NCERT Solution - Application of Derivative JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : NCERT Solution - Application of Derivative JEE Notes | EduRev

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Application of Derivative

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

ANSWER : - The area of a circle (A)with radius (r) is given by,

Now. the rate of change of the area with respect to its radius is given by. NCERT Solution - Application of Derivative JEE Notes | EduRev

1.  when r = 3 cm,

dA/dr = 2π(3) = 6π

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

2. When r = 4 cm.

dA/dr = 2π(4) = 8π

Hence, the area of the circle is changing at the rate of 8π 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?

ANSWER : - Let x be the length of a side, V be the volume, and s be the surface area of the cube.

Then. V = x3 and S = 6x: where x is a function of time t.

It. is siven that  (dV/dt) -8 cm3/s

Then, by usine the chain rule, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of 8/3cm2/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.

ANSWER : - The area of a circle (A) with radius (r) is given by,

A = πr2

Now, the rate of change of area (A) with respect to time (t) is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev [By chain rule]

It is given that,

dr/dt = 3cm/s

∴ dA/dt = 2πr(3) = 6πr

Thus, when r = 10 cm,

dA/dt = 6π(10) = 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?

ANSWER : - Let x be the length of a side and V be the volume of the cube. Then,

NCERT Solution - Application of Derivative JEE Notes | EduRev

It is given that,

dx/dt = 3cm/s

∴ dV/dt = 3x2 (3) = 9x2
 

Thus, when x = 10 cm,

dV/dt = 9(10)2 = 900cm3/s

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 8 cm, how fast is the enclosed area increasing?

ANSWER : - The area of a circle (A) with radius (r) is given by

Therefore, the rate of change of area (A) with respect to time (t) is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev  [By chain rule]

It is given that, dr/dt = 5 cm/s

Thus, when r = 8 cm,

dA/dt = 2π (8) (5) = 80π

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 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?

ANSWER : - The circumference of a circle (C) with radius (r) is given by

C= = 2πr.

Therefore, the rate of change of circumference (C) with respect to time (t) is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev
It is given that  dr/dt = 0.7 cm/s

Hence, the rate of increase of the circumference  is 2π (0.7) = 1.4π cm/s.

Question 7: The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

ANSWER : - Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4 cm/minute, we have:

dx/dt = -5 cm/min and dy/dt = 4 cm/min

(a) The perimeter (Pi of a rectangle is given by.

P = 2(x+y)

NCERT Solution - Application of Derivative JEE Notes | EduRev

(b) The area (A) of a rectangle is given by,

A = x.y

NCERT Solution - Application of Derivative JEE Notes | EduRev

When x = 8 cm andy = 6 cm. 

 

dA/dt = (-5 x 6 +4 x 8) cm2/ min = 2 cm2 / min

Question 8: A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

ANSWER : - The volume of a sphere (V) with radius (r) is given by,

V = 4/3πr3

∴ Rate of change of volume (F) with respect to time (t) is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, when radius =15 cm.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is 1/π cm/s.

Question 9: A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

ANSWER : - The volume of a sphere (V) with radius (r) is given by V = 4/3 πr2

Rate of change of volume (V) with respect to its radius (r) is given by

NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, when radius =10 cm.

dV/dr = 4π(10)= 400π

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

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 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

ANSWER : - Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x maway from the wall.

Then, by Pythagoras theorem, we have:

x2 + y2 = 25 [Length of the ladder = 5 m]

NCERT Solution - Application of Derivative JEE Notes | EduRev

Then, the rate of change of height (y) with respect, to time (t) is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Now., when x = 4 m. wre have:

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

Question 11: A particle moves along the curve . Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

ANSWER : - The equation of the curve is given as:

NCERT Solution - Application of Derivative JEE Notes | EduRev

The rate of change of the position of the particle with respect to time (t) is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

When the y-coordinate of the particle changes 8 times as fast as the

x-coordinate i.e., NCERT Solution - Application of Derivative JEE Notes | EduRev , we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

WhenNCERT Solution - Application of Derivative JEE Notes | EduRev

When NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the points required on the curve are (4,11) and  NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 12: The radius of an air bubble is increasing at the rate of ½ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

ANSWER : - The air bubble is in the shape of a sphere.

