5x − x 2
1π
2
+
π π
4
+
x 2
8
x
π

( )
dA
∴ dx = 5 − 2x 2 + 4
1π π 4
x

( )
=5−x 1+ 2
π
+ 4x
π

∴
d 2A
dx 2 ( )
= − 1+ 2
dA
π π
+ 4 = −1− 4
π

Now, dx = 0

⇒5−x 1+ 2 ( ) π π
+ 4x = 0

π
⇒ 5 − x − 4x = 0

( )
⇒x 1+ 4
π
=5

5 20
⇒x= = π+4
( ) 1+ 4
π

20 d 2A
Thus, when x = π + 4 then <0
dx 2
20
Therefore, by second derivative test, the area is the maximum when length x = π + 4 m
Now,

y = 5 − π+4
20
( ) 2+π
4
=5−
5(2+π)
π+4
10
= π+4m

Hence, the required dimensions of the window to admit maximum light is given by
20 10
length = π + 4 m and breadth = π + 4 m.

Page : 243 , Block Name : Miscellaneous Exercise

Q12 A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the

( )
3
2 2 2
minimum length of the hypotenuse is a 3 + b 3 .

Answer. Let ∆ABC be right-angled at B. Let AB = x and BC = y.
Let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and BC
respectively.

We have,
AC = √x 2 + y 2
Now,
PC = bcscθ
And, AP = asecθ
AC = AP + PC
d ( AC )
∴ dθ
= − bcscθ + asecθ…(1)
d ( AC )
∴ dθ
=0
⇒ asecθtanθ = bcscθcotθ
a sin θ b cos θ
⇒ cos θ ⋅ cos θ = sin θ sin θ

⇒ asin 3θ = bcos 3θ
1 1
⇒ (a) 3 sinθ = (b) 3 cosθ
1

⇒ tanθ =
() b
a
3

1 1
(b)3 (a)3
∴ sinθ = and cosθ =
√ √a + b
2 2 2 2
a3 +b3 3 3

1

It can be clearly shown that
d 2 ( AC )
dθ 2
< 0 when tanθ = () b
a
3

Therefore, by second derivative test, the length of the hypotenuse is the maximum when
1

tanθ =
() b
a
3

1

tanθ = () b
a
3

2 2 2 2

√
b a3 +b3 √
a a3 +b3
AC = 1 + 1
b3 a3

= √
2
a3 + b3 b3 + a3
2

( 2 2

)
( )
3
2 2 2
= a3 + b3

( )
3
2 2 2
Hence, the maximum length of the hypotenuses is a 3 + b 3 .

Page : 243 , Block Name : Miscellaneous Exercise

Q13 Find the points at which the function f given by f (x) = f(x) = (x − 2) 4(x + 1) 3 has
(i) local maxima (ii) local minima (iii) point of in exion

Answer.
The given function is f(x) = (x − 2) 4(x + 1) 3
∴ f ′(x) = 4(x − 2) 3(x + 1) 3 + 3(x + 1) 2(x − 2) 4
= (x − 2) 3(x + 1) 2[4(x + 1) + 3(x − 2)]
= (x − 2) 3(x + 1) 2(7x − 2)
2
Now, f ′(x) = 0 ⇒ x = − 1 and x = or x = 2
7
2 2 2
Now, for values of x close to 7 and to the left of 7 , f ′(x) > 0 Also, for values of x close to and to 7 the right of
2
7
, f ′(x) > 0.
2
Thus, x = 7 is the point of local maxima.
Now, for values of x close to 2 and to the left of 2, f ′(x) < 0 Also, for values of x close to 2 and to the right of
2, f ′(x) < 0 .
Thus, x = 2 is the point of local minima.
Now, as the value of x varies through − 1, f ′(x) does not changes its sign .
Thus, x = -1 is the point of in exion.

