2.

APPLICATIONS OF DERIVATIVES

Let us Study

• Applications of Drivatives to Tangents and Normals • Derivative as a rate measure

• Approximations
• Rolle's Theorem and Lagrange's Mean Value Theorem. • Increasing and Decreasing Functions
• Maxima and Minima

Let us Recall

• Continuous functions.
• Derivatives of Composite, Inverse Trigonometric, Logarithmic, Parametric functions.
• Relation between derivative and slope.
• Higher Order Derivatives.
2.1.1 Introduction :
In the previous chapter we have studied the derivatives of various functions such as composite
functions, Inverse Trigonometric functions, Logarithmic functions etc. and also the relation between
Derivative and slope of the tangent. In this chapter we are going to study various applications of
differentiation such as application to (i) Geometry, (ii) Rate measure (iii) Approximations (iv) Rolle's
Theorem and Lagrange's Mean Value Therorem (v) Increasing and Decreasing functions and (vi)
Maxima and Minima.

Let us Learn

2.1.2 Application of Derivative in Geometry :

In the previous chapter we have studied the relation between derivative and slope of a line or slope
of a tangent to the curve at a given point on it.
Let y = f (x) be a continuous function of x representing a curve in XY- plane and P (x1, y1) be any
point on the curve.
dy
Then = [ f ' (x)](x , y ) represents slope, also called gradient, of the tangent to the curve at
dx (x1, y1)
1 1

P (x1, y1). The normal is perpendicular to the tangent. Hence, the slope of the normal at P will be the negative
of reciprocal of the slope of tangent at P. Let m and m' be the slopes of tangent and normal respectively,

65
 dy 1 dy
then m = and m' = − dy
if ≠ 0.
dx (x1, y1) dx
dx (x1, y1)
(x1, y1)

dy
Equation of tangent at P (x1, y1) is given by y − y1 = m (x − x1) i.e. y − y1 = (x − x1)
dx (x1, y1)

and equation of normal at P (x1, y1) is given by

1
y − y1 = m' (x − x1) where m' = − dy
dx
(x1, y1)

SOLVED EXAMPLES

Ex. 1 : Find the equations of tangent and normal to the curve at the given point on it.
1
(i) y = 2x3 − x2 + 2 at ,2 (ii) x3 + 2x2 y − 9xy = -2 at (2, 1)
2
π
(iii) x = 2 sin3 θ, y = 3 cos3 θ at θ =
4

Solution :

(i) Given that : y = 2x3 − x2 + 2 (ii) Given that : x3 + 2x2 y − 9xy = -2

Differentiate w. r. t. x Differentiate w. r. t. x
dy d dy d dy d
= (2x3 − x2 + 2) = 6x2 − 2x 3x2 + 2 x2 + y (x2) − 9 x + y (x) = 0
dx dx dx dx dx dx
1 1 2 1
Slope of tangent at ,2 =m=6 −2 dy dy
2 2 2 3x2 + 2x2 + 4xy − 9x − 9y = 0
1 dx dx
∴ m = dy dy 9y − 4xy − 3x2
2 (2x2 − 9x) = 9y − 4xy − 3x2 ∴ =
1 dx dx 2x2 − 9x
Slope of normal at , 2 = m' = − 2 Slope of tangent at (2, 1)
2
Equation of tangent is given by dy 9(1) − 4(2)(1) − 3(4) 9 − 8 − 12
=m= =
1 1 2x − 1 dx (2, 1) 2(4) − 9(1) 8− 9
y−2= x− ⇒ 2y − 4 = −11
2 2 2 m = ∴ m = 11
−1
4y − 8 = 2x − 1 ⇒ 2x − 4y + 7 = 0 1
Slope of normal at (2, 1) = m' = −
Equation of normal is given by 11
1 Equation of tangent is given by
y−2=−2 x− ⇒ y − 2 = − 2x + 1 y − 1 = 11(x − 2) ⇒ 11x − y − 21 = 0
2
2x + y − 3 = 0 Equation of normal is given by
1
y−1=− (x − 2) ⇒ 11y − 11 = − x + 2
11
x + 11y − 13 = 0

66
(iii) Given that : y = 3 cos3 θ Now, x = 2 sin3 θ
Differentiate w. r. t. θ Differentiate w. r. t. θ
dy d d dx d d
=3 (cos θ)3 = 9 cos2 θ (cos θ) =2 (sin θ)3 = 6 sin2 θ (sin θ)
dθ dθ dθ dθ dθ dθ
dy dx
∴ = − 9 cos2 θ sin θ ∴ = 6 sin2 θ cos θ
dθ dθ
We know that
dy
dy dθ
9 cos2 θ sin θ 3
= dx
=− =− cot θ
dx 6 sin θ cos θ dθ
2
2
π
Slope of tangent at θ = is
4
3 dy π 3
cot =− π =m=−
2 dx 4 θ=
42
π 2
Slope of normal at θ = = m' =
4 3
π
When, θ =
4
3
π 1 1
x = 2 sin3 =2 =
4 √2 √2
3
π 1 3
y = 3 cos 3
=3 =
4 √2 2√ 2
1 3
∴ The point is P = ,
√ 2 2√ 2
Equation of tangent at P is given by
3 3 1 3 3x 3
y − =− x− ⇒y− =− +
2√ 2 2 √2 2√ 2 2 2√ 2
3x 3
+y− = 0 i.e. 3x + 2y − 3√ 2 = 0
2 √2
Equation of normal is given by

2 1 3 3 2x 2
y − x− ⇒y− = − =
2√ 2 3 √2 2√ 2 2 3√ 2
2x 2 3
−y− + =0
3 3√ 2 2√ 2
i.e. 4√ 2x − 6√ 2y + 5 = 0 . . . [ Multiply by 6√ 2 ]

67
Ex. 2 : Find points on the curve given by y = x3 − 6x2 + x + 3 where the tangents are parallel to the line
y = x + 5.
Solution : Equation of curve is y = x3 − 6x2 + x + 3
Differentiate w. r. t. x
dy d 3
= (x − 6x2 + x + 3) = 3x2 − 12x + 1
dx dx
Given that the tangent is parallel to y = x + 5 whose slope is 1.
dy
∴ Slope of tangent = = 1 ⇒ 3x2 − 12x + 1 = 1
dx
3x (x − 4) = 0 so, x = 0 or x = 4
When x = 0, y = (0)3 − 6(0)2 + (0) + 3 = 3
When x = 4, y = (4)3 − 6(4)2 + (4) + 3 = −25
So the required points on the curve are (0, 3) and (4, −25).

2.1.3 Derivative as a Rate measure :

If y = f (x) is the given function then a change in x from x1 to x2 is generally denoted by
δx = x2 − x1 and the corresponding change in y is denoted by δy = f (x2) − f (x1). The difference quotient
δy f (x2) − f (x1)
= is called the average rate of change with respect to x. This can also be interpreted
δx x2 − x1
geometrically as the slope of the secant line joining the points P ( x1, f (x1)) and Q ( x2, f (x2)) on the graph
of function y = f (x).
Consider the average rate of change over smaller and smaller intervals by letting x2 to approach x1 and
therefore letting δx to approach 0. The limit of these average rates of change is called the instantaneous
rate of change of y with respect to x at x = x1, which is interpreted as the slope of the tangent to the curve
y = f (x) at P ( x1, f (x1)). Therefore instantaneous rate of change is given by
δy f (x2) − f (x1)
lim = xlim
δx → 0
δx 2
→x1 x2 − x1
We recognize this limit as being the derivative of f (x) at x = x1, i.e. f ' (x1). We know that one
interpretation of the derivative f ' (a) is the instantaneous rate of change of y = f (x) with respect x when
x = a. The other interpretation is f (x) at f ' (a) is the slope of the tangent to y = f (x) at (a, f (a)).

SOLVED EXAMPLES
Ex. 1 : A stone is dropped in to a quiet lake and waves in the form of circles are generated, radius of the
circular wave increases at the rate of 5 cm/ sec. At the instant when the radius of the circular
wave is 8 cm, how fast the area enclosed is increasing ?
Solution : Let R be the radius and A be the area of the circular wave.

