CHAPTER: APPLICATION OF DERIVATIVES

SYLLABUS: Applications of derivatives: rate of change of quantities, increasing/decreasing

functions, maxima and minima (first derivative test motivated geometrically and second
derivative test given as a provable tool). Simple problems (that illustrate basic principles and
understanding of the subject as well as real-life situations).

Definations and Formulae:

Derivative as Rate of Change

𝑑𝑦
• Let 𝑦 = 𝑓(𝑥) be a function. Then 𝑑𝑥 denotes the rate of change of 𝑦 𝑤. 𝑟. 𝑡 𝑥.
𝑑𝑦 𝑑𝑦
• The value of at 𝑥 = 𝑥0 i.e (𝑑𝑥 ) i.e. represents the rate of change of 𝑦 𝑤. 𝑟. 𝑡 𝑥 𝑎𝑡
𝑑𝑥 𝑥=𝑥0
𝑥 = 𝑥0
• If two variables x and y are varying with respect to another variable t, i.e., if 𝑥 = 𝑓(𝑡) and
𝑑𝑦
𝑑𝑦𝑑𝑡 𝑑𝑥
y= g(t), then by Chain Rule 𝑑𝑥 = ( 𝑑𝑥 ), provided ≠0
𝑑𝑡
𝑑𝑡
𝑑𝑦
• is positive if 𝑦 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝑎𝑠 𝑥 increases and is negative if
𝑑𝑥
𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝑎𝑠 𝑥 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑠.

Increasing and Decreasing Functions

• A function y = f (x) is said to be increasing on an interval (a, b) if x 1 < x 2 in (a, b) f(x1) ≤

f(x2) for all 𝑥1 , 𝑥2 ∈ (𝑎, 𝑏)
Alternatively, a function y = f (x) is said to be increasing if f ′(x) ≥ 0 for each x in (a, b)
(a) strictly increasing on an interval (a, b) if x 1 < x 2 in (a, b) f(x1) < f(x2) for all x 1 , x 2
(a, b).
Alternatively, a function y = f (x) is said to be strictly increasing if f ′(x) > 0 for each x in
(a, b)
(b) decreasing on (a, b) if x 1 < x 2 in (a, b) f(x1) ≥ f(x 2 ) for all x 1 , x 2 (a, b). Alternatively,
a function y = f (x) ) is said to be decreasing if f ′(x) ≤ 0 for each x in (a, b)
(c) strictly decreasing on (a, b) if x 1 < x 2 in (a, b) f(x 1 ) > f(x 2 ) for all x 1 , x 2 (a, b).

Alternatively, a function y = f (x) is said to be strictly decreasing if f ′(x)<0 for each x in

(a, b)

(d) constant function in (a, b), if f (x) = c for all x (a, b), where c is a constant.
Alternatively , f(x) is a constant function if f ′ (x )= 0.

A point c in the domain of a function f at which either f ′(c) = 0 or f is not differentiable is

called a critical point of f.

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 Maxima and Minima

Definition: Let f be a function defined on an interval I. Then

1) f is said to have a maximum value in I, if there exists a point c in I such that f(c) > f(x), for all
x I. The number f(c) is called the maximum value of f in I and the point c is called a point of
maximum value of f in I.
2) f is said to have a minimum value in I, if there exists a point c in I such that f (c) < f (x), for all
x I. The number f (c), in this case, is called the minimum value of f in I and the point c, in
this case, is called a point of minimum value of f in I.
3) f is said to have an extreme value in I if there exists a point c in I such that f (c) is either a
maximum value or a minimum value of f in I. The number f (c), in this case, is called an extreme
value of f in I and the point c is called an extreme point.

Local Maxima and Local Minima

Definition: Let f be a real valued function and let c be an interior point in the domain of f.
Then
(a) c is called a point of local maxima if there is an h > 0 such that 𝑓 (𝑐) ≥ 𝑓 (𝑥), for all x in
(𝑐 – ℎ, 𝑐 + ℎ), 𝑥 ≠ 𝑐. The value f(c) is called the local maximum value of f.
(b) c is called a point of local minima if there is an h > 0 such that f (c) ≤ f (x), for all x in (c – h,
c + h). The value f(c) is called the local minimum value of f.

Geometrically, the above definition states that if x = c is a point of local maxima of f, then the
graph of f around c will be as shown in Fig.(a) below. Note that the function f is increasing
(i.e., f ′(x) > 0) in the interval (c – h, c) and decreasing (i.e., f ′(x) < 0) in the interval (c, c + h).
This suggests that f ′(c) must be zero,

Similarly ,if x = c is a point of local minima of f, then the graph of f around c will be as shown
in Fig.(b) above. Note that the function f is decreasing (i.e., f ′(x) < 0) in the interval (c – h,

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 c) and increasing (i.e., f ′(x) >0) in the interval (c, c + h). This again suggests that f ′(c) must
be zero,

Theorem: Let f be a function defined on an open interval I. Suppose c I be any point. If f has
a local maxima or a local minima at x = c, then either f ′(c) = 0 or f is not differentiable at c.

Definition: A point c in the domain of a function f at which either f ′(c) = 0 or f is not

differentiable is called a critical point of f.

Theorem: (First Derivative Test) Let f be a function defined on an open interval I. Let f be
continuous at a critical point c in I. Then
1) If f ′(x) changes sign from positive to negative as x increases through c, i.e.,
if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point
sufficiently close to and to the right of c, then c is a point of local maxima.
2) If f ′(x) changes sign from negative to positive as x increases through c, i.e., if f ′(x) < 0 at every
point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to
and to the right of c, then c is a point of local minima.
3) If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima
nor a point of local minima. In fact, such a point is called point of inflection.

Theorem: (Second Derivative Test) Let f be a function defined on an interval I and c I. Let
f be twice differentiable at c. Then
1) x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0 The value f (c) is local maximum
value of f.
2) x = c is a point of local minima if f ′(c) 0 = and f ″(c) > 0 In this case, f (c) is local minimum
value of f.
3) The test fails if f ′(c) = 0 and f ″(c) = 0. In this case, we go back to the first derivative test and
find whether c is a point of local maxima, local minima or a point of inflexion.
Working rule for finding absolute maximum value and/or absolute minimum value
Step 1: Find all critical points of f in the interval, i.e., find points x where either f ′(x) = 0 or f
is not differentiable.
Step 2: Take the end points of the interval.
Step 3: At all these points (listed in Step 1 and 2), calculate the values of f.
Step 4: Identify the maximum and minimum values of f out of the values calculated in Step 3.

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 This maximum value will be the absolute maximum value of f and the minimum value

will be the absolute minimum value of f.

MULTIPLE CHOICE QUESTIONS

S.NO QUESTIONS WITH SOLUTIONS

1 The edge of a cube is increasing at the rate of 0.3 cm/s, the rate of change of its
surface area when edge is 3 cm is
(a) 10.8cm (b) 10.8 cm2 (c) 10.8 cm2/s (d) 10.8 cm/s

dx
Solution: (c) as dt = 0.3 cm/s ,x is edge of a cube.
Surface area S = 6x 2
ds dx dS
Then dt = 12x ⋅ dt = 12x × 0.3 dt = 3.6x
dS
And dt at x = 3 is 3.6 × 3 = 10.8 cm2/s
2 The total revenue in ₹ received from the sale of x units of an article is given by
R(x) =3x2+36x+5. The marginal revenue when x=15 is (in ₹)
(a) 126 (b) 116 (c) 96 (d) 90

Solution: (a), as Rʹ (x)=6x+36

Rʹ(15)=90+36=126
3 The point on the curve 𝑦 = 𝑥 2 where the rate of change of 𝑥 −coordinate is equal
to the rate of change of 𝑦 −coordinate is
1 1 1 1
(a) 2 (b) 4 (c) (2 , 4) (d) (1,1)

𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥
Sol: (C), as y=x2⇒ 𝑑𝑡 = 2𝑥 𝑑𝑡 ; given 𝑑𝑡 = 𝑑𝑡
1
1 = 2𝑥 ⇒ 𝑥 =
2
1 1
Substituting in the equation of curve, we get point as (2 , 4)

4 The interval on which the function f(x)=2x3+9x2+12x-1 is decreasing ,is

(a) (−1, ∞) (b) (−2, −1)
(c) (−∞, −2) (d) [−1,1]

Sol: (b) We have, f(x)=2x3+9x2+12x-1

fʹ(x)=6x2+18x+12=6(x2+3x+2)= 6(𝑥 + 2)(𝑥 + 1)
for f(x) to be decreasing, we must have
𝑓ʹ(𝑥) < 0 ⟹ 6(𝑥 + 2)(𝑥 + 1) ≤ 0
⟹ (𝑥 + 2)(𝑥 + 1) ≤ 0
−2 ≤ 𝑥 ≤ −1.
Hence, f(x) is decreasing on (−2, −1).