Now, the volume of an air bubble (V) with radius (r) is given by,

V = 4/3πr3

The rate of change of volume (V) with respect to time (t) is given by..

NCERT Solution - Application of Derivative JEE Notes | EduRev[By chain rule]

NCERT Solution - Application of Derivative JEE Notes | EduRev

It iis given that  dr/dt = 1/2 cm/s

Therefore , when r = 1 cm,

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

Question 13: A balloon, which always remains spherical, has a variable diameter 3/2 x1 Find the rate of change of its volume with respect to x.

ANSWER : - The volume of a sphere (V) with radius (r) is given by,

V = 4/3πr3

It is given that:

Diameter NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the rate of change of volume with respect to x is as

NCERT Solution - Application of Derivative JEE Notes | EduRevNCERT Solution - Application of Derivative JEE Notes | EduRev

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 radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

ANSWER : - The volume of a cone (V) with radius (r) and height (h) is given by,

V = 1/3πr2h
It is riven that.

h= 1/6r = r=6h

∴ V = 1/3π(6h)2h = 12πh3

The rate of change of volume with respect to time (t) is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

It is also given that NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 15: The total cost (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 17 units are produced.

ANSWER:-Marginal cost is the rate of change of total cost with respect to output.

∴ Marginal cost (MC)  NCERT Solution - Application of Derivative JEE Notes | EduRev

= 0.021x2 - 0,006x + 15
When x = 17, MC = 0.021 (172)- 0.006 (17) + 15
= 0.021(289)-0.006(17)+ 15
= 6.069- 0.102+ 15
= 20.967

Hence, when 17 units are produced, the marginal cost is Rs. 20.967.

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.

ANSWER : - Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

∴ Marginal Revenue (MR)  = dR/dx = 13(2x) +26 = 26x+26

When x = 73
MR = 26(7) + 26 = 182 + 26 = 208
Hence, the required marginal revenue is Rs 208.

Question 17: The rate of change of the area of a circle with respect to its radius r at r = 6 cm is

(A) 10π    (B) 12π   (C) 8π    (D) 11π

ANSWER: - The area of a circle (A) with radius (r) is given by,
A = πr2
Therefore, the rate of change of the area urith respect to its radius r is

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the required rate of change of the area of a circle is 12π cm;. s.

The correct answer is B.

Question 18:

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

Answer 18:

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

∴ Marginal Revenue (MR)= dR/dx= 3(2x) + 36 = 6x + 36

∴ When x = 15,

MR = 6(15) + 36 = 90 + 36 = 126

Hence, the required marginal revenue is Rs 126.

The correct answer is D.

Question 19: Show that the function given by f(x) = 3x  17 is strictly increasing on R.

ANSWER : - Let  be any two numbers in R.

Then, we have:

ANSWER : - Let x1 be x2 any two numbers in R.

Then, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, f is strictly increasing on R.

Question 20: Show that the function given by/(x) = e2x is strictly increasing on R. 

ANSWER : - Let x1 and x2 be any twro numbers in R.

Then, we have:  NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, f is strictly increasing on R.

Question 21: Show that the function given by f(x) = sin x is

(a) strictly increasing in NCERT Solution - Application of Derivative JEE Notes | EduRev (b) strictly decreasing in NCERT Solution - Application of Derivative JEE Notes | EduRev

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

ANSWER : - The given function is f(x) = sinx.

NCERT Solution - Application of Derivative JEE Notes | EduRev

(a) Since for each  NCERT Solution - Application of Derivative JEE Notes | EduRev we have  NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence,/is strictly increasing in NCERT Solution - Application of Derivative JEE Notes | EduRev

(b) Since for each NCERT Solution - Application of Derivative JEE Notes | EduRev, we have NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence,f is strictly decreasing in NCERT Solution - Application of Derivative JEE Notes | EduRev

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

Question 22: Find the intervals in which the function f given by f(x) = 2x2 − 3x is

(a) strictly increasing
 (b) strictly decreasing

ANSWER: - The given function is f(x) = 2x2 - 3x.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Now. the point 3/4 divides the real line into two disjoint intervals i.e.. NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the given function (f) is strictly decreasing in interval NCERT Solution - Application of Derivative JEE Notes | EduRev

In interval  NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the given function (J) is strictly increasing in interval NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 23: Find the intervals in which the function/given by/x) = 2x3 - 3x2 - 36x+ 7 is

(a) strictly increasing    
(b) strictly decreasing

ANSWER: - The given function is  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

The points x = -2 andx = 3 divide the real line into three disjoint intervals i.e..

