# 2nd PUC Maths Question Bank Chapter 6 Application of Derivatives

Students can Download 2nd PUC Maths Chapter 6 Application of Derivatives Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka 2nd PUC Maths Question Bank Chapter 6 Application of Derivatives

### 2nd PUC Maths Application of Derivatives Five Marks Questions and Answers

Derivative a rate measure.

Question 1.
Find the rate of change of the area of a circle with respect to its radius r when
(a) r = 3 cm
(b) r = 4 cm
Let A denotes the area of the circle when its radius is r, then A = πr2.
Now, the rate of change of the area with respect to its radius is given by,

Hence, the area of the circle is changing at the rate of 6 cm2 / cm when its radius is 3 cm.

Hence, the area of the circle is changing at the rate of 8 cm2 / cm when its radius is 4 cm.

Question 2.
The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm ?
Let x be the length of a side (edge), V be the volume and S be the surface area of the cube.
V = x3 and S = 6x2 (∴ Cube has six square faces, each of side x.)

Hence, if the length of the edge of a cube is 12 cm, then the surface area is increasing at the rate of $$\frac { 8 }{ 3 }$$cm2 /s.

Question 3.
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm. .
Let r be the radius of circle and A be the area of circle at any time t. Area of circle A = πr2.
Now, the rate of change of area A with respect to time t is given by

Hence, the rate at which the area of the circle is increasing, when the radius is 10 cm, is 60π cm2/s.

Question 4.
An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Let x be the length of an edge of cube and V be the volume of the cube. Then, V = x2
∴ Rate of change of volume w.r.t time

Hence, the volume of the cube is increasing at the rate of 900 cm3/s when the edge is 10 cm long.

Question 5.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/ s. At the instant when the radius of the circular wave is 8cm, how fast is the enclosed area increasing?
Let r be the radius of the circular wave and A be the area, then A = πr2
Therefore, the rate of change of area (A) with respect to time (t) is given by

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm2/s.

Question 6.
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
At any instant of time t let the radius of the circle be r and its circumference be C, then C = 2πr
⇒ Rate of increase of circumference $$\frac{d \mathrm{C}}{d t}$$ = 2π $$\frac{d \mathrm{r}}{d t}$$ = 2π (0.7) = 1.4π cm/s
Hence, rate of increase of circumference is 1.4π cm/s.

Question 7.
The length x of a rectangle is decreasing at the rate of 5 cm/min and the width y is increasing at the rate of 4 cm/min. When x = 8 cm and y = 6 cm, find the rate of change of
(a) the perimeter.
(b) the area of the rectangle.
At any instant of time t, let length, breadth, perimeter and area of the rectangle are y, P and A respectively, then
P = 2 (x + y) and A = xy …… (i)
It is given that $$\frac{d x}{d t}$$ = -5 cm/min and $$\frac{d y}{d t}$$ = 4 cm/min 6
(- ve sign shows that the length is decreasing)

(a) Now, P = 2(x + y). On differentiating w.r.t. t, we get .

Hence, perimeter of the rectangle is decreasing (- ve sign) at the rate of 2 cm/min.

(b) Here, area of rectangle A = xy. On differentiating w.r.t. t, we get

= 32 – 30 = 2 cm2/min
Hence, area of the rectangle is increasing at the rate of 2 cm2/min.
Note It rate of change is increasing, we take positive sign and if rate of change is decreasing, then we take negative sign.

Question 8.
A balloon which always remains spherical on inflation, is being inflated by pumping in 900 cubic cm of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15cm.
At any instant of time t let the radius of the balloon be r and its volume be V, then Volume of balloon V = (4/3) πr3
The balloon is being inflated at 900 cubic cm/s i.e., the rate of change of volume with respect to time is 900 cm2/s.
On differentiating w.r.t t, we get

Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is $$\frac{1}{\pi}$$cm/s.

Question 9.
A balloon which remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the radius is 10 cm.
Let r be the radius of spherical balloon and V be its volume.
Then, r = 10 cm and V = $$\frac{4}{3}$$ πr3,

Hence, the volume of the balloon is increasing at the rate of 400π cm3/cm.

Question 10.
A ladder 5 m long, is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 m/s. How fast is its height on the wall decreasing when the foot of the ladder is 4m away from the wall?
Let AB = 5m be the ladder and y be the height of the wall at which the ladder touches. Also,
let the foot of the ladder be at B whose distance from the wall is x.
Given that the bottom of ladder is pulled along the ground at 2 cm/s, so $$\frac{d x}{d t}$$ = 2m/ s
As we know that DABC is right angled, so by Pythagoras theorem,

[Negative sign shows that height of ladder on the wall is decreasing at the rate fo $$\frac { 8 }{ 3 }$$cm / s].

Question 11.
A particle move along the curve 6y = x3 + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

when x = 4,then 6y = (-4)3 + 2 ⇒ 6y = 64 + 2 ⇒ y = $$\frac{66}{6}$$ = 11
when x = 4, then 6y = (-4)2 + 2 ⇒ 6y = -64 + 2 ⇒ y = $$\frac{-62}{6}=\frac{-31}{3}$$
Hence, the required points on the curve are (4, 11) and $$\left(-4, \frac{-31}{3}\right)$$

Question 12.
The radius of an air bubble is increasing at the rate of $$\frac { 1 }{ 2 }$$cm / s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
The air bubble is in the shape of a sphere.
Let r be the radius of bubbLe and V be the volume of bubble at any time t.

Hence, the rate at which the volume of the bubble increases in 2π cm3/s.

Question 13
A ballon, which always remains spherical, has a variable diameter 3/2 (2x + 1). Find the rate of change of its volume with respect to x.

Question 14.
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 base. How fast height of the sand cone increasing when the height is 4 cm?
Let r be the radius, h be the height and V be the volume of sand cone at any time t.

Hence, when the height of the sand cone is 4cm, its height. is increasing at the rate of $$\frac{1}{48 \pi}$$ cm / s

Question 15.
The total cost C(x) in rupees associated with the production of x units of an item is given by C (x) = 0.007x3 – 0.003x2 + 15x + 4000.
Find the marginal cost when x = 17 units are produced.
∴ Marginal Cost (MC) = $$\frac{d \mathrm{C}}{d t}$$ = 0.007(3x2) – 0.003(2x) + 15
= 0.021x2 – 0.006x + 15
When x = 17, MC = 0.021(17)2 – 0.006(17) + 15
= 0.021(289) – 0.006(17) + 15 = 6.069 – 0.102 + 15 = 20.967
Hence, when 17 units are produced, the marginal cost is ₹ 20.967.

Question 16.
The total revenue in rupees received from the sale of x units of a product is given by R(x) = 13x2 + 26x + 15. Find the marginal revenue when x = 7.
∴ Marginal Revenue (MR) = $$\frac{d R}{d x}=\frac{d}{d x}$$ (13x2 + 26x + 15)
= 13 × 2x + 26 = 26x + 26
When x = 7, MR = 26(7)+ 26 = 182 + 26 = 208
Hence, the required marginal revenue is ₹ 208.

Question 17.
A man of height 2 metres walks at a uniform speed of 5 km/h away from a lamp post which is 6 metres high. Find the rate at which the length of his shadow increases.
In Fig. let AB be the lamp post, the lamp being at the position B and let MN be the man at a particular time t and let AM = l metres. Then, MS is the shadow of a man. Let MS = s metres.