68
 ∴ A = π·R2
Differentiate w. r. t. t
dA d 2
=π (R )
dt dt
dA dR
= 2πR . . . (I)
dt dt
dR
Given that = 5 cm/sec.
dt
Thus when R = 8 cm, from (I) we get,
dA
= 2π(8) (5) = 80π
dt
Hence when the radius of the circular wave is 8 cm, the area of the circular wave is increasing at
the rate of 80π cm2/ sec.
Ex. 2 : The volume of the spherical ball is increasing at the rate of 4π cc/sec. Find the rate at which the
radius and the surface area are changing when the volume is 288π cc.
Solution : Let R be the radius, S be the surface area and V be the volume of the spherical ball.
4
V = πR3 . . . (I)
3
Differentiate w. r. t. t
dV 4π d
= · (R3)
dt 3 dt
4π dR dV
4π = ·3R2 . . . [Given = 4π cc/sec ]
3 dt dt
dR 1
= 2 . . . (II)
dt R
When volume is 288π cc.
4
i.e. π·R3 = 288π we get, R3 = 216 ⇒ R = 6 . . . [From (I)]
3
dR 1
From (II) we get, =
dt 36
1
So, the radius of the spherical ball is increasing at the rate of cc/sec.
36
Now, S = 4πR2
Differentiate w. r. t. t.
dS d dR
= 4π (R2) = 8πR
dt dt dt
So, when R = 6 cm
dS 1 4π
= 8π(6) =
dt R=6 36 3
4π
∴ Surface area is increasing at the rate of cm2/ sec.
3

69
Ex. 3 : Water is being poured at the rate of 36 m3/sec in to a cylindrical vessel of base radius 3 meters.
Find the rate at which water level is rising.
Solution : Let R be the radius of the base, H be the height and V be the volume of the cylindrical vessel
at any time t. R, V and H are functions of t.
V = πR2 H
V = π(3)2 H = 9π H . . . [ Given : R = 3]
Differentiate w. r. t. t
dV dH
= 9π
dt dt
dH 1 dV
= · . . . (I)
dt 9π dt
Given that,
dV
= 36 m3/sec . . . (II)
dt
dH 1 4
From (I) we get, = · (36) =
dt 9π π
4
∴ Water level is rising at the rate of meter/sec.
π
Ex. 4 : A man of height 180 cm is moving away from a lamp post at the rate of 1.2 meters per second.
If the height of the lamp post is 4.5 meters, find the rate at which (i) his shadow is lengthening.
(ii) the tip of the shadow is moving.
Solution : Let OA be the lamp post, MN be the man, MB = x be the length of shadow and OM = y be
the distance of the man from the lamp post at time t. Given that man is moving away from
the lamp post at the rate of 1.2 meter/sec. x and y are functions of t.
dy dx
Hence = 1.2. The rate at which shadow is lengthening = .
dt dt
B is the tip of the shadow and it is at a distance of (x + y) from the post.
x x+y
= i.e. 45x = 18x + 18y i.e. 27x = 18y
1.8 4.5
2y
∴ x=
3
Differentiate w. r. t. t
dx 2 dy 2
= × = × 1.2 = 0.8 meter/sec.
dt 3 dt 3
rate at which the tip of the shadow is moving is given by
d dx dy
(x + y) = +
dt dt dt
d
∴ (x + y) = 0.8 + 1.2 = 2 meter/sec.
dt
Shadow is lengthening at the rate of 0.8 meter/ sec. and its tip is moving at the rate of 2 meters/sec.

70
2.1.4 Velocity, Acceleration and Jerk :
If s = f (t ) is the desplacement function of a particle that moves along a straight line, then f ' (t ) is
the rate of change of the displacement s with respect to the time t. In other words, f ' (t ) is the velocity
of the particle. The speed of the particle is the absolute value of the velocity, that is, | f ' (t )|.
The rate of change of velocity with respect to time is valled the acceleration of the particle denoted
by a (t ). Thus the acceleration function is the derivative of the velocity function and is therefore the
second derivative of the position function s = f (t ).
dy d 2s
Thus, a = = i.e. a (t ) = v' (t ) = s'' (t ).
dt dt 2
Let us consider the third derivative of the position function s = f (t ) of an object that moves along a
straight line. s''' (t ) = v'' (t ) = a' (t ) is derivative of the acceleration function and is called the Jerk ( j ).
d a d 3s
Thus, j = = . Hence the jerk j is the rate of change of acceleration. It is aptly named because
dt dt 3
a jerk means a sudden change in acceleration, which causes an abrupt movement in a vehicle.

SOLVED EXAMPLES

Ex. 1 : A car is moving in such a way that the Ex. 2 : The displacement of a particle at time t
distance it covers, is given by the equation is given by s = 2t3 − 5t2 + 4t − 3. Find the
s = 4t2 + 3t where s is in meters and t is in time when the acceleration is 14 ft/ sec2, the
seconds. What would be the velocity and the velocity and the displacement at that time.
acceleration of the car at time t = 20 second ? Solution : Displacement of a particle is given by
Solution : Let v be the velocity and a be the s = 2t3 − 5t2 + 4t − 3 . . . (I)
acceleration of the car. Differentiate w. r. t. t.
Distance traveled by the car is given by ds d
Velocity, v = = (2t3 − 5t2 + 4t − 3)
s = 4t2 + 3t dt dt
Differentiate w. r. t. t. ∴ v = 6t2 − 10t + 4 . . . (II)
∴ Velocity of the car is given by dv d
Acceleration, a = = (6t2 − 10t + 4)
ds d dt dt
v= = (4t2 + 3t ) = 8t + 3 . . . (I) ∴ a = 12t − 10 . . . (III)
dt dt
and Acceleration of the car is given by Given : Acceleration = 14 ft/ sec . 2

d dv d ∴ 12t − 10 = 14 ⇒ 12t = 24 ⇒ t = 2
a= = (8t + 3 ) = 8 . . . (II)
dt dt dt So, the particle reaches an acceleration of
Put t = 20 in (I), 14 ft/ sec2 in 2 seconds.
∴ Velocity of the car, vt = 20 = 8(20) + 3 = 163 m/sec. Velocity, when t = 2 is
Put t = 20 in (II), ∴ vt = 2 = 6(2)2 − 10(2) + 4 = 8 ft/ sec.
∴ Acceleration of the car, at = 20 = 8 m/sec2. Displacement when t = 2 is
Note : In this problem, the acceleration is ∴ st = 2 = 2(2)3 − 5(2)2 + 4(2) − 3 = 1 foot.
independent of time. Such a motion is said Hence the velocity is 8 ft/ sec and the
to be uniformly accelerated motion. displacement is 1 foot after 2 seconds.

71
 EXERCISE 2.1

(1) Find the equations of tangents and normals (9) The surface area of a spherical balloon is
to the curve at the point on it. increasing at the rate of 2 cm2/ sec. At what
(i) y = x2 + 2e x + 2 at (0, 4) rate the volume of the balloon is increasing
(ii) x3 + y3 − 9xy = 0 at (2, 4) when radius of the balloon is 6 cm?
(iii) x2 − √ 3xy + 2y2 = 5 at (√ 3, 2) (10) If each side of an equilateral triangle
π increases at the rate of √ 2 cm/ sec, find the
(iv) 2xy + π sin y = 2π at 1,
2 rate of increase of its area when its side of
π π length 3 cm .
(v) x sin 2y = y cos 2x at ,
4 2 (11) The volume of a sphere increase at the rate
π
(vi) x = sin θ and y = cos 2θ at θ = of 20 cm3/ sec. Find the rate of change of its
6
1 surface area when its radius is 5 cm.
(vii) x = √ t , y = t − at t = 4.
√t (12) The edge of a cube is decreasing at the rate of
(2) Find the point on the curve y = √ x − 3 where 0.6 cm/sec. Find the rate at which its volume is
decreasing when the edge of the cube is 2 cm.
the tangent is perpendicular to the line
6x + 3y − 5 = 0. (13) A man of height 2 meters walks at a uniform
speed of 6 km/hr away from a lamp post of 6
(3) Find the points on the curve y = x3 − 2x2 − x
meters high. Find the rate at which the length
where the tangents are parallel to 3x − y + 1 = 0.
of the shadow is increasing.
(4) Find the equations of the tangents to the
(14) A man of height 1.5 meters walks toward a
curve x2 + y2 − 2x − 4y + 1= 0 which are
lamp post of height 4.5 meters, at the rate
parallel to the X-axis. 3
of meter/sec. Find the rate at which
4
(5) Find the equations of the normals to the
(i) his shadow is shortening. (ii) the tip of the
curve 3x2 − y2 = 8, which are parallel to the shadow is moving.
line x + 3y = 4.
(15) A ladder 10 meter long is leaning against a
(6) If the line y = 4x − 5 touches the curve vertical wall. If the bottom of the ladder is
y2 = ax3 + b at the point (2, 3) find a and b. pulled horizontally away from the wall at the
(7) A particle moves along the curve 6y = x3 + 2 rate of 1.2 meters per second, find how fast the
top of the ladder is sliding down the wall when
Find the points on the curve at which
the bottom is 6 meters away from the wall.
y-coordinate is changing 8 times as fast as
(16) If water is poured into an inverted hollow
the X-coordinate.
cone whose semi-vertical angel is 30°, so
(8) A spherical soap bubble is expanding so that its depth (measured along the axis)
that its radius is increasing at the rate of increases at the rate of 1 cm/ sec. Find the
0.02 cm/sec. At what rate is the surface rate at which the volume of water increasing
area is increasing, when its radius is 5 cm? when the depth is 2 cm.