NCERT Solution - Application of Derivative JEE Notes | EduRev

In intervals (-∞, -2) and (3, ∞) , f'(x) is positive while in interval (-2,3), f'(x) is negative.

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

Question 24: Find the intervals 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

ANSWER:-(a) We have,

f(x) = x2 +2x-5

∴ f'(x) = 2x+2

Now,

f'(x) = 0 ⇒ x = -1

Point x = -1 divides the real line into two disioint. intervals i.e.. (-∞, -1) and (-1, ∞) 

In interval  NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ f is strictly decreasing in interval (-∞, -1).
Thus, f is strictly decreasing for x < -1.
In interval NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ f is strictly increasing in interval (-1, ∞) 
Thus, f is strictly deere asing for x < - 1.

(b) We have,

NCERT Solution - Application of Derivative JEE Notes | EduRev

The point  x = -3/2 divides the real line into two disjoint intervals i.e., (-∞, -3/2) and (-3/2, ∞)

In interval  (-∞, -3/2) i.e , when x < -3/2, f'(x) = -6-4x<0

∴ f is strictly increasing for x< -3/2

In interval  (-3/2, ∞) i.e , when x > -3/2, f'(x) = -6-4x<0 

∴ f is strictly decreasing for  x> -3/2
(c) We have

f(x) = -2x3 - 9x2 - 12x +1

∴ f'(x) = -6x2 - 18x - 12 = -6(x2+3x+2) = -6(x+1)(x+2)

Now,

f'(x) = 0 ⇒ x = -1 and x = -2

 Points x = - 1 and x = -2 divide the real line into three disjoint intervals i.e...

(-∞, -2),(-2, -1) and (-1, ∞)

In intervals  (-∞, -2) and (-1, ∞) he., when x < - 2 and x > - 1,

f'(x) = -6(x+1)(x+2) < 0

∴ f is strictly de ere as ing for x < - 2 and x > -1.

Now. in interval (-2, -1) i.e.. when - 2 < x < - 1.

f'(x) = -6(x+1)(x+2) > 0

 ∴ f is strictly increasing for 2 < x < 1

(d) We have.

f(x) = 6-9x = x2

 ∴  f'(x) = -9 -2x

Now, f'

(x) = 0 gives x = -9/2

The point  x = -9/2 divides the real line into two disjoint intervals i.e., (-∞, -9/2) and (-9/2,∞ )

In interval  (-∞, -9/2) i.e., for  NCERT Solution - Application of Derivative JEE Notes | EduRev 

∴ f is strictly increasing for x < -9/2

In interval  (-9/2,∞ ) i.e. for  NCERT Solution - Application of Derivative JEE Notes | EduRev 

∴ f is strictly decreasing for NCERT Solution - Application of Derivative JEE Notes | EduRev

(e) We have.

f(x) = (x+1)3 (x-3)3

f'(x) = 3(x+1)2(x-3)3 +3(x-3)(x+1)3

= 3(x+1)2(x-3)2 [x-3+x+1]

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

=6 (x+1)2(x-3)(x-1)

The points x = -1,x = 1. and x = 3 divide the real line into four disjoint intervals i.e., (-∞, -1) (-1,1), (1,3), and (3, ∞)

In intervals  (-∞, -1) and (-1,1), f'(x) = 6(x+1)2(x-3)2(x-1) <0

∴ f is strictly decreasing in intervals  (-∞, -1) and (-1,1)

In intervals NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ f is strictly increasing in intervals  (1,3), and (3, ∞)

Question 25: Show that NCERT Solution - Application of Derivative JEE Notes | EduRev , is an increasing function of x throughout its domain.