Note that ∆MSN = ∆ASB

or AS = 3s (as MN = 2 and B = 6 (given)
Thus AM = 3s – s = 2s. But AM = l.
So l = 2s.
Therefore $$\frac{d l}{d t}=2 \frac{d s}{d t}$$
Since $$\frac{d l}{d t}$$ = 5 km/h.
Hence the length of the shadow increases at the rate of $$\frac{5}{2}$$ km / h

Question 18.
A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is tan -1(0.5). Water is poured into it at a constant rate of 5 cubic metre per hour. Find the rate at which the level of the’ water is rising at the instant when the depth of water in the tank is 4m.

Question 19.
A car starts from a point P at time t = 0 seconds and stops at point Q. The distance x, in metres, covered by it, in t seconds is given by x = t2 $$\left(2-\frac{t}{3}\right)$$
Find the time taken by it to reach Q and also find the distance between P and Q.
Let v be the velocity of the car at t seconds.

Now v = 0 at P as well as at Q and at P, t = 0. So, at Q, t = 4. Thus, the car will reach the point Q after 4 seconds. Also the distance travelled in 4 seconds is given by

Question 20.
A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/s. Find the rate at which its area is Increasing when radius is 3.2 cm.
Let r be the radius of the given disc and A be its area. Then .
A = πr2

Now approximate rate of increase of radius = dr = $$\frac{d r}{d t}$$ = ∆t = 0.05 cm / s.
Therefore the approximate rate of increase in area is given by

= 2π(3.2)(0.05) = 0.320π cm2 /s (r = 3.2cm)

### 2nd PUC Maths Application of Derivatives Three Marks Quesitions and Answers

(A) Decreasing and Increasing functions.

Question 1.
Show that the function given by f(x) = 3x + 17 is strictly increasing on R.
Let x1 and x2 be any two numbers in R, where x1 < x2.
Then, we have ,
f(x) = 3x + 17
⇒ f(x1) = 3x1 + 17

Hence, f is strictly increasing on R
f(x1) < f(x2)
⇒ 3x1 + 17 < 3x2 + 17
⇒ 3x1 < 3x2
⇒ x1 < x2
∴ f is strictly increasing in R.

Question 2.
Show that the function given by f(x) = e2x is strictly increasing on R.
Let x1 and x2 be any two numbers in R, where x1 < x2
Given,f(x) = e2x
Then, we have x1 < x2 ⇒ 2x1 < 2x2 ⇒ e2x1 < e2x2 ⇒ f(x1) < f(x2)
Hence, f is strictly increasing on R.

Question 3.
Find the intervals in which the function f given by f(x) = x2 – 4x + 6 is
(a) strictly increasing |
(b) strictly decreasing
We have f (x) = x2 – 4x + 6 f'(x) = 2x – 4 .
Therefore, f'(x) = 0 gives x = 2. Now the point x = 2 divides the real line into two disjoint
intervals namely (-∞, 2) and (2, ∞). In the interval (-∞, 2), f'(x) = 2x – 4 < 0 . Therefore, fis strictly decreasing in this interval. Also, in the interval (2, ∞) f'(x) > 0 and so the function f is strictly increasing in this interval.

Question 4.
Find the intervals in which the function/given by f(x) = 2x2 – 3x is
(a) strictly increasing
(b) strictly decreasing.
Given, f(x) = 2x2 – 3x
Differentiating w.r.t, x, f'(x) = 4x – 3
On putting f'(x) = 0, we get 4x – 3 = 0 ⇒ x = 3/4
The point x = 3/4 divides the real line into two disjoint intervals namely

Hence,/(x) is; stictly increasing on $$\left(\frac{3}{4}, \infty\right)$$
and strictly decreasing on $$\left(-\infty, \frac{3}{4}\right)$$

Question 5.
Find the intervals in which the function/given by f(x) = 2x3 – 3x2 – 36x + 7 is.
(a) strictly increasing,
(b) strictly decreasing?
Given, f(x) = 2x3 – 3x2 – 36x + 7
⇒ Differentiating w.r.t. x,
f'(x) = $$\frac{d}{d x}$$(2x3 – 3x2 – 36x + 7)
= 6x2 – 6x – 36
= 6 (x2 – x – 6)
= 6 (x – 3) (x + 2)
⇒ f'(x) = 6(x – 3) (x + 2)
On putting f'(x) = 0, we get 6(x – 3) (x + 2) = 0
⇒ x = 3 and – 2
which divides real line into three intervals namely (-∞,-2), (-2,3) and (3,∞).

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

Question 6.
Find the intervals in which the function f given by f (x) = 4x3 – 6x2 – 72x + 30 is
(a) strictly increasing
(b) strictly decreasing.
we have f (x) = 4x3 – 6x2 – 72x + 30
or f'(x) = 12x2 – 12x – 72
= 12(x2 – x – 6) = 12(x – 3)(x + 2)

Therefore f’(x) = 0 gives x = -2,3. The points x = -2 and x = 3 divides the real line into three disjoint intervals, namely (-∞,-2),(-2,3) and (3,∞).
In the intervals (-∞,-2) and (3,∞), f'(x) is positive while in the interval (-2,3). f”(x) is negative. Consequently, the function is strictly increasing ¡n the intervals (-∞, -2) and (3,∞) while the function is strictly decreasing in the interval (-2,3). However, f is neither increasing nor decreasing in R.

Question 7.
Show that the function given by f(x) = sin x is
(a) strictly increasing in $$\left(0, \frac{\pi}{2}\right)$$.
(b) strictly decreasing in $$\left(\frac{\pi}{2}, \pi\right)$$
(c) neither increasing nor decreasing in (0, π).
The given function is f(x) = sin x
On differentiating w.r.t. x, we get f'(x) = cos x

So, f'(x) is positive and negative in (0, π).
Thus, f(x) is neither increasing nor decreasing in (0, π).

Question 8.
Prove that the function given by f(x) = cosx is
(a) strictly decreasing in (0,π)
(b) strictly increasing in (π,2π) and
(c) neither increasing nor decreasing in (0,2π).
Note that f'(x) = -sinx
(a) Since for each x ∈ (0,π), sin x > 0 >we have f ‘(x) < 0 and so f is strictly decreasing in (0,π)
(b) Since for each x ∈ (π,2π), sin x < 0, we have f'(x) > 0 and so f is strictly increasing in (π, 2π)
(c) Clearly by (a) and (b) above, f is neither increasing nor decreasing in (0,2π).

Question 9.
Find the intervals in the function f is given by
f(x) = sin x + cos x, 0 ≤ x ≤ 2π is strictly increasing or strictly decreasing.
We have
f(x) = sinx + cosx
or f'(x) = cosx-sinx
Now f'(x) = 0 gives sinx = cosx which gives that x = $$\frac{\pi}{4}, \frac{5 \pi}{4}$$ as 0 ≤ x ≤ 2π.
The points x = $$\frac{\pi}{4}$$ and x = $$\frac{5 \pi}{4}$$ divide the interval (0,2π) into three disjoint intervals.