72
2.2.1 Approximations
If f (x) is a differentiable function of x, then its derivative at x = a is given by
f (a + h) − f (a)
f ' (a) = hlim
→0 h
Here we use ≑ sign for approximation.
For a sufficiently small h we have,
f (a + h) − f (a)
f ' (a) ≑
h
i.e. h f ' (a) ≑ f (a + h) − f (a)
∴ f (a + h) ≑ f (a) + h f ' (a)
This is the formula to find the approximate value of the function at x = a + h, when f ' (a) exists.
Let us solve some problems by using this formula.

SOLVED EXAMPLES

Ex. 1 : Find the approximate value of √ 64.1. Ex. 2 : Find the approximate value of (3.98)3.
Solution : Solution :
Let f (x) = √ x . . . (I) Let f (x) = x3 . . . (I)
Differentiate w. r. t. x. Differentiate w. r. t. x.
1 f ' (x) = 3x2 . . . (II)
f ' (x) = . . . (II)
2√ x
Let a = 4, h = − 0.02
Let a = 64, h = 0.1
For x = a = 4, from (I) we get
For x = a = 64, from (I) we get
f (a) = f (4) = (4)3 = 64 . . . (III)
f (a) = f (64) = √ 64 = 8 . . . (III)
For x = a = 4, from (II) we get
For x = a = 64, from (II) we get
1 1 f ' (a) = f ' (4) = 3(4)2 = 48 . . . (IV)
f ' (a) = f ' (64) = =
2√ 64 16 We have, f (a + h) ≑ f (a) + h f ' (a)
∴ f ' (a) = 0.0625 . . . (IV)
f [4 + (− 0.02)] ≑ f (4) + (− 0.02)· f ' (4)
We have, f (a + h) ≑ f (a) + h f ' (a)
f (3.98) ≑ 64 + (− 0.02).(48) ...
f (64 + 0.1) ≑ f (64) + (0.1)· f ' (64) [From (III) and (IV)]

f (64.1) ≑ 8 + (0.1)·(0.0625) . . . f (3.98) ≑ 64 − 0.96

[From (III) and (IV)] ∴ f (3.98) = (3.98)3 ≑ 63.04

≑ 8 + 0.00625

∴ f (64.1) = √ 64.1 ≑ 8.00625

73
Ex. 3 : Find the approximate value of Ex. 4 : Find the approximate value of tan−1(0.99),
sin (30° 30' ). Given that 1° = 0.0175c Given that π ≑ 3.1416.
and cos 30° = 0.866. Solution : Let f (x) = tan−1 x . . . (I)
Solution : Let f (x) = sin x . . . (I) Differentiate w. r. t. x.
Differentiate w. r. t. x. 1
f ' (x) = . . . (II)
f ' (x) = cos x 1 + x2
1 ° Let a = 1, h = −0.01
Now, 30° 30' = 30° + 30' = 30° +
2 For x = a = 1, from (I) we get
π 0.1750 c
π
= + f (a) = f (1) = tan−1 (1) = . . . (III)
6 2 4
π For x = a = 1, from (II) we get
30° 30' = + 0.00875 . . . (II)
6 1
π f ' (a) = f ' (1) = = 0.5 . . . (IV)
Let a = , h = 0.00875 1 + 12
6
π We have, f (a + h) ≑ f (a) + h f ' (a)
For x = a = , from (I) we get
6 f [(1) + (−0.01)] ≑ f (1) + (−0.01)· f ' (1)
π π 1 π
f (a) = f = sin = = 0.5 . . . (III) f (0.99) ≑ − (0.01)·(0.5) . . . [From
6 6 2 4
π (III) and (IV)]
For x = a = , from (II) we get
6 π
≑ − 0.005
π π 4
f ' (a) = f ' = cos = 0.866 . . . (IV)
6 6 3.1416
≑ − 0.005
We have, f (a + h) ≑ f (a) + h f ' (a) 4
≑ 0.7854 − 0.005 = 0.7804
π π π
f + 0.00875c ≑ f + (0.00875)· f '
6 6 6 ∴ f (0.99) = tan−1 (0.99) ≑ 0.7804
f (30° 30' ) ≑ 0.5 + (0.00875)·(0.866) ...
. . . [From (III) and (IV)]
≑ 0.5 + 0.0075775
∴ f (30° 30' ) = sin (30° 30' ) ≑ 0.5075775

Ex. 5 : Find the approximate value of e1.005. Given that e = 2.7183.

Solution : Let f (x) = e x . . . (I) We have, f (a + h) ≑ f (a) + h f ' (a)

Differentiate w. r. t. x.
f (1 + 0.005) ≑ f (1) + (0.005)· f ' (1)
f ' (x) = e x . . . (II)
Let a = 1, h = 0.005 f (1.005) ≑ 2.7183 + (0.005) (2.7183) ...
For x = a = 1, from (I) we get . . . [From (III) and (IV)]

f (a) = f (1) = e1 = 2.7183 . . . (III) f (1.005) ≑ 2.7183 + 0.0135915

For x = a = 1, from (II) we get ≑ 2.7318915
f ' (a) = f ' (1) = e1 = 2.7183 . . . (IV) f (1.005) = e1.005 ≑ 2.73189

74
Ex. 6 : Find the approximate value of Ex. 7 : Find the approximate value of
log10 (998). Given that log10 e = 0.4343. f (x) = x3 + 5x2 − 2x + 3 at x = 1.98.
log x Solution : Let f (x) = x3 + 5x2 − 2x + 3 . . . (I)
Solution : Let f (x) = log10 x =
log 10 Differentiate w. r. t. x.
∴ f (x) = (log10 e)·log x . . . (I) f ' (x) = 3x2 + 10x − 2 . . . (II)
Differentiate w. r. t. x. Let a = 2, h = −0.02
log10 e 0.4343
f ' (x) = x = . . . (II) For x = a = 2, from (I) we get
x
Let a = 1000, h = −2 f (a) = f (2) = (2)3 + 5(2)2 − 2(2) + 3
For x = a = 1000, from (I) we get ∴ f (a) = 27 . . . (III)
f (a) = f (1000) = log10 1000 For x = a = 2, from (II) we get
∴ f (a) = 3log1010 = 3 . . . (III)
f ' (a) = f ' (2) = 3(2)2 + 10(2) − 2
For x = a = 1000, from (II) we get
∴ f ' (a) = 30 . . . (IV)
0.4343
f ' (a) = f ' (1000) = We have, f (a + h) ≑ f (a) + h f ' (a)
1000
∴ f ' (a) = 0.0004343 . . . (IV) f [(2) + (−0.02)] ≑ f (2) + (−0.02)· f ' (2)
We have, f (a + h) ≑ f (a) + h f ' (a) f (1.98) ≑ 27 − (0.02)·(30) . . . [From
f [1000 + (−2)] ≑ f (1000) + (−2) f ' (1000) (III) and (IV)]
f (998) ≑ 3 − (2) (0.0004343) . . . ≑ 27 − 0.6
[From (III) and (IV)]
f (1.98) ≑ 26.4
≑ 3 − 0.0008686
f (998) = log (998 ) ≑ 2.9991314

EXERCISE 2.2

(1) Find the approximate value of given (3) Find the approximate value of
functions, at required points. (i) tan−1 (0.999) (ii) cot−1 (0.999)
3 5
(i) √ 8.95 (ii) √ 28 (iii) √ 31.98 (iii) tan−1 (1.001)
(iv) (3.97)4 (v) (4.01)3 (4) Find the approximate value of
(i) e 0.995 (ii) e 2.1 given that e2 = 7.389
(2) Find the approximate value of
(iii) 3 2.01 given that log 3 = 1.0986
(i) sin (61°) given that 1° = 0.0175c,
(5) Find the approximate value of
√ 3 = 1.732
(i) loge (101) given that loge 10 = 2.3026
(ii) sin (29° 30' ) given that 1° = 0.0175c,
(ii) loge (9.01) given that log 3 = 1.0986
√ 3 = 1.732
(iii) log10 (1016) given that log10 e = 0.4343
(iii) cos (60° 30' ) given that 1° = 0.0175c,
(6) Find the approximate value of
√ 3 = 1.732
(i) f (x) = x3 − 3x + 5 at x = 1.99
(iv) tan (45° 40' ) given that 1° = 0.0175c.
(ii) f (x) = x3 + 5x2 − 7x + 10 at x = 1.12