Page : 243 , Block Name : Miscellaneous Exercise

Q14 Find the absolute maximum and minimum values of the function f given by
f(x) = cos 2x + sinx, x ∈ [0, π]

Answer.

f(x) = cos 2x + sinx
f ′(x) = 2cosx( − sinx) + cosx
= − 2sinxcosx + cosx
Now, f ′(x) = 0
⇒ 2sinxcosx = cosx ⇒ cosx(2sinx − 1) = 0
1
⇒ sinx = 2 or cosx = 0
π π
⇒ x = 6 , or 2 as x ∈ [0, π]

π π
Now, evaluating the value of f at critical points x = 2 and x = 6 and at the end points of the interval [0, π] (i.e., at x
= 0 and x = π), we have:

()
f 6
π π
= cos 2 6 + sin 6 =
π
()
√3 2
2
1 5
+ 2 = 4

f(0) = cos 20 + sin0 = 1 + 0 = 1
f(π) = cos 2π + sinπ = ( − 1) 2 + 0 = 1

()
f 2
π π π
= cos 2 2 + sin 2 = 0 + 1 = 1
5 π
Hence, the absolute maximum value of f is 4 occurring at x = 6 and the absolute minimum value of f is 1 occurring
π
at x = 0, 2 , and π.

Page : 243 , Block Name : Miscellaneous Exercise

Q15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of
4r
radius r is 3 .

Answer. A sphere of xed radius (r) is given. Let R and h be the radius and the height of the cone respectively.

The volume (V) of the cone is given by,
1
V = 3 πR 2h

Now, from the right triangle BCD, we have:
BC = √r 2 − R 2
h = r + √r 2 − R 2

∴∴ V =
1
3 ( √
πR 2 r + r2 − R2 =) 1
3
πR 2r +
1
3
πR 2 r 2 − R 2
√
dV 2 2π R2 ( − 2R)
∴
dR
=
3
πRr +
3 √
πR r 2 − R 2 +
3
⋅
√
2 r2 − R2

2 2π R3
=
3
πRr +
3 √
πR r 2 − R 2 −
√
3 r2 − R2

2 ( )
2πR r 2 − R 2 − πR 3
= πRr +
3 3 r2 − R2
√
2 2πRr 2 − 3πR 3
= πRr +
3
√
3 r2 − R2
dV
Now, =0
dR 2
2πrR 3πR 3 − 2πRr 2
⇒ 3
=
3 r2 − R2
√
√
⇒ 2r r 2 − R 2 = 3R 2 − 2r 2

( ) (
⇒ 4r 2 r 2 − R 2 = 3R 2 − 2r 2 ) 2

⇒ 9R 4 − 8r 2R 2 = 9R 4 + 4r 4 − 12R 2r 2
⇒ 9R 4 − 8r 2R 2 = 9
8r 2
⇒ R2 = 9
8r 2
⇒ R2 = 9

d 2V 2πr ( ) (
3πr 2 − R 2 2πr 2 − 9πR 2 − 2πRr 2 − 3πR 3 ( − 6R) )
Now, 2
= +
dR 3 3

2πr √ ( ) (
3 r 2 − R 2 2πr 2 − 9πR 2 + 2πRr 2 − 3πR 3 (3R) )
= +
3
9 (r − R )
2 2

8r 2 d 2V
Now, when R 2 = 9 , it can be shown that 2 < 0
dR
8r 2
The volume is the maximum when R 2 = 9
8r 2 8r 2 r2

√ √
r 4r
When R 2 = 9 , height of the cone = r + r2 − 9 = r + 9
=r+ 3 = 3 .

Hence, it can be seen that the altitude of the right circular cone of maximum volume
that can be inscribed in a sphere of radius r is 3 .

Page : 243 , Block Name : Miscellaneous Exercise

Q16 Let f be a function de ned on [a, b] such that f ′(x) > 0, for all x ∈ (a, b). Then prove that f is an increasing
function on (a, b).