Question 10.
Find the interval in which the following functions are strictly increasing or decreasing:
(a) x2 + 2x – 5
(b) 10 – 6x – 2x2
(c) – 2x3 – 9x2 – 12x + 1
(d) 6 – 9x – x2
(e) (x + 1)3 (x – 3)3
(a) Let f(x) = x2 + 2x – 5 ⇒ f'(x) = 2x + 2 (By differentiating w.r.t. x)
Putting f'(x) = 0, we get 2x + 2 = 0 ⇒ 2x = -2 => x = — 1
x = -1 divides real line into two intervals namely (-∞,-1) and (-1,∞).

Therefore, f(x) is strictly increasing when x > – 1 and strictly decreasing when x < – 1.

(b) Let f(x) = 10 – 6x – 2x2 ⇒ f'(x) = 0 – 6 – 2.2x = – 6 – 4x
On putting f'(x) = 0, we get – 6 – 4x = 0 ⇒ x = $$\frac{-3}{2}$$
which divides real line into two intervals namely $$\left(-\infty, \frac{-3}{2}\right)$$ and $$\left(\frac{-3}{2}, \infty\right)$$

Hence, f is strictly increasing for x < $$-\frac{3}{2}$$ and strictly decreasign for x > $$-\frac{3}{2}$$

(c) Given, f(x) = – 2x3 – 9x2 – 12x + 1,
⇒ f(x) = – 2.3x2 – 9.2x – 12 = – 6x2 – 18x – 12 = 0
On putting f'(x) = 0, we get – 6x2 – 18x – 12 = 0
⇒ – 6 (x + 2 ) (x + 1) = 0
⇒ x = -2, -1
which divides real line into three intervals (-∞, -2), and (-2,-1) and (-1,∞).

Thereore, f(x) is strictly increasing in -2 < x < -1 and strictly decreasing for x < – 2 and x > – 1.

(d) Given, f(x) = 6 – 9x – x2 ⇒ f'(x) = – 9 – 2x
On putting f'(x) = 0, we get – 9 – 2x = 0 ⇒ x = –$$-\frac{9}{2}$$
which divides the real line in two disjoint intervals $$\left(-\infty, \frac{-9}{2}\right)$$ and $$\left(-\frac{9}{2}, \infty\right)$$

Therefore,f(x) is strictly increasing when x < $$-\frac{9}{2}$$ and strictly decreasing when x < $$-\frac{9}{2}$$

(e) Given, f(x) = (x + 1)3 (x – 3)3
On differentiating, we get
f'(x) = (x + 1)3.3 (x – 3)2.1 + (x – 3)3.3 (x + 1)2.1
= 3(x – 3)2 (x + 1)2 {{x + 1) + (x – 3)} = 3(x – 3)2 (x + 1)2 (2x – 2)
= 6(x – 3)2 (x + 1)2 (x – 1)
On putting f'(x) = 0, we get X = – 1, 1, 3.
Which divides real line into four disjoint intervals namely (-∞,-1), (-1, 1), (1, 3) and (3,∞).

Therefore, f(x) is strictly increasing in (1,3) and (3,∞) and strictly decreasing in (-∞, -1) and (-1,1).

Question 11.
Show that y = log (1+ x) – $$\frac{2 x}{2+2 x}$$, x > -1, ¡s an increasing function of x throughout its domain.
Given y = log (1 + x) – $$\frac{2 x}{2+2 x}$$

∴ y’ > 0 when x > -1 .
Hence, y is an increasing function throughout (x > -1) its domain.

Question 12.
Find the values of x for which y = [x(x – 2)]2 is and increasing function.
Given, y = [x(x – 2)]2
= [x2 – 2x]2
On differentiating wri. x, we get
$$\frac{d y}{d x}$$ = 2(x2 – 2x)$$\frac{d }{d x}$$(x2 – 2x) = 2(x2 – 2x)(2x – 2)= 4x(x – 2)(x – 1)
On putting $$\frac{d y}{d x}$$ = 0,
weget x = 0,1 and 2
which divides real line in two disjoint intervals are
(∞, 0), (0, 1), (1, 2) and (2, ∞).

Question 13.
Prove that y = $$\frac{4 \sin \theta}{(2+\cos \theta)}$$ – θ is an increasing on θ in $$\left[0, \frac{\pi}{2}\right]$$

Question 14.
Prove that the logarithmic function is strictly increasing on (0,∞).
Let f(x) = log x ⇒ f'(x) = $$\frac{1}{x}$$
When x ∈ (0,∞), f'(x) > 0. Therefore, f(x) is strictly increasing in (0,∞).

Question 15.
Prove that the function/given by f(x) = x2 – x + 1 is neither increasing nor decreasing strictly on (-1, 1).
Given, f(x) = x2 – x + 1 f'(x) = 2x – 1
On putting
f'(x) = 0, we get x = 1/2
x = $$\frac { 1 }{ 2 }$$ divides the given interval into two intervals as $$\left(-1, \frac{1}{2}\right)$$ and $$\left(\frac{1}{2}, 1\right)$$

∴ f’ (x) does not have same sign throughtout the interval (-1, 1).
Thus, f(x) is neither increasing nor decreasing strictly in the interval (-1,1).

(B) Tangents and Normals

Question 1.
Find the slope of the tangent to the curve y = 3x4 – 4x at x = 4.
The given curve is y = 3x4 – 4x
Defferentiating w.r.t. x.
$$\frac{d y}{d x}=\frac{d}{d x}$$[3x4 – 4x]
⇒ $$\frac{d y}{d x}$$ = 3 × 4x3 – 4 × 1 = 12x3 – 4
dx dx dx
∴ Slope of tangent at x = 4 is $$\left(\frac{d y}{d x}\right)_{(x=4)}$$ = 12 × (4)3 – 4 = 764

Question 2.
Find the slope of the tangent to the curve y = $$\frac{x-1}{x-2}$$,x ≠ 2 at x = 10.
y = $$\frac{x-1}{x-2}$$,x ≠ 2
On differentiating, we get

Hence, the slope of the tangent at x = 10 is $$-\frac{1}{64}$$

Question 3.
Find the slope of the tangent to curve y = x3 – x + 1 at the point whose x-coordinate is 2.
The given curve is y = x3 – x + 1
⇒ $$\frac{d y}{d x}$$ = 3x2 – 1
∴ Slope of tangent at the given point x = 2 is $$\left(\frac{d y}{d x}\right)_{(x=2)}$$ = 3(2)2 – 1 = 11
Thus, slope of tangent at x = 2 is 11.

Question 4.
Find the slope of the tangent to the curve y = x3 – 3x + 2 at, the point whose x coordinate is 3.
The given curve is y = x3 – 3x + 2
⇒ $$\frac{d y}{d x}$$ = 3x2 – 3 .
∴ Slope of tangent at the point x = 3 is $$\left(\frac{d y}{d x}\right)_{(x=3)}$$ = 3 × (3)2 – 3 = 24
Thus, slope of tangent at x = 3 is 24.