75
2.3.1 Rolle's Theorem or Rolle's Lemma :
If a real-valued function f is continous on [a, b], differentiable on the open interval (a, b) and f (a)
= f (b), then there exists at least one c in the open interval (a, b) such that f ' (c) = 0.
Rolle's Theorem essentially states that any real-valued differentiable function that attains equal
values at two distinct points on it, must have at least one stationary point somewhere in between them,
that is, a point where the first derivative (the slope of the tangent line to the graph of the function) is zero.
Geometrical Significance :
Let f (x) be a real valued
function defined on [a, b]
and it is continuous on [a, b].
This means that we can
draw the graph f (x) between
the values x = a and x = b.
Also f (x) is differentiable
on (a, b) which means the Fig. 2.3.1
graph of f (x) has a tangent
at each point of (a, b). Now the existence of real number c ∈ (a, b) such that f ' (c) = 0 shows that
the tangent to the curve at x = c has slope zero, that is, tangent is parallel to X-axis since f (a) = f (b).

SOLVED EXAMPLES

Ex. 1 : Check whether conditions of Rolle's theorem are satisfied by the following functions.
3
(i) f (x) = 2x3 − 5x2 + 3x + 2, x ∈ 0, (ii) f (x) = x2 − 2x + 3, x ∈ [1, 4]
2
Solution :
(i) Given that f (x) = 2x3 − 5x2 + 3x + 2 . . . (I)
3 3
f (x) is a polynomial which is continuous on 0, and it is differentiable on 0, .
2 2
3
Let a = 0, and b = ,
2
For x = a = 0 from (I) we get,
f (a) = f (0) = 2 (0)3 − 5 (0)2 + 3 (0) + 2 = 2
3
For x = b = from (I) we get,
2
3 3 3 3 2 3 54 45 9
f (b) = f =2 −5 +3 +2 = − + +2
2 2 2 2 8 4 2
3 54 − 90 + 36
f (b) = f = +2=2
2 8
3
So, here f (a) = f (b) i.e. f (0) = f =2
2
Hence conditions of Rolle's Theorem are satified.

76
(ii) Given that f (x) = x2 − 2x + 3 . . . (I)
f (x) is a polynomial which is continuous on [1, 4] and it is differentiable on (1, 4).
Let a = 1, and b = 4
For x = a = 1 from (I) we get,
f (a) = f (1) = (1)2 − 2(1) + 3 = 2
For x = b = 4 from (I) we get,
f (b) = f (4) = (4)2 − 2(4) + 3 = 11
So, here f (a) ≠ f (b) i.e. f (1) ≠ f (4)
Hence conditions of Rolle's theorem are not satisfied.

Ex. 2 : Verify Rolle's theorem for the function Ex. 3 : Given an interval [a, b] that satisfies
f (x) = x2 − 4x + 10 on [0, 4]. hypothesis of Rolle's theorem for the
Solution : function f (x) = x3 − 2x2 + 3. It is known
that a = 0. Find the value of b.
Given that f (x) = x2 − 4x + 10 . . . (I)
Solution :
f (x) is a polynomial which is continuous on
Given that f (x) = x3 − 2x2 + 3 . . . (I)
[0, 4] and it is differentiable on (0, 4).
Let g (x) = x3 − 2x2 = x2 (x − 2)
Let a = 0, and b = 4
From (I), f (x) = g (x) + 3
For x = a = 0 from (I) we get,
We see that g (x) becomes zero for x = 0 and
f (a) = f (0) = (0)2 − 4(0) + 10 = 10
x = 2.
For x = b = 4 from (I) we get,
We observe that for x = 0,
f (b) = f (4) = (4)2 − 4(4) + 10 = 10
f (0) = g (0) + 3 = 3
So, here f (a) = f (b) i.e. f (0) = f (4) = 10
and for x = 2,
All the conditions of Rolle's theorem are
f (2) = g (2) + 3 = 3
satisfied.
∴ We can write that f (0) = f (2) = 3
To get the value of c, we should have
It is obvious that the function f (x) is
f ' (c) = 0 for some c ∈ (0, 4)
everywhere continuous and differentiable as
Differentiate (I) w. r. t. x.
a cubic polynomial. Consequently, it satisfies
f ' (x) = 2x − 4 = 2 (x − 4) all the conditions of Rolle's theorem on the
Now, for x = c, interval [0, 2].
f ' (c) = 0 ⇒ 2 (c − 2) = 0 ⇒ c = 2 So b = 2.
Also c = 2 ∈ (0, 4)
Thus Rolle's theorem is verified.

77
 π 5π
Ex. 4 : Verify Rolle's theorem for the function f (x) = ex (sin x − cos x) on , .
4 4
Solution : Given that, f (x) = ex (sin x − cos x) . . . (I)
We know that ex, sin x and cos x are continuous and differentiable on their domains. Therefore
π 5π π 5π
f (x) is continuous and differentiable on , and , respectively.
4 4 4 4
π 5π
Let a = , and b =
4 4
π
For x = a = from (I) we get,
4
π π
π π π 1 1
f (a) = f = e 4 sin − cos = e4 − =0
4 4 4 √2 √2
5π
For x = b = from (I) we get,
4
5π 5π
5π 5π 5π 1 1
f (a) = f =e 4 sin − cos =e4 − + =0
4 4 4 √2 √2
π 5π

∴ f (a) = f (b) i.e. f =f .
4 4
All the conditions of Rolle's theorem are satisfied.
π 5π
To get the value of c, we should have f ' (c) = 0 for some c ∈ , .
4 4
Differentiate (I) w. r. t. x.
f ' (x) = e x (cos x + sin x) + (sin x − cos x) e x = 2e x sin x
Now, for x = c, f ' (c) = 0 ⇒ 2e c sin c = 0. As e c ≠ 0 for any c ∈ R
sin c = 0 ⇒ c = 0, ± π, ± 2π, ± 3π, . . .
π 5π
It is clearly seen that π ∈ , ∴c=π
4 4
Thus Rolle's theorem is verified.

2.3.2 Lagrange's Mean Value Theorem (LMVT) :

If a real-valued function f is continous on a closed [a, b] and differentiable on the open interval
(a, b) then there exists at least one c in the open interval (a, b) such that
f (b) − f (a)
f ' (c) =
b−a
Lagrange's mean value theorem states, that for any real-valued diffenentiable function which is
continuous at the two end points, there is at least one point at which the tangent is parallel to the the
secant through its end points.

78
Geometrical Significance :
Draw the curve y = f (x) (see Figure 2.3.2) and take the end
points A (a, f (a)) and B (b, f (b)) on the curve, then
f (b) − f (a)
Slope of the chord AB =
b−a
Since by statement of Lagrange's Mean Value Theorem
f (b) − f (a)
f ' (c) =
b−a
Fig. 2.3.2
f ' (c) = Slope of the chord AB.
This shows that the tangent to the curve y = f (x) at the point x = c
is parallel to the chord AB.