5 If at 𝑥 = 1, the function 𝑓(𝑥) = 𝑥 4 − 62𝑥 2 + 𝑎𝑥 + 9 attains its maximum

value on the interval [0, 2]. Then the value of 𝑎 is
(a) 124 b) −124 c) 120 d) −120

Sol: (c)
As fʹ(x) = 4x3-124x+a,

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 Given x=1 is point of maximum
⇒ fʹ(1)=0 ⇒ 4 – 124 + a = 0 ⇒ a = 120
6 The function f(x) = 4 sin3 x – 6 sin2 x + 12 sinx + 100 is strictly
3π π
(a) increasing in (π, 2 ) b) decreasing in ( 2 , π)
π π π
c) decreasing in(− 2 , 2 ) d) decreasing in (0, 2 )

Sol: (b) we have, f(x) = 4 sin3 x – 6 sin2 + 12 sin +100

1 2 3
⇒ fʹ(x)= 12(sin2x-sin x + 1) cos x = {(sin 𝑥 − 2) + 4} cos x
⇒ Sign of fʹ(x) is same as that of cos x
π π
⇒ fʹ(x) < 0 on (0, 2 ) ⇒ f(x) is decreasing on ( 2 , π)

7 Which of the following functions is decreasing in (0, π/2).

(a) 𝑠𝑖𝑛 2𝑥 b) 𝑡𝑎𝑛 𝑥 c) 𝑐𝑜𝑠 𝑥 d) 𝑐𝑜𝑠 3𝑥
𝑑
Sol :(c) We find that 𝑑𝑥 (𝑐𝑜𝑠 𝑥) = −𝑠𝑖𝑛 𝑥 < 0 for all
𝑥 ∈ (0, 𝜋/2)
So, 𝑐𝑜𝑠 𝑥 𝑖𝑠 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑜𝑛 (0, 𝜋/2)

8 The function f(x)=2x3-3x2-12x+4 has

(a) two points of local maximum
(b) two points of local minimum
(c) one maximum and one minimum
(d)no maximum, no minimum

Sol: (c) We have, f(x)=2x3-3x2-12x+4

⇒ fʹ(x)=6x2+18x+12 and fʹʹ (x)=12x-6
At points of local maximum or minimum, we have
⇒ fʹ(x)=0 ⇒ 6(x2-x-2)=0 ⇒ (x-2) (x+1) = 0 ⇒ x = -1, 2
At x = -1, we obtain : fʹʹ (-1) = -18<0. So, x = -1 is a point of local maximum.
At x = 2, we obtain : fʹʹ (2) = 24 - 6 = 18>0. So, x = 2 is a point of local
minimum.
9 The interval on which the function f(x)=x3+6x2+6 is strictly increasing is
(a) (-∞,-4) ∪ (0, ∞) (b) (-∞,-4)
(c) (-4,0) (d) (-∞,0) ∪ (4, ∞)

Sol: (a) We have, f(x) = x3+6x2+6

⇒ fʹ(x) =3x2+12x=3x(x+4)
For f(x) to be increasing, we must have
fʹ(x)>0 ⇒ 3x(x+4)>0 ⇒ x(x+4) > 0 ⇒ x<-4 or, x>0
Hence, f(x) is increasing on (-∞,-4) ∪ (0, ∞).
10 The rate of change of the area of a circle with respect to its radius 𝑟 𝑎𝑡 𝑟 = 6𝑐𝑚
is:
(a) 10π cm2/cm (b) 12π cm2/cm
(c) 8π cm2/cm (d)11π cm2/cm

Sol: (b) Area of circle (𝐴) = 𝜋𝑟 2

𝑑𝐴
⇒ 𝑑𝑟 = 2 𝜋r
𝑑𝐴
⇒ 𝑑𝑟 ]r = 6
= 2π x 6 = 12π

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 11 The total revenue in Rupees received from the sale of x units of a product is
given by R(x) = 3x2+36x+5. The marginal revenue, when 𝑥 = 15 is:
(a) 116 b) 96 c) 90 d) 126

Sol: (d) Total revenue R(x) = 3x2+36x+5

𝑑
Marginal revenue = 𝑑𝑥 𝑅(𝑥) = 6𝑥 + 36 = 6 × 15 + 36 =126

12 The maximum value of the function 𝑓(𝑥) = 5 + 𝑠𝑖𝑛2𝑥 is

(a) 1 b) 6 c) 4 d) −1

Sol: (b)
−1 ≤ 𝑠𝑖𝑛 2𝑥 ≤ 1
⇒ 5 − 1 ≤ 5 + 𝑠𝑖𝑛 2𝑥 ≤ 5 + 1
⇒ 4 ≤ 𝑓(𝑥) ≤ 6
Maximum value of 𝑓(𝑥) = 6

13 The function 𝑓(𝑥) = 𝑥 − 𝑠𝑖𝑛 𝑥 decreases for

(a) 𝑎𝑙𝑙 𝑥 (b) 𝑥 < 𝜋/2
(c) 0 < 𝑥 < 𝜋/4 (d) 𝑛𝑜 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥

Sol: (d)
We have, 𝑓(𝑥) = 𝑥 − 𝑠𝑖𝑛 𝑥
⇒ fʹ(x) = 1 – cos 𝑥 ≥ 0 for all 𝑥 , since −1 ≤ 𝑐𝑜𝑠𝑥 ≤ 1

⇒ 𝑓(𝑥) is increasing for all 𝑥 ∈ 𝑅 ⇒ 𝑓(𝑥) decreases for no value of 𝑥.

14 The absolute maximum value of f(x) = x3 - 3x + 2 in 0 ≤ 𝑥 ≤ 2 is
(a) 4 b) 6 c) 2 d) 0

Sol: (a)
As fʹ(x) = 3x2-3, fʹ(x) = 0 ⇒ x = ±1.
f(0) = 2, f(1) =1 – 3 + 2 =0,
f(-1) = -1 + 3 + 2 = 4, f(2) = 8 – 6 + 2 = 4

CHAPTER VIDEO LINK FOR MCQs SCAN QR CODE FOR

VIDEO

APPLICATION OF DERIVATIVES https://youtu.be/0Zk2RUMBcWU

108
 EXERCISE

1 For the function y = x3+21, the value of x, when y increases 75 times as fast as x,
is
(a) ±3 (b) ±5√3 c) ±5 d) none of these
Answer: c
2 1 𝑥
The maximum value of (𝑥) is
1
1
𝑒 1 𝑒
(a) 𝑒 (b) 𝑒 c) 𝑒 𝑒 d) (𝑒 )
Answer: c
3 The function f(x) = cos x -2px is monotonically decreasing for
1 1
(a) p < 2 b) p > 2 c) p < 2 d) p > 2
Answer: b
4 The maximum value of xy, subject to x+y = 8 is
(a) 8 b) 16 c) 20 d) 24
Answer: b
5 2
A particle moves along the curve 𝑦 = 3 x 3 + 1. The x-coordinates of the points on
the
curve at which y-coordinate is changing twice as fast as x-coordinate is
5 1
(a) 1 b) ±1 c) 3 d) 3
Answer: b

ASSERTION AND REASONING QUESTIONS

In the following questions, a statement of assertion (A) is followed by a statement of

reason(R). Choose the correct answer out of the following choices.(a) Both (A) and (R)
are true and (R) is the correct explanation of (A)
(b) Both (A) and (R) are true and (R) is not the correct explanation of (A)
(c) (A) is true but (R) is false
(d) (A) is false but (R) is true