Answer. Let
x 1, x 2∈(a,b)such that
X 1 < x 2. Consider the sub-interval [x 1, x 2]. Since f (x) is differentiable on (a, b) and
[x 1, x 2]⊂(a,b). Therefore, f(x) is continous on [x 1, x 2] and differentiable on
(x 1, x 2). By the Lagrange’s mean value theorm, there exists c∈(x 1, x 2)such that
f'(c)=f(x 2) − f(x 1)x 2 − x 1 …(1)Since f‘(x) > 0 for all
x∈(a,b), so in particular, f‘(c) > 0f'(c) > 0
⇒f(x 2) − f(x 1)x 2 − x 1 > 0 [Using (1)]
⇒f(x2)-f(x1) > 0 [∵]
X 2 − x 1 > 0 when X 1 < x 2]
⇒f(x 2) > f(x 1)⇒f(x 1) < f(x 2)Since
x1, x2are arbitrary points in
(a,b). Therefore, x 1, x 2
⇒f(x 1) < f(x 2) for all x 1, x 2 ∈(a, b)Hence, f (x) is increasing on (a,b).

Page : 243 , Block Name : Miscellaneous Exercise

2R
Q17 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is .
√3
Also nd the maximum volume

Answer. A sphere of xed radius (R) is given. Let r and h be the radius and the height of the cylinder respectively.

From the given gure, we have h = 2 R 2 − r 2 √
The volume (V) of the cylinder is given by,
√
V = πr 2h = 2πr 2 R 2 − r 2

dV 2πr 2( − 2r)
∴
dr √
= 4πr R 2 − r 2 +
√
2 R2 − r2

2πr 3
= 4πr √ R2 − r2 −
√R 2 − r 2
( )
4πr R 2 − r 2 − 2πr 3
=
√R 2 − r 2
4πrR 2 − 6πr 3
=
√R 2 − r 2
dV
Now, dr = 0 ⇒ 4πrR 2 − 6πr 3 = 0

2R 2
⇒ r2 = 3

) (4πR − 18πr ) − (4πrR − 6πr ) √
( − 2r )
2 2 2 3
d V √ 2 ( ) (
R 2 − r 2 4πR 2 − 18πr 2 − 4πrR 2 − 6πr 3 2 R2 − r2
Now, = 2
= 2 √
2 R2 − r2
dr dr
(R − r ) 2 2

(R − r )(4πR − 18πr ) + r (4πrR − 6πr )
2 2 2 2 2 3

= 3

(R − r ) 2 2 2

4πR 4 − 22πr 2R 2 + 12πr 4 + 4πr 2R 2
= 3

( R2 − r2 ) 2

2R 2 d 2V
2
Now, it can be observed that at r = 3 , dr 2 < 0
2 2R 2
The volume is the maximum when r = 3

√ √
2R 2 2R 2 R2 2R
r 2 = 3 , the height of the cylinder is R2 − 3 = 2 3 =
√3
2R
Hence, the volume of the cylinder is the maximum when the height of the cylinder is .
√3

Page : 243 , Block Name : Miscellaneous Exercise

Q18 Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h
4
and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder is 27 πh 3tan 2α.

Answer. The given right circular cone of xed height (h) and semi-vertical angle (α) can be drawn as:

Here, a cylinder of radius R and height H is inscribed in the cone.
Then, ∴GAO = α, OG = r, OA = h, OE = R, and CE = H.
We have,
r = htana
Now, since △AOG is similar to ΔCEG, we have:
AO CE
OG
= EG
h H
⇒ r = r−R [EG = OG − OE]
h h 1
⇒ H = r (r − R) = htan α (htanα − R) = tan α (htanα − R)

Now, the volume (V) of the cylinder is given by,
πR 2 πR 3
V = πR 2H = tan α (htanα − R) = πR 2h − tan α

dV 3πR 2
∴ dR = 2πRh − tan α

dV
Now, dR = 0
3πR 2
⇒ 2πRh = tan α

⇒ 2htanα = 3R
2h
⇒ R = 3 tanα
d 2V 6πR
Now, = 2πh −
dR 2 tanα
2h
R= tanα
3
d 2V
dR 2
= 2πh −
6π
tanα ( 2h
3 )
tanα = 2πh − 4πh = − 2πh < 0

∴By second derivative test, the volume of the cylinder is the greatest when
2h
R = 3 tanα.