SOLVED EXAMPLES

Ex. 1 : Verify Lagrange's mean value theorem Ex. 2 : Verify Lagrange's mean value theorem
1
for the function f (x) = √ x + 4 on the for the function f (x) = x + on the
x
interval [0, 5]. interval [1, 3].
1
Solution : Given that f (x) = √ x + 4 . . . (I) Solution : Given that f (x) = x + . . . (I)
x
The function f (x) is continuous on the The function f (x) is continuous on the
closed interval [0, 5] and differentiable closed interval [1, 3] and differentiable
on the open interval (0, 5), so the LMVT on the open interval (1, 3), so the LMVT
is applicable to the function. is applicable to the function.
Differentiate (I) w. r. t. x. Differentiate (I) w. r. t. x.
1
1 f ' (x) = 1 − 2 . . . (II)
f ' (x) = . . . (II) x
2√ x + 4 Let a = 1 and b = 3
Let a = 0 and b = 5 1
From (I), f (a) = f (1) = 1 + = 2
1
From (I), f (a) = f (0) = √ 0 + 4 = 2
1 10
f (b) = f (3) = 3 + =
f (b) = f (5) = √ 5 + 4 = 3 3 3
Let c ∈ (1, 3) such that
Let c ∈ (0, 5) such that
f (b) − f (a)
f (b) − f (a) f ' (c) =
f ' (c) = b−a
b−a 10
1 3−2 1 1 3 −2
= = 1 − 2 =
2√ c + 4 5−0 5 c 3−1
4
5 25 9 1 3 2
∴ √ c + 4 = ⇒ c + 4 = ∴ c = ∈ (0, 5) 1− 2 = =
2 4 4 c 2 3
Thus Lagrange's Mean Value Theorem ∴ c2 = 3 ⇒ c = ± √3
is verified.
∴ c = √ 3 ∈ (1, 3) and c = − √ 3 ∉ (1, 3)

79
 EXERCISE 2.3

(1) Check the validity of the Rolle's theorem for (4) If Rolle's theorem holds for the function
the following functions. f (x) = x3 + px2 + qx + 5, x ∈ [1, 3] with
(i) f (x) = x2 − 4x + 3, x ∈ [1, 3] 1
c=2+ , find the values of p and q.
(ii) f (x) = e −x sin x, x ∈ [0, π] √3
(iii) f (x) = 2x2 − 5x + 3, x ∈ [1, 3] (5) Rolle's theorem holds for the function
(iv) f (x) = sin x − cos x + 3, x ∈ [0, 2π] f (x) = (x − 2) log x, x ∈ [1, 2], show that the
(v) f (x) = x2 if 0 ≤ x ≤ 2 equation x log x = 2 − x is satisfied by at least
= 6 − x if 2 ≤ x ≤ 6 one value of x in (1, 2).
x
2 (6) The function f (x) = x (x + 3) e− 2 satisfies all
(vi) f (x) = x 3 , x ∈ [−1, 1]
the conditions of Rolle's theorem on [−3, 0].
(2) Given an interval [a, b] that satisfies
Find the value of c such that f ' (c) = 0.
hypothesis of Rolle's thorem for the function
(7) Verify Lagrange's mean value theorem for
f (x) = x4 + x2 − 2. It is known that a = − 1.
the following functions.
Find the value of b.
(i) f (x) = log x, on [1, e]
(3) Verify Rolle's theorem for the following
(ii) f (x) = (x − 1) (x − 2) (x − 3) on [0, 4]
functions.
11 13
(i) f (x) = sin x + cos x + 7, x ∈ [0, 2π] (iii) f (x) = x2 − 3x − 1, x ∈ − ,
7 7
x
(ii) f (x) = sin , x ∈ [0, 2π] (iv) f (x) = 2x − x2, x ∈ [0, 1]
2
(iii) f (x) = x2 − 5x + 9, x ∈ [1, 4] x−1
(v) f (x) = on [4, 5]
x−3
2.4.1 Increasing and decreasing functions :
Increasing functions :
Definition : A function f is said to be a monotonically (or strictly) increasing function on an interval
(a, b) if for any x1, x2 ∈ (a, b) with if x1 < x2 , we have f (x1) < f (x2).
Consider an increasing function y = f (x) in (a, b). Let h > 0 be a small increment in x then,
x < x + h [ x = x1 , x + h = x2 ]
f (x) < f (x + h) [ f (x1) < f (x2)]
∴ f (x + h) > f (x)
∴ f (x + h) − f (x) > 0
f (x + h) − f (x)
∴ > 0
h
f (x + h) − f (x)
∴ lim ≥ 0
h→0 h Fig. 2.4.1
∴ f ' (x) ≥ 0

80
 If f ' (a) > 0, then in a small δ-neighborhood of a i.e. (a − δ, a + δ), we have f strictly increasing if
f (a + h) − f (a)

>0 for |h| < δ
h
Hence if 0 < h < δ, f (a + h) − f (a) > 0 and f (a − h) − f (a) < 0
Thus for 0 < h < δ, f (a − h) < f (a) < f (a + h)
Decreasing functions :
Definition : A function f is said to be a monotonically (strictly) decreasing function on an interval (a, b)
if for any x1, x2 ∈ (a, b) with x1 < x2 , we have f (x1) > f (x2).
Consider a decreasing function y = f (x) in (a, b). Let h > 0 be a small increment in x then,
x + h > x [ x = x1 , x + h = x2 ]
f (x) > f (x + h) [ f (x1) > f (x2)]
∴ f (x + h) < f (x)
∴ f (x + h) − f (x) < 0
f (x + h) − f (x)
∴ < 0
h
f (x + h) − f (x)
∴ lim ≤ 0
h→0 h Fig. 2.4.2
∴ f ' (x) ≤ 0
If f ' (a) < 0, then in a small δ-neighborhood of a i.e. (a − δ, a + δ), we have f strictly decreasing
f (a + h) − f (a)
because <0 for |h| < δ
h
Hence for 0 < h < δ, f (a − h) > f (a) > f (a + h)
Note : Whenever f ' (x) = 0, at that point the tangent is parallel to X-axis, we cannot deduce that
whether f (x) is increasing or decreasing at that point.

SOLVED EXAMPLES

Ex. 1 : Show that the function f (x) = x3 + 10x + 7 Ex. 2 : Test whether the function
for x ∈ R is strictly increasing.
f (x) = x3 + 6x2 + 12x − 5 is increasing or
Solution : Given that f (x) = x3 + 10x + 7 decreasing for all x ∈ R.
Differentiate w. r. t. x. Solution : Given that f (x) = x3 + 6x2 + 12x − 5
f ' (x) = 3x2 + 10 Differentiate w. r. t. x.
Here, 3x2 ≥ 0 for all x ∈ R and 10 > 0. f ' (x) = 3x2 + 12x + 12 = 3(x2 + 4x + 4)

∴ 3x2 + 10 > 0 ⇒ f ' (x) > 0 f ' (x) = 3(x + 2)2
Thus f (x) is a strictly increasing function. 3(x + 2)2 is always positive for x ≠ −2

∴ f ' (x) ≥ 0 for all x ∈ R
Hence f (x) is an increasing function for all x ∈ R.

81
Ex. 3 : Find the values of x, for which the funciton f (x) = x3 + 12x2 + 36x + 6 is (i) monotonically
increasing. (ii) monotonically decreasing.
Solution : Given that f (x) = x3 + 12x2 + 36x + 6

Differentiate w. r. t. x.

f ' (x) = 3x2 + 24x + 36

= 3(x2 + 8x + 12)

f ' (x) = 3(x + 2) (x + 6)

(i) f (x) is monotonically increasing if f ' (x) > 0

i.e. 3(x + 2) (x + 6) > 0, (x + 2) (x + 6) > 0

then either (x + 2) < 0 and (x + 6) < 0 OR (x + 2) > 0 and (x + 6) > 0

Case (I) : x + 2 < 0 and x + 6 < 0

x < − 2 and x < − 6

Thus for every x < − 6, (x + 2) (x + 6) > 0, hence f is monotonically increasing.

Case (II) : x + 2 > 0 and x + 6 > 0

x > − 2 and x > − 6

Thus for every x > − 2, (x + 2) (x + 6) > 0 and f is monotonically increasing.

∴ From Case (I) and Case (II), f (x) is monotonically increasing if and only if x < − 6 or x > − 2.

Hence, x ∈ (∞,− 6) or x ∈ (− 2, ∞) ⇒ f is monotonically increasing.

(ii) f (x) is said to be monotonically decreasing if f ' (x) < 0

i.e. 3(x + 2) (x + 6) < 0, (x + 2) (x + 6) < 0

then either (x + 2) < 0 and (x + 6) > 0 OR (x + 2) > 0 and (x + 6) < 0

Case (I) : x + 2 < 0 and x + 6 > 0

x < − 2 and x > − 6

Thus for x ∈ (− 6, − 2), f is monotonically decreasing.

Case (II) : x + 2 > 0 and x + 6 < 0

x > − 2 and x < − 6

∴ This case does not arise. . . . [check. why ?]