1 𝑑
Let f(x) be a polynomial function in a degree 6 such that 𝑑𝑥(f(x)) = (𝑥 − 1)3(𝑥 −
3)2 , then
Assertion (A): 𝑓(𝑥) ℎ𝑎𝑠 𝑎 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑎𝑡 𝑥 = 1.
𝑑 𝑑
Reason (R): When 𝑑𝑥(f(x))<0, ∀ 𝑥 ∈ (𝑎 − ℎ, 𝑎) and 𝑑𝑥 (f(x))>0, ∀ 𝑥 ∈ (𝑎, 𝑎 +
ℎ); where ‘h’ is an infinitesimally small positive quantity, then f(x) has a
minimum at x=a, provided f(x) is continuous at x=a.
Sol : (a)
𝑑
(f(x)) = (𝑥 − 1)3(𝑥 − 3)2
𝑑𝑥
Assertion : f(x) has a minimum at x=1 is true as
𝑑 𝑑
(f(x))<0, ∀ 𝑥 ∈ (1 − ℎ, 1) and (f(x))>0, ∀ 𝑥 ∈ (1,1 + ℎ); where, ‘h’ is an
𝑑𝑥 𝑑𝑥
infinitesimally small positive quantity, which is in accordance with the Reason
statement.
2 Let C be the circumference and A be the area of a circle.
Assertion(A): The rate of change of the area with respect to radius is equal to C.
𝐶
Reason(R): The rate of change of the area with respect to diameter is 2.
Sol: (b)

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 Let r be the radius of the circle. Then,
𝑑𝐴
A = 𝜋𝑟 2 and C = 2 𝜋r ⇒ 𝑑𝑟 = 2 𝜋r = C.
So, (A) is true.
Let x be the diameter of the circle. Then,
𝑥 𝑥 𝑑𝐴 𝜋𝑥 𝑑𝐴 𝐶
A = (2)2= 4 𝑥 2 and C = 𝜋𝑥 ⇒ 𝑑𝑥 = 2 ⇒ 𝑑𝑥 = 2
So, (R) is also true but (R) is not a correct explanation for (A).

3 Let the radius, surface area and volume of sphere be r, S and V respectively.
Assertion(A): The rate of change of volume of sphere with respect to its radius is
equal to S.
𝑟
Reason(R): The rate of change of volume of sphere with respect to S is 2.
Sol: (b)
We have,
4 𝑑𝑉 𝑑𝑆 𝑑𝑉
V = 𝜋𝑟 3 and S = 4 𝜋𝑟 2 ⇒ = 4 𝜋𝑟 2and = 8 𝜋r ⇒ =
3 𝑑𝑟 𝑑𝑟 𝑑𝑟
𝑑𝑉 𝑑𝑉/𝑑𝑟 4 𝜋𝑟 2 𝑟
𝑆 , (𝐴)𝑖𝑠 𝑡𝑟𝑢𝑒 𝑎𝑛𝑑 = = = ,(R)is true
𝑑𝑆 𝑑𝑆/𝑑𝑟 8 𝜋r 2
Thus, both (A) & (R) are true but (R) is not a correct explanation for (A).

4 Assertion(A): If the area of a circle increases at a uniform rate, then its perimeter
varies inversely as its radius.
Reason(R): The rate of change of area of a circle with respect to its perimeter is
equal to the radius.

Sol: (a)
Let r be the radius, P be the perimeter and A be the area of a circle. Then,
𝑑𝐴 𝑑𝑃
(𝐴) = 𝜋𝑟 2 and P =2 𝜋r ⇒ = 2 𝜋r and = 2 𝜋
𝑑𝑟 𝑑𝑟
𝑑𝐴 𝑑𝐴/𝑑𝑟 2𝜋r
⇒ 𝑑𝑃 =𝑑𝑃/𝑑𝑟 = =r
2𝜋
So, (R) is true.
𝑑𝐴 𝑑𝐴/𝑑𝑡 𝑑𝑃 1 𝑑𝐴
Now, 𝑑𝑃 = r ⇒ 𝑑𝑃/𝑑𝑡 = 𝑟 ⇒ 𝑑𝑡 = 𝑟 𝑑𝑡
𝑑𝐴
If = constant (=k,say). Then,
𝑑𝑡
𝑑𝑃 𝑘 𝑑𝑃 1
= 𝑟 ⇒ 𝑑𝑡 ∝ 𝑟 ⇒Perimeter varies inversely as the radius.
𝑑𝑡
So, (A) is also true and (R) is a correct explanation for (A).

5 Let f(x) =1-x3- x5

Assertion(A): f(x) is an increasing function
Reason(R): 3x2+5x4>0, for all x ≠ 0.
Sol: (d)
f(x) = 1-x3- x5
⇒ fʹ(x)= -3x2-5x4 = - (3x2+5x4)<0
⸪3x2+5x4>0
⸫ f(x) is decreasing [(A) is false]
x2+x4 always >0 for all x>0
⸫ 3x2+5x4>0, for all x≠0 [R is true]

6 𝐿𝑒𝑡 𝑓(𝑥) = 2𝑠𝑖𝑛3𝑥 + 3𝑐𝑜𝑠3𝑥

5𝜋
Assertion(A): f(x) does not have a maximum or minimum at x = 6
5𝜋
Reason(R): fʹ( 6 ) =0

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 Sol: (c)
f(x) = 2sin3x+3cos3x
fʹ(x)= 6cos3x-9sin3x
for maximum or minimum fʹ(x)=0
5𝜋 5𝜋 5𝜋
fʹ( 6 )= 6cos 2 - 9sin 2
= 0-9 = -9 ≠ 0
5𝜋
⸫ f(x) does not have a maximum at x = 6
⇒ (A) is true
5𝜋
fʹ( 6 ) = −9 ≠ 0 ⸫ (R) is false
⸫ (A) is true but (R) is false.
7 Let f(x) = 2x3-3x2-12x+4
Assertion(A): x = -1 is a point of local maximum
Reason(R): fʹʹ (-1) >0

Sol: (c)
f(x) = 2x3-3x2-12x+4
fʹ(x)= 6x2-6x-12
= 6(x2-x-2)
=6(x-2) (x+1)
fʹ(x) = 0 ⇒ x=2 or x=-1
= fʹʹ (x) = 6(2x-1)
fʹʹ(1) = 6(-2-1) = -18<0
⸫ x = -1 is a point of local maximum, [(A) is true.]
fʹʹ(-1) = -18<0
⸫ (R) is false.

8 𝐿𝑒𝑡 𝑓(𝑥) = 𝑥 + 𝑐𝑜𝑠𝑥

Assertion(A): f(x) is an increasing function on R
Reason(R): −1 ≤ 𝑠𝑖𝑛𝑥 ≤ 1
Sol: (a)
f(𝑥) = 𝑥 + 𝑐𝑜𝑠𝑥
fʹ(x)= 1-sinx ≥0, for all x (⸪-1≤sinx≤1)
⇒ f is an increasing function.
(A)is true
−1 ≤ 𝑠𝑖𝑛𝑥 ≤ 1 ⇒(R)is true
Thus, both (A) & (R) are true and(R) is the correct explanation for (A).

9 Assertion(A) : The function f(x) = x3+5x+1, 𝑥 ∈ ℝ is always increasing

Reason(R): fʹ(x)>0 for 𝑥 ∈ ℝ, for increasing function
Sol: (a)
f(x) = x3+5x+1
fʹ(x)= 3x2+5>0 for all 𝑥 ∈ ℝ
⸫ f(x) is always increasing ((A)is true)
(R)is also true and R is the correct explanation of A.

10 𝐿𝑒𝑡 𝑓(𝑥) = 𝑠𝑖𝑛𝑥

𝜋
Assertion(A) : f(x) is increasing in (0, )
2

𝜋
Reason(R): cos𝜃 is positive for all 𝜃 ∈ (0, 2 )
Sol: (a)

111
 𝜋
f(x) = sinx ⇒ fʹ(x)= cosx>0, for all 𝑥 ∈ (0, 2 )
𝜋
⇒ f is increasing in (0, 2 ) [(A)is true]
(R)is also true and R is the correct explanation of A.