Question 5.
Find the slope of the normal to the curve x = a cos3 θ, y = a sin3 θ at θ = $$\frac{\pi}{4}$$
Given x = a cos3 θ and y = a sin3 θ.
On differentiation x and y both w.r.t. θ,we get

Question 6.
Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos2 θ at θ = $$\frac{\pi}{2}$$
It is given that x = 1 – a sin θ and y = b cos2 θ.
On differentiating x and y both w.r.t θ, we get

Question 7.
Find points at which the tangent to the curve y = x3 – 3x2 – 9x + 7 is parallel to the x- axis.
The equation of the curve is y = x3 – 3x2 – 9x + 7
∴ $$\frac{d y}{d x}$$ = 3x2 – 6x – 9
Now, the tangent is parallel to X-axis, then $$\frac{d y}{d x}$$ = 0
⇒ 3x2 – 6x – 9 = 0 3 ⇒ (x2 – 2x – 3 ) = 0
⇒ (x – 3) (x + 1) = 0 ⇒ x = 3, – 1
When x = 3, then from Eq. (i), we get
y = 33 – (3) (3)2 – 9.3 + 7 = 27 – 27 – 27 + 7 = – 20
Whenx = -1, then from Eq. (i), we get
y = (-1)3 – 3(-1)2 – 9(-1) + 7 = – 1 – 3 + 9 + 7= 12
Hence, the points at which the tangent is parallel to X-axis are (3, 20) and (-1,12).

Question 8.
Find the point on the curve y = (x – 2)2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
If the tangent is parallel to the chord joining the points (2,0) and (4,4), then the slope of the tangent = the slope of the chord

Equation of given curve is y = (x – 2)2
Now, the slope of the tangent to the given curve at a point (x, y) is given by
$$\frac{d y}{d x}$$ = 2(x – 2)
Now, from Eq. (i), we get 2(x – 2) = 2 ⇒ x – 2 = 1 ⇒ x = 3
When x = 3, then y = (3 – 2)2 = 1 [ ∴ Equation of curve is y = (x – 2)2]
Hence, the required point is (3,1).

Question 9.
Find the equations of the tangent and normal to the given curve at the given points.
(i) y = x4 – 6x2 + 13x2 – 10x + 5 at (0, 5)
(ii) y = x4 – 6x3 + 13x2 – 10x + 5 at (1, 3)
(iii) y = x3 at (1,1)
(iv) y = x2 at (0, 0)
(v) x = cos t, y – sin t at t = $$\frac{\pi}{4}$$
(i) The equation of the given curve is
y = x4 – 6x2 + 13x2 – 10x + 5
On differentiating w.r.t. x, we get $$\frac{d y}{d x}$$ – 4x3 – 18x2 + 26x – 10
∴ Slope of tangent at (0, 5) is $$\left(\frac{d y}{d x}\right)_{(0,5)}$$ = 0 – 0 + 0 – 10 = -10
Thus, the slope of the tangent at (0, 5) is – 10. Now, the equation of the tangent is y – 5 = – 10 (x – 0) ⇒ 10x + y – 5 = 0
Again, the slope of normal at (0, 5) is

Therefore, the equation of the normal at (0, 5) is
y – 5 = $$\frac { 1 }{ 10 }$$(x – 10) ⇒ x – 10 + 50 = 0.

(ii) The equation of the given curve is y = x4 – 6x3 + 13x2 – 10x + 5
On differentiating w.r.t. x, we get $$\frac{d y}{d x}$$ – 4x3 – 18x2 + 26x -10
∴ Slope of tangent at (1, 3) is $$\left(\frac{d y}{d x}\right)_{(1,3)}$$ = 4 – 18 + 26 – 10 = 2
Thus, the slope of the tangent at (1,3) is 2. Now, the equation of tangent is
y – 3 = 2 (x – 1) ⇒ y – 3 = 2x – 2 ⇒ y = 2x + 1
Again, the slope of normal at (1,3) is

Hence, the equation of the normal at (1, 3) is
y – 3 = $$\frac { -1 }{ 2 }$$(x – 1) ⇒ 2y – 6 = -x + 1 ⇒ x + 2y – 7 = 0

(iii) The equation of the given curve is y = x3
On differentiating w.r.t. x, we get $$\frac{d y}{d x}$$ = 3x2
∴ Slope of tangent at (1, 1) is $$\left(\frac{d y}{d x}\right)_{(1,1)}$$ = 3(1)2 = 3
Thus, the slope of the tangent at (1,1) is 3 and the equation of the tangent is
y – 1 = 3(x – 1) ⇒ y – 1 = 3x – 3 ⇒ 3 = 3x – 2
Again, the slope of the normal at (1,1)

Hence, the equation of the normal at (1,1) is
y – 1 = $$\frac { -1 }{ 3 }$$(x – 1)
⇒ 3y – 3 = -x + 1 ⇒ x + 3y – 4 = 0.

(iv) The equation of the given curve is y = x2
On differentiating w.r.t. x, we get $$\frac{d y}{d x}$$ = 2x
∴ Slope of tangent at (0, 0) is $$\left(\frac{d y}{d x}\right)_{(0,0)}$$ = 2 × 0 = 0
Hence, the equation of the tangent at (0, 0) having slope 0 is
y – 0 = 0(x – 0) ⇒ y = 0 (i.e., x-axis)

Therefore, the equation of the normal at (x0, y0) = (0,0) is given by x = 0. (i.e., x- axis)

(v) The equation of the given curve is x = cost, y = sin t …. (i)
The point on the curve corresponding to t = $$\frac{\pi}{4}$$ is

Question 10.
Find the sope of the tangent to the curve y = x3 – x at x = 2.
The slope of the tangent at x = 2 is given by:

Question 11.
Find the point at which the tangent to the curve y = $$\sqrt{4 x-3}-1$$ has its slope $$\frac { 2 }{ 3 }$$
Slope of the tangent to the given curve at (x, y) is

Therefore the required point is (3,2).

Question 12.
Find the point on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11.
The equation of the curve is y = x3 – 11x + 5 ….. (i)
the equation of the tangent to the given curve is y = mx + c (in general form)
Given line is y = x – 11
On comparing, we get slope of the tangent, m = 1
Now, the slope of the tangent to the given curve at the point (x, y) is
$$\frac{d y}{d x}$$ = 3x2 – 11
∴ 3x2 – 11 = 1 (Slope of given tangent = 11)
⇒ 3x2 – 11 = 1 ⇒ 3x2 =12 ⇒ x2 = 4 ⇒ x = ± 2
When x = 2, then from Eq. (i), we get y = 23 – 11 × 2 + 5 = -9
When x = -2, then from Eq. (i), we get y = (-2)3 – 11 (-2) + 5 = 19
Equation of tangent at point (2, – 9) is y – (-9) = 1 (x – 2) or y = x – 11.
But the point (-2, 19) does not lie on the line y = x – 11.

Question 13.
Find the equation for all lines having slope -1 that are tangent to the curve y = , x ≠ 1
The slope of the given curve is y = $$\frac{1}{x+1}$$, x ≠ 1 ……. (1)
The slope of the tangent to the given curve at any point (x, y) is given by

For tangents having slope = – 1, we must have

Thus, the points on the given curve at which slope of tangent is – 1, are (2, 1) and (0, -1).
∴ Equation of tangent at (2, 1) is y – 1 = -1 (x – 1) or x + y – 3 = 0
and equation fo tangent at (0, – 1) is y – (- 1) = -19x – 0 or x + y + 1 = 0
Hence, the equations of tfie required lines are x + y – 3 = 0 and x + y + 1 = 0.