2h 1
When R = 3 tanα, H = tan α htanα − 3 tanα = tan α
( 2h
) 1
( )
htan α
3
h
= 3

Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.
Now, the maximum volume of the cylinder can be obtained as:

( )( ) ( 4h 2
)( )
2h 2 h h 4
π 3 tanα 3
=π 9
tan 2α 3
= 27 πh 3tan 2α

Hence, the given result is proved.

Page : 243 , Block Name : Miscellaneous Exercise

Q19 A cylindrical tank of radius 10 m is being lled with wheat at the rate of 314 cubic metre per hour. Then the
depth of the wheat is increasing at the rate of
(A) 1 m/h (B) 0.1 m/h (C) 1.1 m/h (D) 0.5 m/h

Answer. Let r be the radius of the cylinder.
Then, volume (V) of the cylinder is given by
V = π( radius ) 2 × height
= π(10) 2h ( radius = 10m)
= 100πh
Differentiating with respect to time t, we have:
dV dh
dt
= 100π dt
The tank is being lled with wheat at the rate of 314 cubic metres per hour.
dV
dt
= 314m 3 / h
Thus, we have:
dh
314 = 100π dt
dh 314 314
⇒ dt = 100 ( 3.14 ) = 314 = 1
Hence, the depth of wheat is increasing at the rate of 1 m/h.
The correct answer is A.

Page : 243 , Block Name : Miscellaneous Exercise

Q20 The slope of the tangent to the curve x = t 2 + 3t − 8, y = 2t 2 − 2t − 5 at the point
(2, -1) is
22 6 7 −6
(A) 7 (B) 7 (C) 6 (D) 7

Answer.
The given curve is x = t 2 + 3t − 8 and y = 2t 2 − 2t − 5
dx dy
∴ dt = 2t + 3 and dt = 4t − 2
dy dy dt 4t − 2
∴ dx = dt ⋅ dx = 2t + 3

The given point is (2, − 1) .
At x = 2, we have:
t 2 + 3t − 8 = 2
⇒ t 2 + 3t − 10 = 0
⇒ (t − 2)(t + 5) = 0
⇒ t = 2 or t = − 5
At y = − 1, we have:
2t 2 − 2t − 5 = − 1
⇒ 2t 2 − 2t − 4 = 0

(
⇒ 2 t2 − t − 2 = 0 )
⇒ (t − 2)(t + 1) = 0
⇒ t = 2 or t = − 1
The common value of t is 2 .
Hence, the slope of the tangent to the given curve at point (2, -1) is
dy 4(2) −2 8−2 6
dx ] t = 2 = 2 ( 2 ) + 3 = 4 + 3 = 7
The correct answer is B.

Page : 243 , Block Name : Miscellaneous Exercise

Q21 The line y = mx + 1 is a tangent to the curve y 2 = 4x if the value of m is
1
(A) 1 (B) 2 (C) 3 (D) 2

Answer. The equation of the tangent to the given curve is y = mx + 1.
Now, substituting y = mx + 1 in y 2 = 4x, we get:
⇒ (mx + 1) 2 = 4x
⇒ m 2x 2 + 1 + 2mx − 4x = 0
⇒ m 2x 2 + x(2m − 4) + 1 = 0
Since a tangent touches the curve at one point, the roots of equation (i) must be equal.
Therefore, we have:
Discriminant = 0

( )
(2m − 4) 2 − 4 m 2 (1) = 0

⇒ 4m 2 + 16 − 16m − 4m 2 = 0
⇒ 16 − 16m = 0
⇒m=1
Hence, the required value of m is 1 .
The correct answer is A.