EXERCISE

In the following questions, a statement of assertion (A) is followed by a statement of

reason(R). Choose the correct answer out of the following choices.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A)
(b) Both (A) and (R) are true and (R) is not the correct explanation of (A)
(c) (A) is true but (R) is false
(d) (A) is false but (R) is true

1 Assertion (A) :The minimum value of 𝑥 2 – 8x + 17 is 4.

Reason (R) : A function f(x) is minimum at x=c if f ′(𝑐) = 0 and f ′ ′(𝑐) is
positive.
Answer: d
2 Assertion (A) :The absolute minimum value of 𝑥 3 – 18x2 -96x in [0,9] is 0.
Reason (R) : A function always attains absolute minimum in the interval
[𝑎, 𝑏] 𝑎𝑡 𝑥 = 𝑎
Answer: c
3 Assertion (A) : Let f(x)=𝑒 𝑥 is an increasing function ∀𝑥 ∈ 𝑅.
Reason (R) : If f ′(𝑥) ≤ 0 then f(x) is an increasing function.
Answer: c
4 Assertion (A) : f(x) =logx is defined for all 𝑥 ∈ (0, ∞).
Reason (R) : If f ′(𝑥) > 0 then f(x) is strictly increasing function.
Answer: b
5 Assertion (A) : f(x) = sin 2x+3 is defined for all real values of x.
Reason (R) : Minimum value of f(x) is 2 and Maximum value is 4.
Answer: b

2 MARK QUESTIONS

1 Radius of variable circle is changing at the rate of 5cm/s. What is the radius of the
circle at a time when its area is changing at the rate of 100cm2 /s ?

Solution: The area A of a circle with radius r is given by A = 𝜋𝑟 2

𝑑𝐴 𝑑𝑟 10
= 2 𝜋r ⇒ 100 = 2 𝜋r x 5 ⇒ r = cm.
𝑑𝑡 𝑑𝑡 𝜋
𝑑𝑟 𝑑𝐴
[𝑑𝑡 = 5𝑐𝑚/𝑠, = 100cm2 /s]
𝑑𝑡
10
Hence, radius of the circle is r = 𝜋 cm.
2 The side of an equilateral triangle is increasing at the rate of 0.5 cm/s. Find the
rate of increase of its perimeter.
𝑑𝑎
Sol: Let the side of the triangle be ‘a’ then 𝑑𝑡 = 0.5 𝑐𝑚/𝑠
Perimeter of the triangle, P =3a
𝑑𝑃 𝑑𝑎
⇒ 𝑑𝑡 = 3 𝑑𝑡 =3 x 0.5 cm/s = 1.5 cm/s

112
3 If the rate of change of volume of a sphere is equal to the rate of change of its
radius, then find the radius.
𝑑𝑉 𝑑𝑟
Sol: Given, =
𝑑𝑡 𝑑𝑡
𝑑 4 𝑑𝑟
⇒ ( 𝜋𝑟 3 ) = 𝑑𝑡
𝑑𝑡 3
𝑑𝑟 𝑑𝑟
⇒ 4𝜋𝑟 2 . 𝑑𝑡 = 𝑑𝑡
2
⇒ 4𝜋𝑟 = 1
1 1
⇒ 𝑟 2 = 4𝜋 ⇒ 𝑟 = 2 𝜋 units
√

4 3
A balloon which always remain spherical has a variable diameter (2𝑥 + 1).
2
Find the rate of change of its volume with respect to x.
3
Sol: Diameter of the balloon = 2 (2𝑥 + 1)
3
⸫ r = radius of the balloon = 4 (2𝑥 + 1)
4 3
Volume of the balloon, 𝑉 = 3 𝜋[4 (2𝑥 + 1)]3
9
= 16 𝜋(2𝑥 + 1)3
𝑑𝑉 9 27
⇒ 𝑑𝑥 = 16 𝜋.3(2𝑥 + 1)2 . 2 = 8
𝜋(2𝑥 + 1)2

5 x and y are the sides of two squares such that 𝑦 = 𝑥 − 𝑥 2 . Find the rate of change
of the area of Second Square with respect to the area of the first square.
Sol: The area 𝐴1 of square of side x is given by 𝐴1 = 𝑥 2
and area 𝐴2 of square of side y is given by 𝐴2 = y 2 =(𝑥 − 𝑥 2 )2
𝑑𝐴1 𝑑𝐴2
= 2𝑥, = 2(𝑥 − 𝑥 2 ) (1 − 2𝑥)
𝑑𝑥 𝑑𝑥

𝑑𝐴2 𝑑𝐴2 𝑑𝐴1 2(𝑥−𝑥 2 )(1−2𝑥)

= ÷ =
𝑑𝐴1 𝑑𝑥 𝑑𝑥 2𝑥
= (1 − 𝑥) (1 − 2𝑥) = 1 − 3𝑥 − 2𝑥 2

6 Find the intervals in which 𝑓(𝑥) = 𝑥 2 − 2𝑥 + 15 is strictly increasing or strictly

decreasing.
Sol: We have,
𝑓(𝑥) = −𝑥 2 − 2𝑥 + 15
⇒ 𝑓ʹ(𝑥)= −2𝑥 − 2 = -2(𝑥 + 1)
For 𝑓(𝑥) to be increasing, we must have
𝑓ʹ(𝑥)>0
⇒ -2(𝑥 + 1)>0
⇒ 𝑥 + 1<0
⇒ 𝑥 < −1 ⇒ 𝑥 ∈ (−∞, −1)
Thus, 𝑓(𝑥) is increasing on the interval (−∞, −1).
For 𝑓(𝑥) to be decreasing, we must have
𝑓ʹ(𝑥) < 0
⇒ -2(𝑥 + 1)<0
⇒ 𝑥 + 1>0
⇒ 𝑥 > −1 ⇒ 𝑥 ∈ (−1, ∞)
So, 𝑓(𝑥) is decreasing on (−1, ∞).
7 Show that the function 𝑓(𝑥) = (𝑥 3 − 6𝑥 2 + 12𝑥 + 18) is an increasing function
on R.
Sol: 𝑓(𝑥) = (𝑥 3 − 6𝑥 2 + 12𝑥 + 18)
⇒ 𝑓ʹ(𝑥) = 3𝑥 2 + 12𝑥 + 12
= 3(𝑥 2 + 4𝑥 + 4) = 3(𝑥 − 2)2 ≥ 0 for all 𝑥 ∈ 𝑅.
Thus, 𝑓ʹ(𝑥) ≥ 0 for all 𝑥 ∈ 𝑅.
Hence, 𝑓(𝑥) is an increasing function on R.

113
8 Find the intervals on which the function 𝑓(𝑥) = (5 + 36𝑥 + 3𝑥 2 − 2𝑥 3 ) is
increasing.
Sol: 𝑓(𝑥) = (5 + 36𝑥 + 3𝑥 2 − 2𝑥 3 )
⇒ 𝑓ʹ(𝑥) = 36 + 6𝑥 − 6𝑥 2
= -6(𝑥 2 − 𝑥 − 6) = -6(x+2)(x-3)
𝑓(𝑥) is increasing
⇒ 𝑓ʹ(𝑥) ≥ 0
⇒ -6(𝑥 + 2)(𝑥 − 3) ≥ 0
⇒ (𝑥 + 2)(𝑥 − 3) ≤ 0
⇒ −2 ≤ 𝑥 ≤ 3
⇒ 𝑥 ∈ [−2,3].
⸫ 𝑓(𝑥) is increasing on[−2,3].

9 Find the maximum and the minimum values of the function 𝑓(𝑥) = 𝑥 + 2, 𝑥 ∈
(0,1).
Sol: 𝑓(𝑥) = 𝑥 + 2
𝑓ʹ(𝑥) = 1
so for no value of x, 𝑓ʹ(𝑥) = 0.
So 𝑓(𝑥) has no critical points.
Hence, 𝑓(𝑥) has neither local maximum nor local minimum.

10 Amongst all pairs of positive numbers with sum 24, find those whose product is
maximum.
Sol: Let the numbers be x and (24 − 𝑥).
Let 𝑃 = 𝑥(24 − 𝑥) = (24𝑥 − 𝑥 2 )
𝑑𝑃 𝑑2 𝑃
Then, 𝑑𝑥 = (24 − 2𝑥)and 𝑑𝑥 2 = -2.
𝑑𝑃
Now, 𝑑𝑥 = 0 ⇒ (24 − 2𝑥) = 0 ⇒ x = 12.
𝑑2 𝑃
Thus,{𝑑𝑥 2 } = -2<0.
𝑥=12
⸫ x=12 is a point of maximum.
Hence, the required numbers are 12 and 12.