Question 14.
Find the equation of all lines having slope 2 which are tangents to the curve y = $$\frac{1}{x-3}$$, x ≠ 3.
The equation of the given curve is y = $$\frac{1}{x-3}$$, x ≠ 3.
The slope of the tangent to the given curve at any point (x, y) is given by

For tangents having slope 2, we must have

Which is not possible as square of a real number cannot be negative.
Hence, there is no tangent to the given curve having slope 2.

Question 15.
Find the equation of all lines having slope 0 which are tangents to the curve y = $$\frac{1}{x^{2}-2 x+3}$$
The equation of the given curve is y = $$\frac{1}{x^{2}-2 x+3}$$ …….. (i)
The slope of the tangent to the given curve at dny point (JC, y) is given by

For tangents having slope 0, we must have $$\frac{d y}{d x}$$ = 0 .

∴ The equation of tangent to the given curve at point $$\left(1, \frac{1}{2}\right)$$ having slope = 0 is

Hence, the equation of the required line is y = $$\frac { 1 }{ 2 }$$

Question 16.
Find points on the curve $$\frac{x^{2}}{9}+\frac{y^{2}}{16}$$ = 1 at which the tangents are
(a) parallel X-axis
(b) parallel to Y-axis.
The equation of the given curve is $$\frac{x^{2}}{9}+\frac{y^{2}}{16}$$ = 1 ……. (i)
On differentiating both sides w.r.t. x, we get

(a) For tangent parallel to X-axis, we must have, $$\frac{d y}{d x}$$ = 0

When x = 0, then from Eq (i), we get

Hence, the points on Eq. (i) at which the tangents are parallel to X-axis are (0, 4) and (0, -4).

(b) For tangent parallel to Y-axis, we must have, $$\frac{d y}{d x}$$ = 0

Hence, the points on Eq. (i) at which the tangents are parallel to Y-axis are (3, 0) and (-3,0).

Question 17.
Find the equation of the tangent line to the curve y = x2 – 2x +7 which is
(a) parallel to the line 2x – y + 9 = 0.
(b) perpendicular to the line 5y – 15x = 13.
The equation of the given curve is y = x2 – 2x + 7 …….. (i)
On differentiating wtx, we get $$\frac{d y}{d x}$$ = 2x – 2

(a) The equation of the line is 2x – y + 9 = 0 ⇒ y = 2x + 9
This is of the form y = mx + c
∴ Slope of the line is m = 2
If a tangent is parallel to the line 2x – y + 9 = 0, then the slope of the tangent is equal to the slope of the line.
Therefore, we have $$\frac{d y}{d x}$$ = m ⇒ 2x – 2 = 2 ⇒ 2x = 4 ⇒ x = 2
When x = 2, then from Eq (i), we get y = 22 – 2 × 2 + 7 = 7
∴ The point on the given curve at which tangent is parallele to given line is (2,7) and the equation of the tangent is
y – 7 = 2(x – 2) ⇒ 2x – y + 3 = 0
Hence, the equation of the tangent line to the given curve which is parallel to line 2x – y + 9 = 0 is y – 2x – 3 = 0.

(b) The equation 6f the line is 5y – 15x = 13 ⇒ y = 3x + $$\frac{13}{5}$$
This is of the form y = mx + c
Slope of the line is 3.
If a tangent is perpendicular to the fine 5y – 15x = 13, then the slope of the tangent = $$-\frac{1}{3}$$

When x = $$\frac{5}{6}$$ then fipm Eq (i), we get

∴ The point on the given curve at which tangent is perpendicular to given line is $$\left(\frac{5}{6}, \frac{217}{36}\right)$$ and the equation of the tangent is

Hence, the equation of the tangent line to the given curve which is perpendicular to line 5y – 15x = 13 is 12x + 36y – 227 = 0.

Question 18.
Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = – 2 are parallel.
The equation of the given curve is y = 7xx3 + 11 …… (i)

It is observed that the slopes of the tangents at the points where x = 2 and x = -2 are equal.
Hence, the two tangents are parallel.

Question 19.
Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinate of the point.
The equation of the given curve is y = x3 …. (1)
∴ $$\frac{d y}{d x}$$ = 3x2 The slope of the tangent at the point (x, y) is given by $$\left(\frac{d y}{d x}\right)_{(x, y)}$$ = 3x2
When the slope of the tangent is equal to the y-coordinate of the point, then y = 3x2.
⇒ 3x2 = x3 (∵ y = x3 given) ⇒ x2 (3 – x) = 0 ⇒ x = 0 or = 3
When x = 0, then from Eq. (i), we get y = 0, = 0 ∴ The required point is (0, 0).
When x = 3, then from Eq. (i), we get y = 33 = 27
Hence, the required points are (0, 0) and (3,27).

Question 20.
For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.

Therefore, the slope of the tangent at point (x, y) is 12x2 – 10x4.
The equation of the tangent at (x, y) is given by .
y – y = (12x2 – 10x4) (x – x) ……. (i)
When the tangent passes through the origin (0,0),
then X = Y = 0
Therefore, Eq. (i) reduced to
– y = (12x2 – 10x4) (-x) ⇒ y = 12x3 – 10x5
Also we have y = 4x3 – 2x5
∴ 12x3 – 10x5 = 4x3 – 2x5
⇒ 8x5 – 8x3 = 0 x5 – x3 = 0
⇒ x3(x2 – 1) = 0 ⇒ x = 0, 1
When x = 0, y = 4(0)3 – 2(0)5 = 0.
When x = 1, y = 4(1)3 – 2(1)5 = 2
When x = – 1, y = 4 (-1)3 – 2 (-1)5 = – 2
Hence, the required points are (0, 0), (1,2) and (-1, -2).

(C) Approximations

Question 1.
Using differentials, find the approximate value of each of the following upto 3 places of decimal.

Question 2.
Find the approximate value of f(2.01), where f(x) = 4x2 + 5x + 2.
Consider f(x) = 4x2 + 5x + 2 ⇒ f'(x) = 8x + 5
Let x = 2 and Δx = 0.01
Also, f(x + ΔX) = f(x) + Δxf'(x)
∴ f(x + Δx) = (4x2 + 5x + 2) + (8x + 5)Δx
⇒ f (2 + 0.01) = (4 × 22 + 5 × 2 + 2) + (8 × 2 + 5)(0 01) (as x = 2, Δx = 0.01)
= 28 + 21 × 0.01 = 28.21 ⇒ f(2.01) = 28.21.

Question 3.
Find the approximate value of f(5.001), where f(x) = x3 – 7x2 + 15.
Consider f(x) = x3 – 7x2 + 15. ⇒ f'(x) = 3x2 – 14x
Let x = 5 and Δx = 0.001
Also, f(x + Δx) = f(x) + Δxf'(x)
Therefore, f(x + Δx) = (x3 + 7x2 + 15) + Δx(3x2 + 14x)
⇒ f(5.001) = (53 – 7 × 52 + 15) + (3 × 52 -14 × 5)(0.001) (as x = 5, Δx = 0.001)
= 125 – 175 + 15 +(75 – 70) (0.001) = -35 + (5) (0.001)
= -35 + 0.005 = -34.995.