ANSWER:- We have.

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Since x > - 1. pointx =0 divides the domain (-1 ∞) in two disjoint intervals i.e.. -1 < x < 0 & x > 0.

When -1 < x < 0. we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, funetion f is increasing throughout this domain.

Question 26: Find the values of x for which  y = [x(x-2)]2 is an increasing function.

ANSWER:- We have.

NCERT Solution - Application of Derivative JEE Notes | EduRev

The points x = 0.x = 1. andx = 2 divide the real line into four disjoint intervals

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ y is strictly decreasing in intervals  (-∞ , 0) and (1,2)

However, in intervals (0. 1) and (2, ∞), dy/dx > 0

∴ y is strictly increasing in intervals (0,1) and (2, ∞).
∴ y is strictly increasing for 0 < x < 1 and x > 2.

Question 27: Prove that  NCERT Solution - Application of Derivative JEE Notes | EduRev is an increasing function of θ in [0, π/2]

ANSWER:- We have.

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

In interval  NCERT Solution - Application of Derivative JEE Notes | EduRev we have cos NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Tli ere fore.}' is strictly increasing in interval (0, π/2)

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

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

Question 28: Prove that the logarithmic function is strictly increasing on (0, ∞).

ANSWER : -

The given function is f (x) = logx.

∴ f'(x) = 1/x

It is clear that for x > 0,  f'(x) = 1/x >0 .

Hence f(x) = log x is strictly increasing in interval (0;∞).

Question 29: Prove that the function/given by f(x) = x2 - x + 1 is neither strictly increasing nor strictly decreasing on (-1. 1).

ANSWER: - The given function is f(x) = x2 - x + 1.

NCERT Solution - Application of Derivative JEE Notes | EduRev

The point 1/2 divides the interval (-1. 1) into two disjoint intervals i.e.. NCERT Solution - Application of Derivative JEE Notes | EduRev

Now. in interval  NCERT Solution - Application of Derivative JEE Notes | EduRev

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

However, in interval  NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, f is strictly increasing in interval  (1/2,1)

Hence. f is neither strictly increasing nor decreasing in interval (-1.1).

Question 30: Which of the following functions are strictly decreasing on (0,π/2) ?

(A) cos x    
(B) cos2x    
(C) cos 3x    
(D) tanx

ANSWER :

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev is strictly decreasing in interval NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev is strictly decreasing in interval NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev 

NCERT Solution - Application of Derivative JEE Notes | EduRev

The point  NCERT Solution - Application of Derivative JEE Notes | EduRev divides the interval NCERT Solution - Application of Derivative JEE Notes | EduRev 1 into two disjoint intervals i.e., 0 NCERT Solution - Application of Derivative JEE Notes | EduRev

Now, in interval  NCERT Solution - Application of Derivative JEE Notes | EduRev
f3 is strictly decreasing in interval NCERT Solution - Application of Derivative JEE Notes | EduRev

However, in interval  NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ f3 is strictly increasing in interval NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, fis neither increasing nor decreasing in interval NCERT Solution - Application of Derivative JEE Notes | EduRev.

  NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ f4 is strictly increasing in interval NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, functions cos x and cos 2.x are strictly decreasing in NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the correct answers are A andB.

Question 31: On which of the following intervals is the function f given by NCERT Solution - Application of Derivative JEE Notes | EduRev strictly decreasing?

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
(D) ) None of these

ANSWER:-We have,

NCERT Solution - Application of Derivative JEE Notes | EduRev

In interval  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, function fis strictly increasing in interval (0,1).

In interval  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, function f is strictly increasing in interval  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ f is strictly increasing in intervalNCERT Solution - Application of Derivative JEE Notes | EduRev .

Hence, function f is strictly decreasing in none of the intervals.

The correct answer is D.

Question 32: Find the least value of a such that the function f given f(x) = x2  ax 1  is strictly increasing on (1, 2).

Answers : We have

NCERT Solution - Application of Derivative JEE Notes | EduRev

Now, function f will be increasing in  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, we have to find the least value of a such that

NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, the least value of a for/to be increasing on (1? 2) is given by.