11 Find the local maxima and local minima, if any of the function f, given by f(x) =
𝜋
sin x + cos x, 0 < x < 2 .
Sol: f(x) = sin x + cos x
𝑓ʹ(𝑥) = cos x – sin x,
for points of local maxima or minima
𝜋
𝑓ʹ(𝑥) = 0 ⇒ cos x – sin x =0⇒ tan x = 1 ⇒ 𝑥 = 4
𝜋
fʹʹ (x) = – sin x – cos x, fʹʹ ( 4 )<0
𝜋
⸫𝑥 = is a point of local maximum
4
𝜋 𝜋 𝜋
Local maximum value = f( 4 ) = 𝑠𝑖𝑛 4 + 𝑐𝑜𝑠 4
1 1
= + = √2
√2 √2
12 Find the interval/s in which the function 𝑓 ∶ ℝ → ℝ defined by (𝑥) = 𝑥𝑒 𝑥 , is
increasing.

Sol: 𝑓(𝑥) = 𝑥𝑒 𝑥 ⇒ 𝑓ʹ(𝑥) = 𝑒 𝑥 (𝑥 + 1)

For 𝑓(𝑥) to be increasing, 𝑓ʹ(𝑥)≥0
⇒ 𝑒 𝑥 (𝑥 + 1) ≥ 0 ⇒ 𝑥 ≥ −1 as 𝑒 𝑥 > 0, ∀ 𝑥 ∈ ℝ

114
 Hence, 𝑓(𝑥) increases in [−1, ∞).
13 1
If 𝑓(𝑥) = 4𝑥 2 +2𝑥+1; 𝑥 ∈ ℝ, then find the maximum value of 𝑓(𝑥).
1
Sol: We have 𝑓(𝑥) = 4𝑥 2 +2𝑥+1
1 1 3
Let 𝑔(𝑥) = 4𝑥 2 + 2𝑥 + 1 = 4(𝑥 2 + 2𝑥 4 + 16) + 4
1 2 3 3
= 4(𝑥 + 4) + 4 ≥ 4
3
⇒minimum value of 𝑔(𝑥) = 4.

4
⸫ maximum value of 𝑓(𝑥) = 3.
14 Find the maximum profit that a company can make, if the profit function is given
by 𝑃(𝑥) = 72 + 42𝑥 − 𝑥 2 , where x is the number of units and P is the
profit in rupees.

Sol: 𝑃(𝑥) = 72 + 42𝑥 − 𝑥 2

𝑃ʹ(𝑥) =42 – 2x , 𝑃ʹʹ(x) = −2
For maxima or minima,
𝑃ʹ(𝑥) = 0 ⇒ 42 – 2x = 0 ⇒ x = 21
𝑃ʹʹ(x) = −2 < 0
So, 𝑃(𝑥) is maximum at x = 21.
The maximum value of 𝑃(𝑥)
= 𝑃(21) =72+(42 x 21)- (21)2 = 513
i.e., the maximum profit is Rs.513.
15 Check whether the function 𝑓 ∶ ℝ → ℝ defined by 𝑓(𝑥) = 𝑥 3 + 𝑥, has any
critical point/s or not. If yes, then find the point/s.

Sol: 𝑓(𝑥) = 𝑥 3 + 𝑥, for all 𝑥 ∈ ℝ.

𝑓ʹ(𝑥) = 3𝑥 2 + 1 > 0 for all 𝑥 ∈ ℝ, (𝑥 2 ≥ 0 )
⇒ 𝑓ʹ(𝑥) ≠ 0
Hence, no critical point exists.

EXERCISE

1 The total cost C(x) associated with the production of x units of an item is given by
C(x) = = 0.005𝑥 3 − 0.02𝑥 2 − 30𝑥 + 5000.
Find the marginal cost when 3 units are produced, where by marginal cost we
mean the instantaneous rate of change of total cost at any level of output.
Answer: 29.985
2 Find the intervals in which the function 𝑓(𝑥) = 2𝑥 3 − 3𝑥 2 − 36𝑥 + 7 is strictly
increasing & decreasing.
Answer: Increasing. in (−∞, −𝟐) ∪ (𝟑, ∞). Decreasing. in (−𝟐 , 𝟑).

3 A ladder, 5-meter-long, standing on a horizontal floor, leans against a vertical

wall. If the top of the ladder slides downwards at the rate of 10cm/sec, find the
rate at which the angle between the floor and the ladder is decreasing when lower
end of ladder is 2 meters from the wall.
𝟏
Answer: 𝟐𝟎 radians/sec
4 Let x and y be the radii of two circle such that 𝑦 = 𝑥 2 + 1. Find the rate of
change of circumference of second circle w.r.t the circumference of the first
circle.
Answer: 𝟐𝒙

115
5 Find the least value of the function 𝑓(𝑥) = 𝑥 3 − 18𝑥 2 + 96𝑥 in the interval
[0,9].
Answer: 0

3 MARK QUESTIONS

1 A man 1.6 m tall walks at the rate of 0.5 m/s away from a lamp post, 8 meters
high. Find the rate at which his shadow is increasing and the rate with which the
tip of shadow is moving away from the pole.
Solution: Let AB be the lamp post and CD the height of the man.
Let distance of the man from the lamp post be x m
and from tip of shadow be y m. A
𝑑𝑥
= 0.5 m/s
𝑑𝑡
In similar triangles ABO ans CDO
8 𝑥+𝑦
=
1.6 𝑦
1 8
⇒ 5y = x+y ⇒ y = 4 𝑥 m
𝑑𝑦 1 𝑑𝑥 0.5
⸫ 𝑑𝑡 = 4 𝑑𝑡 = 4 = 0.125 m/s C
1
⸫ Rate at which shadow is increasing 0.125 m/s .
B
𝑑 x 6 Dy O
Rate of change of tip of shadow = (𝑥 + 𝑦) m m m
𝑑𝑡
𝑑𝑥 𝑑𝑦
= 𝑑𝑡 + 𝑑𝑡
= 0.5+0.125 = 0.625 m/s

2 The area of an expanding rectangle is increasing at the rate of 48𝑐𝑚2 /𝑠. The
length of the rectangle is always equal to square of breadth. At what rate, the
length is increasing at the instant when breadth is 4.5 cm?
Solution: Let the length of the rectangle be l and its breadth b.
3
Then 𝑙 = 𝑏 2 ⇒ 𝐴 = 𝑙. √𝑙 = 𝑙 2
𝑑𝐴 3 𝑑𝑙
= 2 √𝑙. 𝑑𝑡
𝑑𝑡

3 𝑑𝑙
⇒ 48 = 2 x 4.5 x 𝑑𝑡 …… ( b= √𝑙 )
𝑑𝑙 320 64
⇒ 𝑑𝑡] = = = 7.11 cm/s
𝑏=4.5 𝑐𝑚 45 9

3 Sand is pouring from a pipe at the rate of 12 cm3/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?
Solution: Let h be the height, V the volume and r the radius of the base of cone at
the time t.
1
Given ℎ = 6 𝑟
⇒ 𝑟 = 6ℎ
𝑑𝑉
𝑑𝑡
= 12 𝑐𝑚3 /𝑠
1
Volume of the cone, 𝑉 = 3 𝜋𝑟 2 h h
1
= 𝜋(6ℎ)2 h = 12 𝜋ℎ3
3
𝑑𝑉 𝑑ℎ
⸫ 𝑑𝑡 = 12 𝜋 . 3ℎ2 . 𝑑𝑡
r
𝑑ℎ 𝑑ℎ 1
⇒ 12 = 36 𝜋h2 𝑑𝑡 ⇒ 𝑑𝑡
= 3𝜋h2

116
 𝑑ℎ 1 1
⇒ ] =
𝑑𝑡 ℎ=4 3𝜋𝑋16
= 48𝜋 𝑐𝑚/𝑠.

4 Find the intervals in which the function 𝑓(𝑥) = 2𝑥 3 − 9𝑥 2 + 12𝑥 + 15

is strictly increasing.