Question 4.
Find the approximate change in the volume V of a cube of side x metre caused by increasing the side by 1%.
The volume (V) of a cube of side x is given by V = x3 ⇒ $$\frac{d V}{d x}$$ = 3x2
(Differentiate wr.t. x)

= 3x2 (0.01x) = 0.03x3 (as Δx = 1% of x = 0.01x) = 0.03 x3 m3

Question 5.
Find the approximate change in the surface area of a cube of side x metre caused by decreasing the side by 1%.
We know that the surface area of a cube is given by S = 6x2

= (12x) Δx = 12x (-0.01)x (as Δx = – 1% of x = – 0.01x)
= – 0.12 x2 m2
Hence, the approximate change in the surface area of the cube is of change is negative, then it is decreasing.

Question 6.
If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.
Let r be the radius of the sphere and Δr be the error in measuring radius.
Then, r = 7m and Δr – 0.02m.
Now, volume of a sphere is given by V = $$\frac { 1 }{ 2 }$$πr3

Hence, the approximate error in calculating the volume is 3.92 πm3

Question 7.
If the radius of sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.
Let r be the radius of the sphere and Δr be the error in measuring radius.
Then, r = 9m and Δr = 0.03m.

Hence, the approximate error in calculating the surface area is 2.16 πm3.

Question 8.
If f(x) = 3x2 + 15x + 5, then the approximate value of f(3.02) is
(a) 47.66
(b) 57.66
(c) 67.66
(d) 77.66
Cosnider f(x) = 3x2 + 15x + 5 ⇒ f'(x) = 6x + 15
Let x = 3 and Δx = 0.02. Also, f(x + Δx) = f(x) + Δxf'(x).
⇒ f(x + Δx) = (3x2 + 15x + 5) + (6x + 15)Δx
⇒ f(3.02) = 3 × 32 + 15 × 3 + 5 + (6 × 3 + 15) (0.02) (as x = 3, Δx = 0.02)
= 27 + 45 + 5 + (18 + 15) (0302) = 77 + 33(0.02) = 77 + 0.66
⇒ f(3.02) = 77.66
Hence, the approximate value of f(3.02) is 77.66. ∴ The correct option is (d).

Question 9.
Find the approximate change in the volume of a cube of side x metre caused by increasing the side by 3%.
We know that the volume V of a cube of side x is given by V = x3
⇒ $$\frac{d V}{d x}$$ = 3x2
Let x be change in side = 3% ofx = 0.03x
Now, change in volume, ΔV = $$\left(\frac{d V}{d x}\right)$$ Δx
= (3x2)Δx = (3x2)(0.03x) (as Δx = 3% of x is 0.03x)
= 0.09x3 m3

(D) Maxima and Minima

Question 1.
Find the maximum and minimum values, if any, of the following functions given by.
(a) f(x) = (2x – 1)3 + 3
(b) f(x) = 9x2 + 12x + 2
(c) f(x) = -(x – 1)2 + 10
(d) g (x) = x3 + 1.
(a) Given, function is f(x) = (2x – 1)3 + 3
It can be observed that (2x -1)2 ≥ 0 for every x ∈ R
[Since, it is perfect square for any real number]
Therefore, f'(x) = (2x – 1)2 + 3 ≥ 3 for every x ∈ R
The minimum value of f is attained when 2x – 1 = 0

For any value of x, value of f(x) will always be greater than 3, hence function f does not have particular maximum value.

(b) Given, function is f(x) = 9x2 + 12x + 2 = 9x2 + 12x + 4 – 2
[we add and subtract 2 for making it perfect square]
= (9x2 + 6x + 6x + 4) – 2 = [3x (3x + 2) + 2 (3x + 2)] – 2
= [(3x + 2)(3x + 2)] – 2 = (3x + 2)2 – 2
It can be observed that (3x + 2)2 ≥ 0 for every x ∈ R
Therefore, f(x) = (3x + 2)2 – 2 ≥ – 2 for every x ∈ R
The minimum value of f is attained when 3x + 2 = 0

For any value of x, f(x) ≥ – 2, hence function f does not have a particular maximum value.

(c) Given function is f(x) = -(x – 1)2 + 10
It can be observed that (x + 1)2 ≥ 0 for all x ∈ R
Therefore, f(x) = -(x + 1)2 + 10 ≤ 10 for every x ∈ R
The minimum value of f is attained when (x – 1) = 0
i.e., (x – 1) = 0 ⇒ x = 1
∴ Maximum value of f = f(1) = -(1 – 1)2 + 10 = 10
For any value of x, f(x) ≤ 10, hence function f does not have a particular minimum value.
Here, g(x) = x3 + 1, g’ (x) = 3x2, g”(x) = 6x
For maxima or minima put g'(x) = 0
⇒ 3x2 = 0 ⇒ x = 0
At x = 0, g”(0) = × 0 = 0
Hence at x = 0, g (x) is neither fnaxima nor minima. It is a point of inflexion.

Question 2.
Find the local maxima and local minima, if any of the following function. Also, find the local maximum and the local minimum values, as the case may be as follows:
(i) f(x) = x2
(ii) g(x) = x3 – 3x
(iii) h(x) = sin x + cos x, 0 < x < $$\frac{\pi}{2}$$
(iv) f(x) = sin x – cos x, 0 < x < 2π
(v) f(x) = x3 – 6x2 + 9x + 15
(vi) g(x) = $$\frac{x}{2}+\frac{2}{x}$$, x > 0
(i) Given function is f(x) = x2
∴ Differentiating w.r.t. x, f'(x) = 2x and again differentiating w.r.t. x, we get f”(x) = 2
For maxima or minima, put f'(x) = 0
∴ 2x = 0 ⇒ x = 0
Thus, x = 0 is the only critical point which could be possibly the point of local maxima or local minima of f.
We have f”(0) = 2 > 0 (which is positive)
Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is f(0) = x2 = 02 = 0.

(ii) Given function is g (x) = x3 – 3x
∴ g (x) = 3x2 – 3 and g”(x) = 6x
For maxima or minima, put g’(x) = 0
∴ 3x2 – 3 ⇒ x2 = 1 ⇒ x = +1
Thus, we expect extremum only at two points -1 and 1. ]
At x = – 1, g” (-1) = 6 (-1) = – 6 < 0
∴ g has a local maxima at x = – 1 and local maximum value = g(-1)
= (-1)3 – 3(-1) = -1 + 3 = 2
At x = 1, g”(1) – 6 × 1 = 6 > 0
g has a local minima at x = 1 and local minimum value = g(1)
= 13 – 3 × 1 = – 2

(iii) Given function is h (x) = sinx + cos x, 0 < x < $$\frac{\pi}{2}$$
∴ h'(x) = cos x – sin x and h”(x) = – sin x – cos x
For maxima or minima, put h'(x) = 0

Therefore, by second derivative test, x = $$\frac{\pi}{4}$$ is a point of local maxima and local maximum value of h at x = $$\frac{\pi}{4}$$ is

(vi) Given function is f(x) = sin x – cos x 0 < x < 2 π
∴ f'(x) = cos x + sin x and f”(x) = – sin x + cos x
For maxima or minima, put f'(x) = 0

(v) Given function is f(x) = x3 – 6x2 + 9x + 15
⇒ f'(x) = 3x2 – 12x + 9 and f”(x) = 6x – 12
For maxima or minima, put f'(x) = 0
⇒ 3x2 – 12x + 9 = 0
⇒ 3(x2 – 4x + 3) = 0 ,
⇒ 3(x2 – 3x – x + 3) = 0
⇒ 3[x(x – 3) – 1(x – 3)] = 0
⇒ 3(x – 1) (x – 3) = 0
⇒ x = 1 or x = 3
At x = 1, f”(1) = 6 × 1 – 12 = 6 – 12 = -6 < 0
∴ x = 1 is a point of maxima.
Maximum value = f(1) = (1)3 – 6(1)2 + 9(1) + 15 = 1- 6 + 9 + 15 = 19
At x = 3, f”(3) = 6 × 3 – 12 = 18 – 12 = 6 > 0
x = 3 is a point of minima
Minimum value = f(3) = 33 – 6 × 32 + 9 × 3 + 15
= 27 – 54 + 27 + 15 = 15.