82
2.4.2 Maxima and Minima :
Maxima of a function f (x) : A function f (x) is said to have a maxima at x = c if the value of the function
at x = c is greater than any other value of f (x) in a δ-neighborhood of c. That is for a small δ > 0 and for
x ∈ (c − δ, c + δ) we have f (c) > f (x). The value f (c) is called a Maxima of f (x). Thus the function f (x)
will have maxima at x = c if f (x) is increasing in c − δ < x < c and decreasing in c < x < c + δ.
Minima of a function f (x) : A function f (x) is said to have a minima at x = c if the value of the function
at x = c is less than any other value of f (x) in a δ-neighborhood of c. That is for a small δ > 0 and for
x ∈ (c − δ, c + δ) we have f (c) < f (x). The value f (c) is called a Minima of f (x). Thus the function f (x)
will have minima at x = c if f (x) is decreasing in c − δ < x < c and increasing in c < x < c + δ.
If f ' (c) = 0 then at x = c the function is neither increasing nor decrasing, such a point on the curve
is called turning point or stationary point of the function. Any point at which the tangent to the graph
dy
is horizontal is a turning point. We can locate the turn points by looking for points at which = 0.
dx
At these points if the function has Maxima or Minima then these are called extreme values of the
function.
Note : The maxima and the minima of a function are not necessarily the greatest and the least values
of the function in the whole domain. Actually these are the greatest and the least values of the
function in a small interval. Hence the maxima or the minima defined above are known as local
(or relative) maximum and the local (or relative) minimum of the function f (x).

To find the extreme values of the function let us use following tests.

2.4.3 First derivative test :

A function f (x) has a maxima at x = c if
(i) f ' (c) = 0
(ii) f ' (c − h) > 0 [ f (x) is increasing for values of x < c ]
(iii) f ' (c + h) < 0 [ f (x) is decreasing for values of x > c ]
where h is a small positive number.
A function f (x) has a minima at x = c if
(i) f ' (c) = 0
(ii) f ' (c − h) < 0 [ f (x) is decreasing for values of x < c ]
(iii) f ' (c + h) > 0 [ f (x) is increasing for values of x > c ]
where h is a small positive number.
Note : If f ' (c) = 0 and f ' (c − h) > 0, f ' (c + h) > 0 or f ' (c − h) <0, f ' (c + h) < 0 then f (x) in neither
maxima nor minima. In such a case x = c is called a point of inflexion. e.g. f (x) = x3 , f (x) = x5
in [−2, 2].

83
 SOLVED EXAMPLES

Ex. 1 : Find the local maxima or local minima of f (x) = x3 − 3x.

Solution : Given that f (x) = x3 − 3x . . . (I)
Differentiate (I) w. r. t. x.
f ' (x) = 3x2 − 3 = 3 (x2 − 1) . . . (II)
For extreme values, f ' (x) = 0
3x2 − 3 = 0 i.e. 3 (x2 − 1) = 0
i.e. x2 − 1 = 0 ⇒ 0 ⇒ x2 = 1 ⇒ x = ± 1
The turning points are x = 1 and x = −1
Let's consider the turning point, x = 1
Let x = 1 − h for a small, h > 0, from (II) we get,
f ' (1 − h) = 3 [(1 − h)2 − 1] = 3 (1 − 2h + h2 − 1) = 3h (h − 2)
∴ f ' (1 − h) < 0 . . . [ since, h > 0, h − 2 < 0 ]
∴ f ' (x) > 0 for x = 1 − h ⇒ f (x) is decreasing for, x > 1.
Now for x = 1 + h for a small, h > 0, from (II) we get,
f ' (1 + h) = 3 [(1 + h)2 − 1] = 3 (1 + 2h + h2 − 1) = 3 (h2 + 2h)
∴ f ' (1 + h) > 0 . . . [ since, h > 0, h2 + 2h > 0 ]
∴ f ' (x) < 0 for x = 1 + h ⇒ f (x) is increasing for, x < 1.
∴ f ' (x) < 0 for 1 − h < x < 1
∴ f ' (x) > 0 for 1 < x < 1 + h.
∴ x = 1 is the point of local minima.
Minima of f (x), is f (1) = 13 − 3 (1) = −2
Now, let's consider the turning point, x = −1
Let x = −1 − h for a small, h > 0, from (II) we get,
∴ f ' (−1 − h) = 3 [(−1 − h)2 − 1] = 3 (1 + 2h + h2 − 1) = 3 (h2 + 2h)
∴ f ' (−1 − h) > 0 . . . [ since, h > 0, h2 + 2h > 0 ]
∴ f ' (x) > 0 for x = −1 − h ⇒ f (x) is increasing for, x < −1.
Now for x = −1 + h for a small, h > 0, from (II) we get,
∴ f ' (−1 + h) = 3 [(− 1 + h)2 − 1] = 3 (1 − 2h + h2 − 1) = − 3h (2 − h)
∴ f ' (−1 + h) < 0 . . . [ since, h > 0, 2 − h > 0 ]
∴ f ' (x) < 0 for x = − 1 + h ⇒ f (x) is decreasing for, x > −1.
∴ f ' (x) > 0 for −1 − h < x < −1
∴ f ' (x) > 0 for −1 < x < −1 + h.
∴ x = − 1 is the point of local maxima.
Maxima of f (x), is f (−1) = (−1)3 − 3(−1) = −1 + 3 = 2
Hence, Maxima of f (x) is 2 and Minima of f (x) is −2

84
2.4.4 Second derivative test :
A function f (x) has a maxima at x = c if f ' (c) = 0 and f '' (c) < 0
A function f (x) has a minima at x = c if f ' (c) = 0 and f '' (c) > 0
Note : If f '' (c) = 0 then second derivative test fails so, you may try using first derivative test.

L2 L1 L3

L1 L3 L2

Fig. 2.4.4 (a) Fig. 2.4.4 (b)

Maxima at A : Consider the slopes of the tangents (See Fig 2.4.4a) Slope of L1 is +ve, slope of L2 = 0
and slope of L3 is −ve. Thus the slope is seen to be decreasing if there is a maximum at A.

Minima at A : Consider the slopes of the tangents (See Fig 2.4.4b) slope of L1 is −ve, slope of L2 = 0
and slope of L3 is +ve. Thus the slope is seen to be increasing if there is a minima at A.

SOLVED EXAMPLES

Ex. 1 : Find the local maximum and local minimum value of f (x) = x3 − 3x2 − 24x + 5.
Solution : Given that f (x) = x3 − 3x2 − 24x + 5 . . . (I)
Differentiate (I) w. r. t. x. For maximum of f (x), put x = −2 in (I)

f ' (x) = 3x2 − 6x − 24 . . . (II)
f (−2) = (−2)3 − 3(−2)2 − 24 (−2) + 5 = 33.
For extreme values, f ' (x) = 0 For x = 4, from (III) we get
3x2 − 6x − 24 i.e. 3 (x2 − 2x − 8) = 0 f '' (4) = 6(4) − 6 = 18 > 0
i.e. x2 − 2x − 8 = 0 i.e. (x + 2) (x − 4) = 0 ∴ At x = 4, f (x) has a minimum value.

⇒ x + 2 = 0 or x − 4 = 0 ⇒ x = −2 or x = 4 For minima of f (x), put x = 4 in (I)
The stationary points are x = −2 and x = 4. f (4) = (4)3 − 3 (4)2 − 24 (4) + 5 = −75
Differentiate (II) w. r. t. x. ∴ Local maximum of f (x) is 33 when x = −2

f '' (x) = 6x − 6 . . . (III) and
For x = −2, from (III) we get, Local minimum of f (x) is −75 when x = 4.

f '' (−2) = 6 (−2) − 6 = −18 < 0
∴ At x = −2, f (x) has a maximum value.

85
Ex. 2 : A wire of length 120 cm is bent in the form
of a rectangle. Find its dimensions if the
area of the rectangle is maximum.
Solution : Let x cm and y cm be the length and
the breadth of the rectangle. Perimeter of
rectangle = 120 cm.
∴ 2 (x + y) = 120 so, x + y = 60
∴ y = 60 − x . . . (I)
Let A be the area of the rectangle
∴ A = xy = x (60 − x) = 60x − x2. . . [From (I)] Fig. 2.4.5
Differentiate w. r. t. x. After leaving the margins, length of the
dA printing space is (x − 1) m and breadth of
= 60 − 2x . . . (II)
dx the printing space is ( y − 1.5) m.
dA
For maximum area =0 Let A be the area of the printing space
dx 24
i.e. 60 − 2x = 0 ⇒ x = 30 A = (x − 1) ( y − 1.5) = (x − 1) − 1.5
x
Differentiate (II) w. r. t. x. 24
= 24 − 1.5x − + 1.5 . . . [ From (I)]
d2A x
= − 2 . . . (III)
dx2 24
A = 25.5 − 1.5x − . . . (II)
For, x = 30 from (III) we get, x
d2A Differentiate w. r. t. x.
2 =−2<0 dA 24
dx x = 30 = − 1.5 + 2 . . . (III)
dx x
When, x = 30, Area of the rectangle is dA
For maximum printing space =0
maximum. dx
Put x = 30 in (I) we get y = 60 − 30 = 30 24
i.e. − 1.5x + 2 = 0 ⇒ 1.5x2 = 24 ⇒ x = ± 4, x ≠ −4
∴ Area of the rectangle is maximum if length x
∴ x = 4
= breadth = 30 cm. Differentiate (III) w. r. t. x.
d2A 48
Ex. 3 : A Rectangular sheet of paper has it area 24 = − 3 . . . (IV)
dx 2 x
sq. meters. The margin at the top and the For, x = 4, from (IV) we get,
bottom are 75 cm each and at the sides 50 cm d2A 48
2 =− 3<0
each. What are the dimensions of the paper, dx x = 4 (4)
if the area of the printed space is maximum ? When, x = 4 Area of the rectangular
Solution : Let x m and y m be the width and the printing space is maximum.
24
length of the rectangular sheet of paper Put x = 4 in (I) we get y = =6
4
respectively. Area of the paper = 24 sq. m. ∴ Area of the printing space is maximum
24
∴ xy = 24 ⇒ y = . . . (I) when width and length of sheet are 4 meter
x
and 6 meter respectively.