-a/2 =1

-a/2 =1 ⇒ a = -2

Hence, the required value of a is -2.

Question 33: Let I be any interval disjoint from (−1, 1). Prove that the function f given by f(x) = x 1/x.  is strictly increasing on I.

ANSWER : - We have,

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

The points x = 1 andx = -1 divide the real line in three disjoint intervals i.e.,  NCERT Solution - Application of Derivative JEE Notes | EduRev

In interval (-1,1). it is observed that:

NCERT Solution - Application of Derivative JEE Notes | EduRev

In intervals  NCERT Solution - Application of Derivative JEE Notes | EduRev , it is observed that:

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ f is strictly increasing on  (-∞, 1) and (1, ∞) 

Question 34: Prove that the function f given by f(x) = log sin x is strictly increasing on (0,π/2)  and strictly decreasing on (π/2, π)

Answer:-

NCERT Solution - Application of Derivative JEE Notes | EduRev

In interval   NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ f  is strictly decreasing in NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 35: Prove that the function f given by f (x) = log cos x is strictly decreasing on  NCERT Solution - Application of Derivative JEE Notes | EduRev  and strictly increasing on NCERT Solution - Application of Derivative JEE Notes | EduRev

ANSWER We have, 

NCERT Solution - Application of Derivative JEE Notes | EduRev
∴ f is strictly decreasing on NCERT Solution - Application of Derivative JEE Notes | EduRev

In interval  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ f is strictly increasing on NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 36: Prove that the function given by NCERT Solution - Application of Derivative JEE Notes | EduRev is increasing in R.

ANSWER:-We have,

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, f'(x) is always positive in R.
Hence, the given function (f) is increasing in R.

Question 37: The interval in which y= x2 e-x is increasing is

(A) (-∞,∞)    
 (B) (-2,0)    
 (C) (2,∞)    
 (D) (0,2)

ANSWER:- We have.

NCERT Solution - Application of Derivative JEE Notes | EduRev

The points x = 0 and x = 2 divide the real line into three disjoint intervals i.e.;

NCERT Solution - Application of Derivative JEE Notes | EduRev

In interv als  NCERT Solution - Application of Derivative JEE Notes | EduRev is always positive.

∴ f is decreasing on (-∞,0) and (2, ∞)

In interval (0,2),  NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ f is strictly increasing on (0.2).
Hence f is strictly increasing in interval (0,2).
The correct answer is D.

Quetion 38: Find the slope of the tangent to the curve y = 3x4 - 4x at x = 4. 

ANSWER: - The given curve is y = 3x4 - 4x.
Then, the slope of the tangent to the given curve at x = 4 is given by;

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 39: Find the slope of the tangent to the curve NCERT Solution - Application of Derivative JEE Notes | EduRev , x ≠ 2 at x = 10.

ANSWER : - The riven curve is  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, the slope of the tangent atx = 10 is given by;

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the slope of the tangent at x = 10 is -1/64

Quetion 40: Find the slope of the tangent to curve y = x3 - x + 1 at the point whose x-coordinate is 2. 

ANSWER : - The given curve is y = x3 - x + 1

NCERT Solution - Application of Derivative JEE Notes | EduRev

The slope of the tangent to a curve at  NCERT Solution - Application of Derivative JEE Notes | EduRev

It is given that x0 = 2.
Hence, the slope of the tangent at the point where the x-coordinate is 2 is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Quetion 41: Find the slope of the tangent to the curve y = x3 - 3x + 2 at the point whose x-coordinate is 3.

ANSWER : - The given curve is y = x3 - 3x + 2

NCERT Solution - Application of Derivative JEE Notes | EduRev

The slope of the tangent to a curve at NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the slope of the tangent at the point where the x-coordinate is 3 is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 42: Find the slope of the norm al to the curve x = a cos3θ, y = a sin3θ at NCERT Solution - Application of Derivative JEE Notes | EduRev

ANSWER: - It is given thatx = a cos3θ, and y = a sin3θ.