Solution:We have,
𝑓(𝑥) = 2𝑥 3 − 9𝑥 2 + 12𝑥 + 15
⇒ 𝑓ʹ(𝑥) = 6𝑥 2 − 18𝑥 + 12 = 6(𝑥 2 − 3𝑥 + 2)
(i) For 𝑓(𝑥) to be increasing, we must have
𝑓ʹ(𝑥) > 0
⇒ 6(𝑥 2 − 3𝑥 + 2) > 0
⇒ 𝑥 2 − 3𝑥 + 2 > 0 [∵ 6 > 0 ∴ 6(𝑥 2 − 3𝑥 + 2) > 0 ⇒ 𝑥 2 − 3𝑥 + 2 > 0]
⇒ (𝑥 − 1) (𝑥 − 2) > 0 (See fig)
⇒ 𝑥 < 1 or 𝑥 > 2
⇒ 𝑥 ∈ (−∞, 1) ∪ (2, ∞).
So, 𝑓(𝑥) is increasing on (−∞, 1) ∪ (2, ∞).
+ +
- 1 2 ∞
∞ Fig. Signs of 𝑓ʹ(𝑥) for different values of x

5 Show that 𝑓(𝑥) = 𝑡𝑎𝑛−1 (𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥) is a strictly increasing function on the
𝜋
interval (0, 4 ).
Solution: 𝑓(𝑥) = 𝑡𝑎𝑛−1 (𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥)
1 𝑑
⇒ 𝑓ʹ(𝑥) = 1+(𝑐𝑜𝑠𝑥+𝑠𝑖𝑛𝑥)2 . 𝑑𝑥 (𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥)

(−𝑠𝑖𝑛𝑥+cos 𝑥)
=
1+𝑐𝑜𝑠2 x+𝑠𝑖𝑛2 x+2 sin x cos x

𝑐𝑜𝑠 𝑥−𝑠𝑖𝑛 𝑥
=
(2+sin 2𝑥)
𝜋
Now, when 0 <x<4 , we have 𝑐𝑜𝑠𝑥 > 𝑠𝑖𝑛𝑥 &𝑠𝑖𝑛2𝑥 > 0
⸫ (𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥) > 0 and (2 + 𝑠𝑖𝑛2𝑥) > 0.
𝜋
⸫ 𝑓ʹ(𝑥) > 0 for all x when 0 <x<4
𝜋
Hence, 𝑓(𝑥) is strictly increasing in (0, 4 ).

𝜋
6 Separate [0, 2 ] into subintervals in which 𝑓(𝑥) = sin 3x is (a) increasing (b)
decreasing.
Solution: 𝑓(𝑥) = sin 3x ⇒ 𝑓ʹ(𝑥) = 3cos 3x
𝜋 3𝜋
Also, 0 ≤ 𝑥 ≤ 2 ⇒ 0 ≤ 3𝑥 ≤ 2
(a) 𝑓(𝑥) is increasing
⇒ 𝑓ʹ(𝑥) ≥ 0
⇒ 3 cos 3𝑥 ≥ 0 ⇒ cos 3𝑥 ≥ 0
𝜋
⇒ 0 ≤ 3𝑥 ≤ 2
𝜋
⇒0≤𝑥≤ 6
𝜋
⇒ 𝑥 ∈ [0, 6 ].
𝜋
⸫ 𝑓(𝑥) is increasing on [0, 6 ].
(b) 𝑓(𝑥) is decreasing
⇒ 𝑓ʹ(𝑥) ≤ 0
⇒ 3 cos 3𝑥 ≤ 0 ⇒ cos 3𝑥 ≤ 0

117
 𝜋 3𝜋
⇒ 2 ≤ 3𝑥 ≤ 2
𝜋 𝜋
⇒6≤𝑥≤ 2
𝜋 𝜋
⇒𝑥 ∈ [ 6 , 2 ].

7 Show that 𝑓(𝑥) = 2𝑥 + 𝑐𝑜𝑡 −1 𝑥 + 𝑙𝑜𝑔(√1 + 𝑥 2 − 𝑥) is increasing in R.

Solution: We have,
𝑓(𝑥) = 2𝑥 + 𝑐𝑜𝑡 −1 𝑥 + 𝑙𝑜𝑔(√1 + 𝑥 2 − 𝑥)
−1 1 1
⇒ 𝑓ʹ(𝑥) = 2 + (1+𝑥 2 ) + √1+𝑥 2 ( 2
.2𝑥 − 1)
−𝑥 2√1+𝑥
1 1 𝑥−√1+𝑥2
= 2 - 1+𝑥 2 + √1+𝑥 2 . √1+𝑥 2
−𝑥
1 1
= 2 - 1+𝑥 2 - √1+𝑥 2
2+2𝑥 2 −1−√1+𝑥 2 1+2𝑥 2 −√1+𝑥 2
= =
1+𝑥 2 1+𝑥 2
For increasing function, 𝑓ʹ(𝑥) ≥ 0
1+2𝑥 2 −√1+𝑥 2
⇒ ≥0
1+𝑥 2
2
⇒ 1 + 2𝑥 ≥ √1 + 𝑥 2
⇒ (1 + 2𝑥 2 )2 ≥ 1 + 𝑥 2
⇒ 1 + 4𝑥 4 + 4𝑥 2 ≥ 1 + 𝑥 2
⇒ 4𝑥 4 + 3𝑥 2 ≥ 0
⇒ 𝑥 2 (4𝑥 2 + 3) ≥ 0
which is true for any real value of x.
Hence, 𝑓(𝑥) is increasing in R.

𝜋
8 Prove that 𝑓(𝑥) = sin x + √3 cos 𝑥 has maximum, value at 𝑥 = 6 .
Solution:We have, 𝑓(𝑥) = sin x + √3 cos 𝑥
⸫ 𝑓ʹ(𝑥) = cos 𝑥 + √3(−𝑠𝑖𝑛𝑥)
= cos 𝑥 − √3𝑠𝑖𝑛𝑥
For 𝑓ʹ(𝑥) = 0, ⇒ cos 𝑥 − √3𝑠𝑖𝑛𝑥=0
1 𝜋
⇒ cos 𝑥 = √3𝑠𝑖𝑛𝑥 ⇒ tan 𝑥 = 3 = tan 6
√
𝜋
⇒𝑥=
6
Again, differentiating 𝑓ʹ(𝑥), we get
𝑓ʹʹ(𝑥) = −sin 𝑥 - √3 cos 𝑥
𝜋 𝜋 𝜋
𝑓ʹʹ (6 ) = −sin 6 - √3 cos 6
1 √3
= − 2 − √3 . 2
1 3
= − 2 − 2 = −2 < 0
𝜋
Hence 𝑥 = is the point of local maxima.
6

9 Find the local maximum and the local minimum values of the function f(x) =
−3 4 45
𝑥 −8𝑥 3 − 2 𝑥 2 + 105
4
Solution: 𝑓ʹ(𝑥) = −3𝑥 3 − 24𝑥 2 − 45𝑥
= −3𝑥(𝑥 2 + 8𝑥 + 15)
= −3𝑥(𝑥 + 3)(𝑥 + 5)
𝑓ʹʹ(𝑥) = −9𝑥 2 − 48𝑥 − 45
𝑓ʹ(𝑥) = 0 ⇒ −3𝑥(𝑥 + 3)(𝑥 + 5) = 0
⇒ 𝑥 = 0, 𝑥 = −3, 𝑥 = −5

118
 𝑓ʹʹ(0) = -45<0
So x=0 is a point of local maxima
. 𝑓ʹʹ(−3)=+18 > 0
So x=-3 is a point of local minima
𝑓ʹʹ(−5)=−30 < 0
So x=-5 is a point of local maxima

10 A telephone company in a town has 500 subscribers on its list and collects
fixed charges of Rs.300 per subscriber per year. The company proposes to
increase the annual subscription and it is believed that for every increase of
Rs.1 per one subscriber will discontinue the service. Find what increase will
bring maximum profit?
Solution: Consider that company increases the annual subscription by Rs. x
So, x subscribers will discontinue the service
⸫ Total revenue of company after the increment is given by
R(x) = (500-x)(300+x)
=15 x 104 + 500𝑥 − 300𝑥 − 𝑥 2
= −𝑥 2 + 200𝑥 + 150000
On differentiating both sides w.r.t x, we get
Rʹ (x) = −2𝑥 + 200
Now, Rʹ (x) = 0
⇒ 2x = 200 ⇒ x=100
⸫ Rʹʹ (x) = -2<0
So, R(x) is maximum when x=100
Hence, the company should increase the subscription fee by Rs.100, so that it
has maximum profit.