86
Ex. 4 : An open box is to be cut out of piece of Ex. 5 : Two sides of a triangle are given, find the
square card of side 18 cm by cutting of angle between them such that the area of
equal squares from the corners and turning the triangle is maximum.
up the sides. Find the maximum volume of Solution : Let ABC be a triangle. Let the given
the box. sides be AB = c and AC = b.
Solution : Let the side of each of the small squares
cut be x cm, so that each side of the box to
be made is (18 − 2x) cm. and height x cm.

Fig. 2.4.7

Let ∆ be the area of the triangle.

1
∆ = bc sin A . . . (I)
Fig. 2.4.6 2
Differentiate w. r. t. A.
Let V be the volume of the box. d∆ bc
V = Area of the base × Height = cos A . . . (II)
dA 2
= (18 − 2x)2 x = (324 − 72x + 4x2) x d∆
For maximum area =0
V = 4x3 − 72x2 + 324x . . . (I) dA
bc π
Differentiate w. r. t. x i.e. cos A = 0 ⇒ cos A = 0 ⇒ A =
2 2
dV
= 12x2 − 144x + 324 . . . (II) Differentiate (II) w. r. t. A.
dx
dV d2∆ bc
For maximum volume =0 2 = − sin A . . . (III)
dx dA 2
i.e. 12x2 − 144x + 324 = 0 ⇒ x2 − 12x + 27 = 0 π
For, A = from (III) we get,
(x − 3) (x − 9) = 0 ⇒ x − 3 = 0 or x − 9 = 0 2
∴ x = 3 or x = 9, but x ≠ 9 ∴ x = 3 d2∆ bc π -bc
2 π =− sin = <0
Differentiate (II) w. r. t. x dA A = 2 2 a
2
d2V π
= 24x − 144 . . . (III)
When, A = Area of the triangle is
dx2 2
For, x = 3 from (III) we get, maximum.
d2V Hence, the area of the triangle is maximum
2 = 24 (3) − 144 = − 72 < 0
dx x = 3 π
when the angle between the given sides .
Volume of the box is maximum when x = 3. 2
Maximum volume of the box π
Note : sin A is maximum (=1), when A =
= (18 − 6)2 (3) = 432 c.c. 2

87
Ex. 6 : The slant side of a right circular cone is l. Show that the semi-vertical angle of the cone of
maximum volume is tan−1 (√ 2 ).
Solution : Let x be the height of the cone and r be the radius of the base.
So, r2 = l2 − x2 . . . (I)
Let V be the volume of the cone.
1 π
V = πr2x = (l2 − x2) x
3 3
π
∴ V = (l2 x − x3)
3
Differentiate w. r. t. x
dV π
= (l2 − 3x2) . . . (II)
dx 3
dV
For maximum volume =0
dx
π l2
i.e. (l2 − 3x2) = 0 ⇒ x2 =
3 3
l l l l l
x=± ⇒x= or x = − is the stataionary point but, x ≠ − ∴ x=
√3 √3 √3 √3 √3
Differentiate (II) w. r. t. x
d2V
2 = − 2πx . . . (III)
dx
l
For, x = from (III) we get,
√3
d2V 2πl
2 l =− <0
dx x = √ 3 √3
l
Volume of the cone is maximum when height of the cone is x = .
√3
l l 2 l √2
Put x = in (I) we get, r = l2 − =
√3 √3 √3
Let α be the semi-vertical angle.
l √2
r
Then tan α = = √3
l
= √2
x √3

∴ α = tan−1 (√ 2 )

Ex. 7 : Find the height of a covered box of fixed volume so that the total surface area of the box is
minimum whose base is a rectangle with one side three times as long as the other.
Solution : Given that, box has a rectangular base with one side three times as long as other.
Let x and 3x be the sides of the rectangular base.
Let h be the height of the box and V be its volume.

88

V = (x) (3x) (h) = 3x2h . . . [Observe that V is constant]
Differentiate w. r. t. x.
dV dh d
= 3x2 +h (3x2)
dx dx dx
dh dh 2h
∴ 3x2 + 6xh = 0 ⇒ =− . . . (I)
dx dx x
Let S be the surface area of the box.
∴ S = (2 × 3x2) + (2 × 3xh) + (2 × xh) = 6x2 + 8xh
Differentiate w. r. t. x.
dS dh d
= 12x + 8 x +h (x)
dx dx dx
dS 2h
= 12x + 8 x − + h . . . [ from (I) ]
dx x
= 12x + 8(−2h + h)
dS
∴ = 12x − 8h . . . (II)
dx
For minimum surface area
dS 3x
= 0 ⇒ 12x − 8h = 0 ⇒ h =
dx 2
Differentiate (II) w. r. t. x.
d2S dh 2h 16h
2 = 12 − 8 = 12 − 8 − = 12 + . . . (III) . . . [ from (I) ]
dx dx x x
Both x and h are positive, from (III) we get,
d2S 16h
2 = 12 + >0
dx x
3
Surface area of the box is minimum if height = × shorter side of base.
2

EXERCISE 2.4

(1) Test whether the following functions are (ii) f (x) = 3 + 3x − 3x2 + x3
increasing or decreasing. (iii) f (x) = x3 − 6x2 − 36x + 7
(i) f (x) = x3 − 6x2 + 12x − 16, x ∈ R
(3) Find the values of x for which the following
(ii) f (x) = 2 − 3x + 3x2 − x3, x ∈ R
1 functions are strictly decreasing -
(iii) f (x) = x − , x ∈ R and x ≠ 0 (i) f (x) = 2x3 − 3x2 − 12x + 6
x
(2) Find the values of x for which the following 25
(ii) f (x) = x +
functions are strictly increasing - x
(i) f (x) = 2x3 − 3x2 − 12x + 6 (iii) f (x) = x − 9x2 + 24x + 12
3

89
(4) Find the values of x for which the function (14) Find the largest size of a rectangle that can be
f (x) = x3 − 12x2 − 144x + 13 inscribed in a semi circle of radius 1 unit, So
(a) Increasing (b) Decreasing that two vertices lie on the diameter.
(5) Find the values of x for which (15) An open cylindrical tank whose base is a
f (x) = 2x3 − 15x2 − 144x − 7 is circle is to be constructed of metal sheet so
(a) strictly increasing as to contain a volume of πa3 cu. cm of water.
(b) strictly decreasing Find the dimensions so that sheet required is
x minimum.
(6) Find the values of x for which f (x) = is
x2 + 1 (16) The perimeter of a triangle is 10 cm. If one of
(a) strictly increasing the side is 4 cm. What are the other two sides
(b) strictly decreasing of the triangle for its maximum area ?
1 (17) A box with a square base is to have an open
(7) Show that f (x) = 3x + increasing in
3x top. The surface area of the box is 192 sq.cm.
1 1 1
, 1 and decreasing in , . What should be its dimensions in order that
3 9 3
the volume is largest ?
(8) Show that f (x) = x − cos x is increasing for (18) The profit function P (x) of a firm, selling x
all x. items per day is given by
(9) Find the maximum and minimum of the P (x) = (150 − x)x − 1625. Find the number
following functions - of items the firm should manufacture to get
(i) y = 5x3 + 2x2 − 3x maximum profit. Find the maximum profit.
(ii) f (x) = 2x3 − 21x2 + 36x − 20 (19) Find two numbers whose sum is 15 and when
(iii) f (x) = x − 9x + 24x
3 2 the square of one number multiplied by the
16 cube of the other is maximum.
(iv) f (x) = x2 + 2
x (20) Show that among rectangles of given area,
log x the square has the least perimeter.
(v) f (x) = x log x (vi) f (x) =
x (21) Show that the height of a closed right circular
(10) Divide the number 30 in to two parts such cylinder, of a given volume and least surface
that their product is maximum. area, is equal to its diameter.
(11) Divide the number 20 in to two parts such (22) Find the volume of the largest cylinder that
that sum of their squares is minimum. can be inscribed in a sphere of radius 'r' cm.
(12) A wire of length 36 meter is bent in the form 2x
(23) Show that y = log (1 + x) − , x > −1 is
of a rectangle. Find its dimensions if the area 2+x
of the rectangle is maximum. an increasing function on its domain.
4 sin θ
(13) A ball is thrown in the air. Its height at any (24) Prove that y = − θ is an increasing
2 + cos θ
time t is given by h = 3 + 14t − 5t2. Find the π
maximum height it can reach. function of θ ∈ 0, .
2