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
Tli ere fore, the slope of the tangent at  NCERT Solution - Application of Derivative JEE Notes | EduRev is given by..

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the slope of the normal at  NCERT Solution - Application of Derivative JEE Notes | EduRev is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 43: Find the slope of the norm al to the curve x = 1 - a sin θ, y = b cos2θ at   θ = π/2

ANSWER: - It is given thatx = 1 - a sin θ and y = b cos2θ.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, the slope of the tangent at  NCERT Solution - Application of Derivative JEE Notes | EduRev is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the slope of the normal at  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 44:Find points at which the tangent to the curve yx3 − 3x2 − 9x  7 is parallel to the x-axis.

ANSWER : - The equation of the given curve is 

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

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

Question 45: Find a point on the curve y = (x - 2)2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).

ANSWER: - If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then the slope of the tangent = the slope of the chord.

The slope of the chord is  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Now, the slope of the tangent to the given curve at a point (x, y) is given by.

Since the slope of the tangent = slope of the chord, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the required point is (3. 1).

Question 46: Find the point on the curve yx3 − 11x  5 at which the tangent is yx − 11.

ANSWER : - The equation of the given curve is yx3 − 11x  5.

The equation of the tangent to the given curve is given as yx − 11 (which is of the form ymx   c).

∴Slope of the tangent = 1

Now, the slope of the tangent to the given curve at the point (xy) is given by,  

Then, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the required points are (2.-9) and (-2,19).

Question 47: Find the equation of all lines having slope -1 that are tangents to the curve NCERT Solution - Application of Derivative JEE Notes | EduRev

ANSWER : - The equation of the given curve is NCERT Solution - Application of Derivative JEE Notes | EduRev

The slope of the tangents to the given curve at any point (x. y) is given by.
NCERT Solution - Application of Derivative JEE Notes | EduRev
If the slope of the tangent is -1. then we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

When x = 0,y = -1 and when x = 2, y = 1.

Thus, there are two tangents to the given curve having slope −1. These are passing through the points (0, −1) and (2, 1).

∴The equation of the tangent through (0, −1) is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ The equation of the tangent through (2,1) is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Henee; the equations of the required lines are y + x + 1 = 0 and y - x - 3 = 0.

Question 48: Find the equation of all lines having slope 2 which are tangents to the curve

NCERT Solution - Application of Derivative JEE Notes | EduRev

ANSWER : - The equation of the given curve is NCERT Solution - Application of Derivative JEE Notes | EduRev

The slope of the tangent to the given curve at any point (x; y) is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev
If the slope of the tangent is 2, then we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

This is not possible since the L.H.S. is positive while the R.H.S. is negative.

Hence, there is no tangent to the given curve having slope 2.

Question 49: Find the equations of all lines having slope 0 which are tangent to the curve

NCERT Solution - Application of Derivative JEE Notes | EduRev

ANSWER : - The equation of the given curve is NCERT Solution - Application of Derivative JEE Notes | EduRev

The slope of the tangent to the given curve at any point (x. y) is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev

If the slope of the tangent is 0. then we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴The equation of the tangent through  NCERT Solution - Application of Derivative JEE Notes | EduRev is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev
Hence, the equation of the required line is  y = 1/2.

Question 50: Find points on the curve NCERT Solution - Application of Derivative JEE Notes | EduRev at which the tangents are

(i) parallel to x-axis    (ii) parallel to y-axis

ANSWER.: - The equation of the given curve is NCERT Solution - Application of Derivative JEE Notes | EduRev

On differentiating both sides with respect to x, we have:
NCERT Solution - Application of Derivative JEE Notes | EduRev

(i) The tangent is parallel to the x-axis if the slope of the tangent is he.,   - 16x/9y = 0 which is possible

if x = 0.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the points at which the tangents are parallel to the x-axis are (0. 4) and (0, - 4).

(ii) The tangent is parallel to the y-axis if the slope of the normal is 0. which gives NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the points at which the tangents are parallel to the y-axis are (3, 0) and (- 3, 0).