Page : 244 , Block Name : Miscellaneous Exercise

Q22
The normal at the point (1, 1) on the curve 2y + x 2 = 3 is
(A) x + y = 0 (B) x − y = 0
(C) x + y + 1 = 0 (D) x − y = 1

Answer. The equation of the given curve is 2y + x 2 = 3.
Differentiating with respect to x, we have:
2dy
dx
+ 2x = 0
dy
⇒ dx = − x
dy
∴ dx ] ( 1 , 1 ) = − 1
The slope of the normal to the given curve at point (1, 1) is
−1
dy =1
dx ] ( 1.1 )
Hence, the equation of the normal to the given curve at (1, 1) is given as:
⇒ y − 1 = 1(x − 1)
⇒y−1=x−1
⇒x−y=0
The correct answer is B.

Page : 244 , Block Name : Miscellaneous Exercise

Q23
The normal to the curve x 2 = 4y passing (1, 2) is
(B) x − y = 3
(A) x + y = 3
(D) x − y = 1
(C) x + y = 1

Answer. The equation of the given curve is x 2 = 4y.
Differentiating with respect to x, we have:
dy
2x = 4 ⋅ dx
dy x
⇒ dx = 2
The slope of the normal to the given curve at point (h, k) is given by,
−1 2
dy = − h
dx ] ( h , k )
∴Equation of the normal at point (h, k) is given as:
−2
y − k = h (x − h)
Now, it is given that the normal passes through the point (1, 2).
Therefore, we have:
−2 2
2 − k = h (1 − h) or k = 2 + h (1 − h)
Since (h, k) lies on the curve x 2 = 4y, we have h 2 = 4k
h2
⇒k= 4
From equation (i), we have:

h2 2
4
= 2 + h (1 − h)
h3
⇒ 4 = 2h + 2 − 2h = 2

⇒ h3 = 8
⇒h=2
h2
∴k= 4 ⇒k=1
Hence, the equation of the normal is given as:
−2
⇒ y − 1 = 2 (x − 2)

⇒ y − 1 = − (x − 2) .
⇒x+y=3
The correct answer is A .

Page : 244 , Block Name : Miscellaneous Exercise

Q24 The points on the curve 9y 2 = x 3 , where the normal to the curve makes equal intercepts with the axes are

( )
(A) 4, ± 3
8
(B) 4, 3
−8

( )
(C) 4, ± 8
3
(D)
( ) ± 4, 8
3

Answer. The equation of the given curve is 9y 2 = x 3
Differentiating with respect to x, we have:
dy
9(2y) dx = 3x 2
dy x2
⇒ dx = 6y

The slope of the normal to the given curve at point x 1, y 1 is ( )
−1 6y 1
dx ] ( x , y , ) = −
x1
2

The equation of the normal to the curve at x 1, y 1
is ( )
− 6y 1
y − y1 =
x1
2 (x − x 1 )
⇒ x 21 y − x 21 y 1 = − 6xy 1 + 6x 1y 1

⇒ 6xy 1 + x 21 y = 6xy 1 + x 21 y 1
6xy 1 x 21 y
⇒ 2 + 2 =1
6x 1y 1 + x 1 y 1 6x 1y 1 + x 1 y 1
x y
⇒ + =1
(
x1 6 + x1 ) (
y1 6 + x1 )
It is given that the normal makes equal intercepts with the axes. Therefore, We have:

(
x1 6 + x1 ) (
y1 6 + x1 )
∴ 6
= x1
x1 y1
⇒ 6 = x
1
2
⇒ x 1 = 6y 1

Also, the point x 1, y 1 ( ) lies on the curve, so we have
3
9y 2 = x 1
From (i) and (ii), we have:

9
()
x 21 2
6
3
x 41
3
= x1 ⇒ 4 = x1 ⇒ x1 = 4

From (ii), we have:
2
9y 1 = (4) 3 = 64

2 64
⇒ y1 = 9
8
⇒ y1 = ± 3 .

Hence, the required points are ( 8
4, ± 3 )
The correct answer is A.

Page : 244 , Block Name : Miscellaneous Exercise