90
 Let us Remember

֍ Equations of tangent and Normal at P (x1, y1) respectively are given by

dy
y − y1 = m (x − x1) where m =
dx (x1, y1)

1 dy
y − y1 = m' (x − x1) where m' = − dy
, if ≠0
dx
dx (x1, y1)
(x1, y1)

֍ Approximate value of the function f (x) at x = a + h is given by f (a + h) ≑ f (a) + h f ' (a)

֍ Rolle's theorem : If real-valued function f is continous on a closed [a, b], differentiable on the
open interval (a, b) and f (a) = f (b), then there exists at least one c in the open interval (a, b)
such that f ' (c) = 0.
֍ Lagrange's Mean Value Theorem (LMVT) : If a real-valued function f is continous on a
closed [a, b] and differentiable on the open interval (a, b) then there exists at least one c in the
f (b) − f (a)
open interval (a, b) such that f ' (c) =
b−a
֍ Increasing and decreasing functions :
(i) A function f is monotonically increasing if f ' (x) > 0.
(ii) A function f is monotonically decreasing if f ' (x) < 0.
(iii) A function f is increasing if f ' (x) ≥ 0.
(iv) A function f is decreasing if f ' (x) ≤ 0.
֍ (i) First Derivative test :
A function f (x) has a maxima at x = c if
(i) f ' (c) = 0
(ii) f ' (c − h) > 0 [ f (x) is increasing for values of x < c ]
(iii) f ' (c + h) < 0 [ f (x) is decreasing for values of x > c ]
where h is a small positive number.
A function f (x) has a minima at x = c if
(i) f ' (c) = 0
(ii) f ' (c − h) < 0 [ f (x) is decreasing for values of x < c ]
(iii) f ' (c + h) > 0 [ f (x) is increasing for values of x > c ]
where h is a small positive number.
(ii) Second Derivative test :
A function f (x) has a maxima at x = c if f ' (c) = 0 and f " (c) < 0.
A function f (x) has a minimum at x = c if f ' (c) = 0 and f " (c) > 0.

91
 MISCELLANEOUS EXERCISE 2

(I) Choose the correct option from the given alternatives :

(1)
If the function f (x) = ax3 + bx2 + 11x − 6 satisfies conditions of Rolle's theorem in [1, 3] and
1
f ' 2 + = 0, then values of a and b are respectively.
√3
(A) 1, −6 (B) −2, 1 (C) −1, −6 (D) −1, 6
x2 − 1
(2) If f (x) = x2 + 1 , for every real x, then the minimum value of f is -
(A) 1 (B) 0 (C) −1 (D) 2
(3) A ladder 5 m in length is resting against vertical wall. The bottom of the ladder is pulled along
the ground away from the wall at the rate of 1.5 m/ sec. The length of the higher point of
ladder when the foot of the ladder is 4.0 m away from the wall decreases at the rate of
(A) 1 (B) 2 (C) 2.5 (D) 3
(4) Let f (x) and g (x) be differentiable for 0 < x < 1 such f (0) = 0, g (0) = 0, f (1) = 6. Let there
exist a real number c in (0, 1) such that f ' (c) = 2g' (c), then the value of g (1) must be
(A) 1 (B) 3 (C) 2.5 (D) −1
(5) If f (x) = x3 − 6x2 + 9x + 18, then f (x) is strictly decreasing in -
(A) (−∞, 1) (B) [3, ∞) (C) (−∞, 1] U [3, ∞) (D) (1, 3)
(6)
If x = − 1 and x = 2 are the extreme points of y = α log x + βx2 + x then
1 1 1 1
(A) α = −6, β = (B) α = −6, β = − (C) α = 2, β = − (D) α = 2, β =
2 2 2 2
(7) The normal to the curve x2 + 2xy − 3y2 = 0 at (1, 1)
(A) Meets the curve again in second quadrant. (B) Does not meet the curve again.
(C) Meets the curve again in third quadrant. (D) Meets the curve again in fourth quadrant.
x
(8) The equation of the tangent to the curve y = 1 − e 2 at the point of intersection with Y-axis is
(A) x + 2y = 0 (B) 2x + y = 0 (C) x − y = 2 (D) x + y = 2
(9) If the tangent at (1, 1) on y2 = x (2 − x)2 meets the curve again at P then P is
9 3
(A) (4, 4) (B) (−1, 2) (C) (3, 6) (D) ,
4 8
(10) The appoximate value of tan (44° 30' ) given that 1° = 0.0175.
(A) 0.8952 (B) 0.9528 (C) 0.9285 (D) 0.9825

92
(II) (1)
If the curves ax2 + by2 = 1 and a' x2 + b' y2 = 1 intersect orthogonally, then prove that
1 1 1 1
− = − .
a b a' b'
(2) Determine the area of the triangle formed by the tangent to the graph of the function y = 3 − x2
drawn at the point (1, 2) and the cordinate axes.
(3)
Find the equation of the tangent and normal drawn to the curve y4 − 4x4 − 6xy = 0 at the
point M (1, 2).
(4) A water tank in the form of an inverted cone is being emptied at the rate of 2 cubic feet per
second. The height of the cone is 8 feet and the radius is 4 feet. Find the rate of change of the
water level when the depth is 6 feet.
(5) Find all points on the ellipse 9x2 + 16y2 = 400, at which the y-coordinate is decreasing and the
x-coordinate is increasing at the same rate.
2
(6) Verify Rolle's theorem for the function f (x) = on [−1, 1].
e + e− x
x

(7) The position of a particle is given by the function s(t ) = 2t2 + 3t − 4. Find the time t = c in the
interval 0 ≤ t ≤ 4 when the instantaneous velocity of the particle equals to its average velocity
in this interval.

(8) Find the approximate value of the function f (x) = √ x2 + 3x at x = 1.02.

2
(9) Find the approximate value of cos −1 (0.51) given π = 3.1416, = 1.1547.
√3
(10) Find the intervals on which the function y = xx, (x > 0) is increasing and decreasing.
x
(11) Find the intervals on the which the function f (x) = , is increasing and decreasing.
log x
(12) An open box with a square base is to be made out of a given quantity of sheet of area a2, Show
a3
the maximum volume of the box is .
6√ 3
(13) Show that of all rectangles inscribed in a given circle, the square has the maximum area.
(14) Show that a closed right circular cyclinder of given surface area has maximum volume if its
height equals the diameter of its base.
(15) A window is in the form of a rectangle surmounted by a semi-circle. If the perimeter be 30 m,
find the dimensions so that the greatest possible amount of light may be admitted.
(16) Show that the height of a right circular cylinder of greatest volume that can be inscribed in a
right circular cone is one-third of that of the cone.
(17) A wire of length l is cut in to two parts. One part is bent into a circle and the other into a
square. Show that the sum of the areas of the circle and the square is least, if the radius of the
circle is half the side of the square.

93
(18) A rectangular sheet of paper of fixed perimeter with the sides having their length in the ratio
8 : 15 converted in to an open rectangular box by folding after removing the squares of equal
area from all corners. If the total area of the removed squares is 100, the resulting box has
maximum valume. Find the lengths of the sides of rectangular sheet of paper.
(19) Show that the altitude of the right circular cone of maximum volume that can be inscribed in
4r
a shpere of radius r is .
3
(20) Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of
2R
radius R is . Also find the maximum volume.
√3
(21) Find the maximum and minimum values of the function f (x) = cos2 x + sin x.

v v v

94