EXERCISE
1 Find two positive numbers whose sum is 16 and sum of whose cubes is minimum.
Answer: 8, 8
2 2𝑥
Show that y = log(1+x) - 2+𝑥 , , x> −1 is an increasing function of x throughou its
domain .
3 Show that the function 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 3𝑥, 𝑥 ∈ 𝑅 is increasing on R.
4 Find the intervals in which the function 𝑓(𝑥)=2𝑥 3 − 9𝑥 2 + 12𝑥 − 15 is
(i) increasing. (ii) decreasing

Answer: : (−∞, 𝟏) ∪ (𝟐, ∞)

5 The total revenue received from the sale of 𝑥 units of a product is given by
R(x) = 3𝑥 2 + 36𝑥 + 5 in rupees. Find the marginal revenue when x=5, where by
marginal revenue we mean the rate of change of total revenue with respect to the
number of items sold at an instant. At what value of x is R(x) minimum

Answer: 66

5 MARK QUESTIONS

1 Find the intervals in which the function f(x) =3x4 -4x3-12x2+5 is increasing or
decreasing.
Solution:

119
 𝑓(𝑥) = 3𝑥 4 − 4𝑥 3 − 12𝑥 2 + 5
𝑓 ′ (𝑥) = 12𝑥 3 − 12𝑥 2 − 24𝑥
= 12𝑥(𝑥 2 − 𝜘 − 2)
=12𝑥(𝑥 − 2)(𝑥 + 1)
𝑓 1 (𝑥) = 0 ⇒ 𝑥 = 0,2, −1

Intervals Sign of f’(x) Nature of f(x)

(-∞,-1) -ve decreasing

(-1,0) +ve increasing

(0,2) -ve decreasing
(2,∞) +ve increasing

Hence f is increasing in (−1,0) ∪ (2, ∞) and decreasing in (−∞, −1) ∪ (0,2).

2 Show that the surface area of a closed cuboid with a square base and given volume
is minimum, when it is a cube.

Solution: Let x be side of the square base and y be the height of the cuboid.
Volume (V)= x.x.y=x2y …….. (i)
𝑉
y=𝑥 2
Surface area (S)=2(x.x+x.y+x.y)
𝑣
=2𝑥 2 + 4𝜘𝑦 = 2𝑥 2 + 4𝑥 𝑥 2
4𝑣 𝑑𝑠 4𝑣
S=2𝑥 2 + 𝑥 ⇒ 𝑑𝑥 = 4𝑥 − 𝑥 2
For minimum surface area,
𝑑𝑠 4𝑣
= 0 ⇒ 4𝑥 − 2 = 0 ⇒ 𝑥 3 = 𝑣
𝑑𝑥 𝑥
3
x= √𝑣
𝑑2𝑠 8𝑣
= 4 +
𝑑𝑥 2 𝑥3
2
𝑑 𝑆 8𝑣
2
=4+ >0
𝑑𝑥 𝑣
3
For x= √𝑣, surface area is minimum
x3 = V
x3 = x2y [from (i)]
x=y cuboid is a cube.

3 Prove that the volume of the largest cone that can be inscribed in a sphere of
8
radius R is 27 of the volume of the sphere.
Solution: Let a cone of base radius x and height y be inscribed in a sphere of
radius R.
R2=(y-R)2 +x2
x2 = 2Ry-y2 [in right triangle OAB]…(i)
1
Volume of the cone, V=3 π x2y

120
 1
𝜋𝑦(2𝑅𝑦 − 𝑦 2 )
3
1
= 𝜋(2𝑅𝑦 2 − 𝑦 3 ) [from (i)]……(ii)
3
𝑑𝑉 1
= 𝜋(4𝑅𝑦 − 3𝑦 2 )
𝑑𝑦 3
For maximum volume,
𝑑𝑉
=0
𝑑𝑦
⇒4Ry=3y2
4𝑅
⇒y = 3
𝑑2𝑣 𝜋
= (4𝑅 − 6𝑦)
𝑑𝑦 2 3
𝑑2𝑣 4𝑅
< 0, for y =
𝑑𝑦 2 3
1 4𝑅 2 4𝑅 3
Vmax. = π[2𝑅 ( ) −( ) ]
3 3 3
3 3
1 32𝑅 64𝑅
= 𝜋[ − ]
3 9 27
32𝜋𝑅 3
= 𝑐𝑚3
81
8 4
= ( 𝛱𝑅 3 )
27 3
8
(𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑒)
27
4 Show that the height of the cylinder of maximum volume that can be inscribed in
2𝑅
a sphere of radius R is . Also, find the maximum volume.
√3
Solution:
Let x be radius of base and y height of a cylinder which is inscribed in a sphere of
radius R.
4𝑥 2 + 𝑦 2 = 4𝑅 2 …….(i)
Volume of cylinder
4𝑅 2 −𝑦 2
V=𝜋𝑥 2 𝑦 = 𝜋𝑦 ( )
4
𝜋
= 4 (4𝑅 2 𝑦 − 𝑦 3 )
[from(i)] ……(ii)
𝑑𝑣 𝜋
= (4𝑅 2 − 3𝑦 2 )
𝑑𝑦 4
𝑑𝑉
For maximum volume, 𝑑𝑦 = 0
2𝑅
Y=
√3
𝑑 2 𝑣 −3𝜋𝑦
=
𝑑𝑦 2 2
𝑑2 𝑣 2𝑅
< 0 for y=
𝑑𝑦 2 √3
Now substituting the value of y in equation (ii), we get maximum volume
4𝜋𝑅 3
= 3√3 cubic units.
5 Two men A and B start with velocities v at the same time from the junction of two
roads inclined at 450 to each other. If they travel by different roads, then find the
rate at which they are being separated.

121
 Solution: Let two men start from the point C with velocity v each at the same
time.
Also, ∠𝐵𝐶𝐴 = 450
Since, A and B are moving with same velocity v, so they will cover same distance
in same time.
Therefore, 𝛥𝐴𝐵𝐶 is an isosceles triangle with
AC=BC.
Now, draw CD⊥AB.
Let at any instant t, the distance between them is AB.
Let AC=BC=x and AB=y
In ∆ACD and ∆DCB,
∠CAD=∠ CBD
∠CDA=∠ CDB=900
∠ACD=∠ DCB
1
∠ACD= 2 𝑋∠ ACB
1
∠ ACD= 2 𝑋450
𝜋
∠ ACD= 8
𝜋 𝐴𝐷
Sin 8 = 𝐴𝐶
𝜋 𝑌/2
Sin 8 = 𝑥
𝑦
= x sin 𝜋8
2
𝜋
Y=2x.sin8
Now, differentiating both sides w.r.t. t ,we get
𝑑𝑦
𝑑𝑡
= 2.sin 𝜋8.𝑑𝑥
𝑑𝑡

𝜋 𝑑𝑥
=2.sin .v …………. [v= ]
8 𝑑𝑡
√2−√2 𝜋 √2−√2
=2v. …………. [sin 8 = ]
2 2
=√2 − √2 v unit/s
Which is the rate at which A and B are being separated.

EXERCISE

1 𝑥4
Find the intervals in which the function f(x)= 4 − 𝑥 3 − 5𝑥 2 + 24𝑥 + 12 is
(i) strictly increasing (ii) strictly decreasing.
Answer: (i) (−𝟑, 𝟐) ∪ (𝟒, ∞). (ii) (−∞, −𝟑) ∪ (𝟐, 𝟒).
2 The length of the sides of an isosceles triangle are 9+x2 , 9+x2 and 18-2x2 units.
Calculate the area of the triangle in terms of x and find the value of x which makes
the area maximum.
Answer: 𝑥 = √3
3 A rectangle is inscribed in a semicircle of radius r with one of its sides on the
diameter of the semicircle. Find the dimensions of the rectangle, so that its area is
maximum. Also find maximum area.
𝑟
Answer: √2r ,
√2
4 If the sum of a side and the hypotenuse of a right- angled triangle be given, show
that the area of the triangle will be maximum if the angle between the given side
and the hypotenuse be 600.

122
 5 Show that the semi-vertical angle of a right circular cone of given total surface
1
area and maximum volume is 𝑠𝑖𝑛−1 3.
6 Show that the surface area of a closed cuboid with a square base and given volume
is minimum, when it is a cube.
7 If the length of three sides of a trapezium other than the base are equal to 10 cm ,
then find the maximum area of the trapezium .
8 Find the maximum area of an isosceles triangle inscribed in the ellipse
𝑥2 𝑦2
+ = 1 with its vertex at one end of the major axis.
16 9
Answer: 9√3 sq.units

CASE STUDY QUESTIONS

Read the following and answer the questions given below.