Question 51: Find the equations of the tangent and normal to the given curves at the indicated points:

(i) yx4 − 6x3  13x2 − 10x  5 at (0, 5)
 (ii) yx4 − 6x3  13x2 − 10x  5 at (1, 3)
 (iii) yx3 at (1, 1) 
 (iv) yx2 at (0, 0)
 (v) x= cos t, y = sin t at 
  t = π/4

ANSWER: - (i) The equation of the curve is y = x4 - 6x3 + 13x2 - 10x + 5. On differentiating with respect to x, we get:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, the slope of the tangent at (0, 5) is -10. The equation ofthe tangent is given as:

y-5 = -10(x-0)

⇒ y-5 = -10x
⇒10x+y =5

The slope of the normal at (0, 5) is  NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, the equation of the normal at (0. 5) is given as:

NCERT Solution - Application of Derivative JEE Notes | EduRev

(ii) The equation of the curve is  y= x4 - 6x3+13x2 -10x +5 

On differentiating with respect to x, we get:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, the slope of the tangent at (1. 3) is 2. The equation of the tangent is given as:

y-3 =2(x-1)

⇒y-3 = 2x-2

⇒ y = 2x+1
The slope of the normal at (1,3) is  NCERT Solution - Application of Derivative JEE Notes | EduRev

There fore, the equation of the normal at (1. 3) is given as:

NCERT Solution - Application of Derivative JEE Notes | EduRev

(iii) The equation of the curve is yx3.

On differentiating with respect to x, we get:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, the slope of the tangent at (1. 1) is 3 and the equation of the tangent is given as:

y-1 = 3(x-1)

⇒ y = 3x-2

The slope of the normal at (1. 1) is NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, the equation of the normal at (1. 1) is given as:

NCERT Solution - Application of Derivative JEE Notes | EduRev

⇒3y-3 = -x+1

⇒ x+3y-4 =0


(iv) The equation of the curve is y = x2.
On differentiating with respect to x. we get:

NCERT Solution - Application of Derivative JEE Notes | EduRev
Thus, the slope of the tangent at (0. 0) is 0 and the equation of the tangent is given as:

y-0 = 0(x-0)

⇒ y=0

The slope of the normal at (0, 0) is  NCERT Solution - Application of Derivative JEE Notes | EduRev,  which is not defined. 

Therefore, the equation of the normal at   (xoyo) = (0,0) is given by x = x0 - 0

(v) The equation of the curve is x = cos = sin t.

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ The slope of the tangent at NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, the equation of the tangent to the given curve at.   NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

The slope of the normal at  NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, the equation of the normal to the riven curve at  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 53: Find the equation of the tangent line to the curve yx2 − 2x  7 which is

(a) parallel to the line 2x − y  9 = 0   
 (b) perpendicular to the line 5y − 15x = 13

ANSWER : - The equation of the given curve is y = x2 - 2x + 7 

On differentiating with respect to x. we get:

dy/dx = 2x-2

(a) The equation of the line is 2x - y + 9 = 0.
2x-y + 9 = 0 ⇒ y = 2x+9
This is of the form y = mx + c.
∴ Slope of the line = 2
If a tangent is parallel to the line 2x - y + 9 = 0, then the slope of the tangent is equal to the slope of the line.
Therefore, we have:

2= 2x-2

⇒2x = 4

⇒ x=2

Now, x =2

⇒ y=4 -4+7 = 7
Thus, the equation of the tangent passing through (2,7) is given by

y-7 = 2(x-2)

⇒ y-2x-3 =0

Hence, the equation of the tangent line to the given curve (which is parallel to line 2x-y + 9 = 0) is y - 2x - 3 = 0  

(b) The equation of the line is 5y - 15x = 13.

NCERT Solution - Application of Derivative JEE Notes | EduRev

This is of the form y = mx + c.

∴ Slope of the line = 3

If a tangent is perpendicular to the line 5y − 15x = 13, then the slope of the tangent

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, the equation of the tangent passing through NCERT Solution - Application of Derivative JEE Notes | EduRev  is given by.

Question 54: Show that the tangents to the curve y = 7x3  11 at the points where x = 2 and x = −2 are parallel.

ANSWER : - The equation of the given curve is y = 7x3  11.