Question 3.
Prove that the following functions do not have maxima or minima:
(a) f(x) = ex
(b) g (x) = log x
(c) It (x) = x3 + x2 + x + 1
(a) Given function is f(x) = ex ⇒ f’ (x) = ex
Now, if f'(x) = 0, then ex = 0. But, the exponential function can never assume 0 for any value of x.
Therefore, there does not exist any x ∈ R such that f'(x) = 0
Hence, function f does not have maxima or minima.

(b) Given function is g(x) = log x ⇒ g (x) = $$\frac{1}{x}$$
Since, log x is defined for a positive number x, then g'(x) > 0 for any x. Therefore, there does not exist any x ∈ R such that g'(x) = 0
Hence, function g does not have maxima or minima.

(c) Given function is h(x) = x3 + x2 + x + 1
⇒ h(x) = 3x2 + 2x + 1 = 0
Now, put h'(x) = 0 ⇒ 3x2 + 2x + 1 = 0

Therefore, there does not exist any x ∈ R such that h'(x) – 0
Hence, function h does not have maxima or minima.

Question 4.
Find the absolute maximum value and the absolute minimum value of the following functions in the given integer
(a) f(x) = x3, x ∈ [-2,2]
(b) f(x) = sin x + cos x, x ∈ [0,π]
(c) f(x) = 4x – $$\frac{1}{2}$$x2, x ∈ $$\left[-2, \frac{9}{2}\right]$$
(d) f(x) = (x – 1)2 + 3x, x ∈ [-3, 1]
(a) Given function is f(x) = x3 ⇒ f'(x) = 3x2
For maxima or minima put f'(x) = 0 ⇒ 3x2 = 0 ⇒ x = 0 ∈ [-2,2]
Now, we evaluate the value of f at critical point x = 0 and at end points of the interval [-2,2],
At x = 0, f(0) = 03 = 0
At x = – 2, f(-2) = (-2)3 = -8
At x = 2, f(2) = (2)3 = 8
Thus, absolute maximum value is 8 at x = 2 and absolute minimum value is -8 at x = -2.

(b) Given function is f (x) = sin x + cosx, ⇒ f'(x) = cos x – sin x
For maxima o rminima put f’(x) = 0 = cos x – sin x = 0 ⇒ $$\frac{\sin x}{\cos x}$$ = 1
⇒ tan x = 1 ⇒ x = $$\frac{\pi}{4}$$ ∈ [0, π]
Now, we evaluate the value of f at critical point x = $$\frac{\pi}{4}$$ and at the end points of the interval [0, π]

At x = 0, f(0) = sin 0 + cos 0 = 0 + 1 = 1
At x = π, f(π) = sin π + cos π = 0 – 1 = -l
Thus, absolute maximum value is √2 at x = $$\frac{\pi}{4}$$ and absolute minimum value is – 1 at x
= – π.

(c) Given function is f(x) = 4x – $$\frac { 1 }{ 2 }$$x2 ⇒ f'(x) = 4 – $$\frac { 1 }{ 2 }$$(2x) = 4 – x
For maxima or minima put f'(x) = 0,4 – x = 0 ⇒ x = 4 ∈ $$\left[-2, \frac{9}{2}\right]$$
Now, we evaluate the value of f at critical point x = 4 and at end points of the interval $$\left[-2, \frac{9}{2}\right]$$

Thus, absolute maximum value is 8 at x = 4 and absolute minimum value is – 10 at x = – 2.

(d) Given function is f(x) = (x – 1)2 + 3
∴ f(x) = 2(x – 1)
For maxima or minima put f’ (x) = 0 ⇒ 2(x – 1) = 0 ⇒ x = 1
Now, we evaluate the value of f at critical point x = 1 and at end points of the interval [-3,1].
At x = 1, f(1) = (1 – 1)3 + 3 = 3
At x = – 3, f(-3) = (-3 – 1)2 = 3 = 19
Thus, absolute maximum value is 19 at x = -3 and absolute minimum value is 3 at x = 1.

Question 5.
Find both the maximum value and the minimum value of 3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3].
Let f(x) = 3x4 – 8x3 + 12x2 – 48x + 25
⇒ f'(x) = 12x3 – 24x2 + 24x – 48 = 12 (x3 – 2x2 + 2x – 4)
= 12 {x2(x – 2) + 2(x – 2)} = 12 (x – 2) (x2 + 2)
For maxima or minima put f’(x) = 0
⇒ 12 (x – 2) (x2 + 2) = 0 ⇒ if x – 2 = 0 ⇒ x = 2 ∈ [0, 3]
and if, x2 + 2 = 0 ⇒ x2 = -2 ⇒ x = √-2
Hence, the only real root is x = 2 which is considered as critical point,
Now, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0,3].
At x = 2, f(2) = 3 × 24 – 8 × 23 + 12 × 22 – 48 × 2 + 25
= 48 – 64 + 48 – 96 + 25 = -39
At x = 0, f(0) = 0 – 0 + 0 – 0 + 23 = 25
At x = 3, f(3) = 3 × 34 – 8 × 33 + 12 × 34 – 48 × 3 + 25
= 243 – 216 + 108 – 144 + 25 = 16
Hence, we can conclude that the absolute maximum value of f is 25 at x = 0 and the absolute minimum value of f is – 39 at x = 2.

Question 6.
Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].
Let f(x) = 2x3 – 24x + 107
⇒ f'(x) = 6x2 – 24 = 6(x2 – 4) = 6(x + 2) (x – 2) .
For maxima or minima put f'(x) = 0 ⇒ 6(x + 2) (x – 2) = 0 ⇒ x = 2, -2
We first consider the interval [1, 3],
So, we have to evaluate the value of f at the critical point x = 2 ∈ [1,3] and at the end points of [1,3].
At x = 1, f(1) = 2 × 13 – 24 × 1 + 107 = 85
At x = 2, f(2) = 2 × 23 – 24 × 2 + 107 = 75
At x = 3, f(3) = 2 × 33 – 24 × 3 + 107 = 89
Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3.
Next, we consider the function in the interval [-1, -3], then evaluate the value of f at the critical point
x = -2 ∈ [-3,-1] and at the end points of the interval [-3,-1].
At x = – 1, f(- 1) = 2(-1)3 – 24(-1) + 107 = -2 + 24 + 107 = 129
At x = – 2, f(- 2) = 2(- 2)3 – 24(- 2) + 107 = – 16 + 48 + 107 = 139
At x = – 3, f(-3) = 2(-3)3 – 24(-3) + 107 = -54 + 72 + 107 = 125
Hence, the absolute maximum value of f(x) in the interval [-3, -1] is 139 occurring at x = -2:

Question 7.
It is given that at x = 1, the function x4 – 62x2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.
Let f(x) = x4 – 62x2 + ax + 9 ⇒ f(x) = 4x3 – 124x + a
It is given that function f attains its maximum value on the interval [0, 2] at x = 1.
∴ f'(1) = 0
⇒ 4 × 13 -124 × 1 + o = 0
⇒ 4 – 124 + a = 0 ⇒ a = 120
Hence, the value of a is 120.
Hence, the value of a is 120.