1 Dr. Ritham residing in Delhi went to see an apartment of 3BHK in Noida. The
window
of the house was in the form of a rectangle surmounted by a semicircular
opening
having a perimeter of the window 10 m as shown in figure.

(i). If x and y represent the length and breadth of the rectangular region, then the
relation between the variable is
𝑥 𝑥
a) x + y + 2 = 10 b) x + 2y + 2 = 10
𝑥
c) x + 2y + π 2 = 10 d) 2x+ 2y = 10
(ii) The area of the window (A) expressed as a function of x is
𝜋𝑥 3 𝑥2 −𝑥 2 𝜋𝑥 2
a) 𝐴 = 𝑥 − − b) 𝐴 = 5𝑥 − −
8 2 2 8
𝑥2 3𝑥 2 𝑥2 𝜋𝑥 2
c) A= 5 𝑥 − 2 − 8 d) A= 5x + 2 + 8
(iii) Dr. Ritam is interested in maximizing the area of the whole window. For this
to
Happen the value of x should be
20 20 20 20
(a) 𝜋 b) 4−𝜋 c ) 2+𝜋 d) 4+𝜋
Solution:
(i) Since perimeter of window = x + y + y + perimeter of semicircle
1 𝑥 𝑥
= x+ 2y + 2 x 2 π x 2 [𝐻𝑒𝑟𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓𝑠𝑒𝑚𝑖𝑐𝑖𝑟𝑐𝑙𝑒 𝑖𝑠 2]
𝜋𝑥
= x + 2y + 2

123
 Option ( c ) is correct.

1 𝑥 2
(ii) A = x x y + 2 π (2)

𝜋𝑥 2 𝑥 𝜋𝑥 𝜋𝑥 2
= 𝑥𝑦 + = 𝑥 (5 − 2 − )+
8 4 8

𝑥2 𝜋𝑥 2 𝜋𝑥 2 𝑥2 𝜋𝑥 2
= 5𝑥- - + = 5𝑥 - -
2 4 8 2 8

Option ( b ) is correct.

(iii) For maximum value of A

𝑑𝐴
= 0
𝑑𝑥

𝜋𝑥 𝜋𝑥
⇒5 – 𝑥 - =0 ⇒𝑥+ =5
4 4

⇒4𝑥 + π 𝑥 = 20 ⇒ 𝑥(4 + π) = 20
20
⇒ 𝑥 = 4+𝜋

Option (d) is correct

2 Read the following passage and answer the questions given below:
The relation between the height of the plant (‘y” in cm ) with respect to its
exposure
1
to the sunlight is governed by the following equation y = 4x - 2 x 2 , where ‘ x ‘ is
the
number of days exposed to the sunlight for x ≤ 3.

(i) Find the rate of growth of the plant with respect to the number of days
exposed to the sunlight.
(ii) Does the rate of growth of the plant increase or decrease in the first three
days? What will be the height of the plant after 2 days?
Solution:
(i)The rate of growth of the plant with respect to the number of days exposed to
𝑑𝑦
sunlight is given by 𝑑𝑥 = 4 – x.
𝑑𝑦
(ii) Let rate of growth be represented by the function g (x) = 𝑑𝑥 .
𝑑 𝑑𝑦
Now, g ՜՜(x) = 𝑑𝑥 (. 𝑑𝑥 ) = -1 < 0
g(x ) decreases.
So the rate of growth of the plant decreases for the first three days.
1
Height of the plant after 2 days is y = 4 x 2 - 2 (2) 2 = 6 cm.

124
3 A rectangular hall is to be developed for a meeting of farmers in an agriculture
college to aware them for new techniques in cultivation. It is given that the floor
has a fixed
perimeter P as shown below. And if x & y represents the length & breadth of the
rectangular region.

Answer the following question.

(i) The area of the rectangular region ‘A’ expressed as a function of x is
1 1 1
(a) 2 ( P + x2 ) (b) 2 ( P x – 2x2 ) (c) 2 ( P x + 2x2 ) (d) P x
– 2x 2

(ii) Principal of agriculture college is interested in maximizing the area of floor

‘A’ . For this to happen the value of x should be
𝑃 2𝑃 𝑃
(a) P (b) 2 (c) 3
(d) 4
Solution:
(i) A = x.y
𝑃−2𝑥
= x. 2
𝑠𝑖𝑛𝑐𝑒, 𝑔𝑖𝑣𝑒𝑛 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 𝑃
𝑃𝑥−2𝑥 2 2( 𝑥 + 𝑦 ) =
= [ ]
2 𝑃 𝑃−2𝑥
𝑌 = 2 –𝑥 = 2

Option ( b) is correct.
𝑃𝑥−2𝑥 2 𝑑𝐴 𝑃−4𝑥
(ii) A = 2 ⇒ = 2
𝑑𝑥
For maximum or minimum value of x
𝑑𝐴 𝑃−4𝑥
=0 ⇒ =0
𝑑𝑥 2
𝑃
⇒P – 4x =0 ⇒ x = 4

𝑑2 𝐴 𝑃
Also , 𝑑𝑥 2 = −2 < 0 𝑎𝑡 𝑥 = 4
∴ 𝐴 𝑖𝑠 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
Option (d) is correct.

EXERCISE

1 Q1.Read the following and answer the questions given :

On the request of villagers, a construction agency designs a tank with the
help of an architect. Tank consists of rectangular base with rectangular sides,
open at the top so that its depth is 2 m and volume is 8 m 3 as shown below:

125
 (i) If x and y represents the length and breadth of its rectangular base, then the
relation between the variables is
𝑥
(a) x + y = 8 (b) x . y = 4 (c) x + y = 4 (d) 𝑦 = 4
(ii) If construction of tank cost ₹70 per sq. meter for the base and ₹45 per square
meter or sides, then making cost ‘ C’ expressed as a function of x is
4 4
(a ) C = 80 + 80 ( 𝑥 + 𝑥 ) (b) C = 280x +280 ( 𝑥 + 𝑥 )
4 𝑥
( c ) C = 280 + 180( 𝑥 + ) (d) C = 70x + 70 (𝑥 + )
𝑥 4
(iii) The owner of a construction agency is interested in minimizing the cost ‘C’ of
whole tank, for this to happen the value of x should be
(a) 4 m (b) 3 m (c) 1 m (d) 2 m
Answer: (𝒊)𝒃 (𝒊𝒊) 𝒄 (𝒊𝒊𝒊) 𝒅

2 Read the following text and answer the following questions on the basis of the
same:
In a residential society comprising of 100 houses, there were 60 children between
the ages of 10 – 15 years. They were inspired by their teachers to start composting
to ensure that biodegradable waste is recycle, For this purpose, instead of each child
doing it for only his/her house, children convinced the Residents welfare association
to do it as a society initiative. For this they identified a square area in the local park.
Local authorities charged amount of ₹ 50 per square meter for space so that there is
no misuse of the space and Resident welfare association takes it seriously.
Association hired a labourer for digging out 250 m3 and he charged ₹400 x (depth)2.
Association will like to have minimum cost.

i).Let side of square plot is x m and its depth is h meters, then cost C for the pit is
50 12500 250 250
(a) ℎ + 400h2 (b) ℎ + 400h2 (c) ℎ + h2 (d) ℎ +400h2
𝑑𝑐
ii).Value of h (in m ) for which 𝑑ℎ = 0 is
(a) 1.5 (b) 2 (c) 2.5 (d) 3
iii). Value of x (in m) for minimum cost is
5
(a)5 (b)10√3 (c)5√5 (d) 10
Answer: (i) b (ii) c (iii) d

126
3 Read the following text and answer the following questions, on the basis of the
same:
P(𝑥) = -5x2 +125x + 37500 is the total profit function of a company, where 𝑥 is
the production of the company.

(i) What will be the production when the profit is maximum?

(a) 37,500 (b) 12.5 (c) -12.5 (d) -37,500
(ii) What will be the maximum profit?
(a) ₹38,28,125 (b) ₹38,28,.25 (c) ₹39,000 (d) None of these
(iii) Check in which interval the profit is strictly increasing.
(a) (12.5, ∞ ) (b) for all real numbers
(c) for all positive real numbers (d) (0,12.5)
Answer: (i) b (ii) b (iii) d

127