Question 8.
Find two numbers whose sum is 24 and whose product is large as possible
Let one number be x. Then, the other number is (24 – x). (∵ Sum of the two numbers is 24)
Let y denotes the product of the two numbers. Thus, we have
y = x (24 – x) = 24x – x2
On differentiating twice w.r.t. x, we get

∴ By second derivative test, x = 12 is the point of local mxima of y. Thus, the product of the number is maximum when the numbers are 12 and 24 – 12 -= 12.
Hence, the numbers are 12 and 12.

Question 9.
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.
Let the two numbers be x, y and P = xy3
Given x + y = 60 ⇒ x = 60 – y
On putting this value in P = xy3, we get
P = (60 – y)y3 ⇒ P = 60y3 – y4
On differentiating twice w.r.t. y, we get

⇒ P has a local maxima at y = 45
∴ By second derivative test, x = 45 is a point of local mxima of P. Thus, the function xy3 is maximum when y = 45 and x = 60 – 45 = 15.
Hence, the required numbers are 15 and 45.

Question 10.
Find two positive numbers x and y such that their siun is 35 and the product is x2y5 is maximum.
Let the numbers be JC and y and P = x2y5 then x + y = 35
⇒ x = 35 – y, ∴ P = (35 – y)2y5
On differentiating twice w.r.t. y, we get
$$\frac{d P}{d y}$$ = (35 – y)25y4 + y52(35 – y)(-1)
= y4(35 – y)[5(35 – y) – 2y] = y4 (35 – y)(175 – 5y – 2y)
= y4(35 – y)(175— 7y) = (35y4 – y5) (175 – 7y)
and $$\frac{d^{2} P}{d y^{2}}$$ = (35y4 – y5(-7) + (175 – 7y)(4 × 35 × y3 – 5y4)
= -7y4(35 – y) + 7(25 – y) × 5y3(28 – y)
= -7y4(35 – y) + 35y3(25 – y) (28 – y)
For maximua put $$\frac{d \mathrm{P}}{d y}$$ = 0
⇒ y4(35 – y)(175 – y) = 0 ⇒ y = 0, 35 – y = 0, 175 – 7y = 0
⇒ y = 0, y = 25, y = 35
When y = 0, x = 35 – 0 = 35 and the product x2y5 will be 0.
When y = 35 and x = 35 – 35 = 0. This will make the product x2y5 equal to 0.
∴ y = 0 and y = 35 cannot be the possible value of y.
When y = 25,
$$\left(\frac{d^{2} P}{d y^{2}}\right)_{y=25}$$ = -7 × (25)4 × (35 – 25) + 35 × (25)3 × (25 – 25)(28 – 25)
= -7 × 390625 × 10 + 35 × 15625 × 0 × 3
= -27343750 + 0 = -27343750 < 0
∴ By second derivative test, P will be the maximum wheny = 25 and x = 35 – 25 = 10. Hence, the required numbers are 10 and 25.

Question 11.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Let one number is x. Then, the other number will be (16 – x).
Let the sum of the cubes of these numbers be denoted by S.
Then, S = x3 + (16 – x)3
On differentiating w.r.t. x, we get
$$\frac{d S}{d x}$$ = 3x2 + 3(16 – x)2 (-1) = 3x2 – 3(16 – x)2
⇒ $$\frac{d^{2} S}{d x^{2}}$$ = 6x + 6(16 – x) = 96
For minima put $$\frac{d S}{d x}$$ = 0
⇒ 3x2 – 3(16 – x)2 = 0 ⇒ x2 – (256 + x2 – 32x) = 0
⇒ 32x = 256 ⇒ x = 8
At x = 8 $$\left(\frac{d^{2} S}{d x^{2}}\right)_{x=8}$$ = 96 > 0
∴ By second derivative test, x = 8 is the point of local minima of S.
Thus, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 – 8 = 8.
Hence, the required numbers are 8 and 8.

Question 12.
A square piece of tin of side 18 cm is to be made into a box without top, by cutting – off square from each corner and folding up the flaps of the box. What should be the side of the square to be cut off so that the volume of the box is maximum possible.
Let the side of the square to be cut-off be x cm (0 < x < 9).
Then, the length and the breadth of the box will be (18 – 2x) cm each and the height of the box is x cm.
Let V the volume of the open box formed by folding up the flaps, then
V = x (18 – 2x) (18 – 2x)
= 4x (9 – x)2 = 4x (81 + x2 – 1 8x)
= 4 (x3 – 18x2 + 81x)
On differentiating twice w.r.t. x, we get

$$\frac{d V}{d x}$$ = 4(3x2 – 36x + 81) = 12(x2 – 12x + 27) = 0
and $$\frac{d^{2} V}{d x^{2}}$$ = 12(2x – 12) = 24(x – 6)
For maxima put $$\frac{d V}{d x}$$ = 0 ⇒ 12(x2 – 12x + 27) = 0
⇒ x2 – 12x + 27 = 0 = 0 ⇒ (x – 3) (x – 9) = 0 ⇒ x = 39
But x = 9 is not possible.
∵ 2x = 2 × 9 = 18
which is equal to side of square piece.
At x = 3, $$\left(\frac{d^{2} V}{d x^{2}}\right)_{x=3}$$ = 24(3 – 6) = -72 < 0
∴ By second derivative test, x = 3 is the point of local minima.
Hence, if we cut off the side 3 cm from each comer of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.

Question 13.
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting-off square from each corner and folding up the flaps. What should be the side of the square to be cut-off so that the volume of the box is maximum?
Let the side of the square to be cut-off be x cm. Then, the hight of the box is x, the length is 45 – 2x and the breadth is 24 – 2x.
Let V be the corresponding volume of the open box then,
V = x (24 – 2x) (45 – 2x)
⇒ V = x (4x2 – 138x + 1080) = 4x3 – 138x2 + 1080x
On differentiating twice w.r.t. x, we get

$$\frac{d \mathrm{V}}{d x}$$ = 12x2 – 276x + 1080x dx
and $$\frac{d^{2} V}{d x^{2}}$$ = 24x – 276
For maxima put $$\frac{d \mathrm{V}}{d x}$$ = 0
⇒ 12x2 – 276x + 1080 = 0
⇒ x2 – 23x + 90 = 0
⇒ (x- 18) (x – 5) = 0 ⇒ x = 5, 18
It is not possible to cut-off a square of side 18 cm from each comer of the rectangular sheet. Thus, x cannot be equal to 18.
At x = 5, $$\left(\frac{d^{2} V}{d x^{2}}\right)_{x=5}$$ = 24 × 5 – 276 = 120 – 276 = -156 < 0
∴ By second derivative test, x = 5 is the point of maxima.
Hence, the side of the square to be cut-off to make the volume of the box maximum possible is 5cm