The slope of the tangent to a curve at (x0y0) is.
Therefore, the slope of the tangent at the point where x = 2 is given by,

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 55: Find the points on the curve yx3 at which the slope of the tangent is equal to the y-coordinate of the point.

ANSWER : - The equation of the given curve is yx3.

When the slope of the tangent is equal to the y-coordinate of the point, then y = 3x2.

Also, we have yx3.

∴3x2x3

⇒ x2 (x − 3) = 0

⇒ x = 0, x = 3

When x = 0, then y = 0 and when x = 3, then y = 3(3)2 = 27.

Hence, the required points are (0, 0) and (3, 27).

Question 56: For the curve y = 4x3 − 2x5, find all the points at which the tangents passes through the origin.

ANSWER : - The equation of the given curve is y = 4x3 − 2x5

∴ dy/dx = 12x2 - 10x4

Therefore, the slope of the tangent at a point (x,y) is 12x2- 10x4.
The equation of the tangent at (x. y) is given by.

Y-y = (12x2 - 10x4) (X-x)     ...(1)
When the tangent, passes through the origin (0, 0). thenX = Y= 0.

Therefore enuatinn (1) redures to

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the required points are (0. 0). (1.2), and (-1,-2).

Question 57: Find the points on the curve x2   y2 − 2x − 3 = 0 at which the tangents are parallel to the x-axis.

ANSWER : - The equation of the given curve is x2   y2 − 2x − 3 = 0.

On differentiating with respect to x, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Now. the tangents are parallel to the x-axis if the slope of the tangent is 0.

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

Question 58: Find the equation of the normal at the point (am2am3) for the curve ay2x3.

ANSWER : - The equation of the given curve is ay2x3.

On differentiating with respect to x, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

The slope of a tangent to the curve at  NCERT Solution - Application of Derivative JEE Notes | EduRev

=> The slope of the tangent to the given curve at (am2., am3) is

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ Slope of normal at NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the equation of the normal at (am2, am3) is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 59: Find the equation of the normals to the curve yx3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

ANSWER : - The equation of the given curve is yx3 + 2x + 6.

The slope of the tangent to the given curve at any point (xy) is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ Slope of the normal to the given curve at any point 

NCERT Solution - Application of Derivative JEE Notes | EduRev

The equation of the given line is x - 14y + 4 = 0. 
NCERT Solution - Application of Derivative JEE Notes | EduRev (which is of the form y = mx + c)

∴ Slone of the given line = NCERT Solution - Application of Derivative JEE Notes | EduRev

If the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line.

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, there are two normals to the given curve with slope  NCERT Solution - Application of Derivative JEE Notes | EduRev and passing through the points (2, 18) and (-2.-6).

Thus, the equation of the normal through (2,18) is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

And. the equation of the normal through (—2, -6) is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the equations of the normals to the given curve (which are parallel to the given line) are NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 60:Find the equations of the tangent & normal to the parabola y2=4ax at the point (at2, 2at).

ANSWER : - The equation of the given parabola is y2 = 4ax.

On differentiating y2 = 4ax with respect to x, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

.-.The slope of the tangent at NCERT Solution - Application of Derivative JEE Notes | EduRev
Then, the equation of the tangent at  NCERT Solution - Application of Derivative JEE Notes | EduRev is given by..

NCERT Solution - Application of Derivative JEE Notes | EduRev

Now, the slope of the normal at  NCERT Solution - Application of Derivative JEE Notes | EduRev is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, the equation of the normal at (at1.2at) is given as:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 61: Prove that the curves xy2 and xy = k cut at right angles if 8k2 = 1. [Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.]

ANSWER : - The equations of the given curves are given as

Putting xy2 in xyk, we get:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, the point of intersection of the given curves is  NCERT Solution - Application of Derivative JEE Notes | EduRev

Differentiating x = y2 with respect to x, we have: 
NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, the slope of the tangent to the curve  NCERT Solution - Application of Derivative JEE Notes | EduRev

On differentiating xy = k with respect to .x, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Slope of the tangent to the curve xy = kat NCERT Solution - Application of Derivative JEE Notes | EduRev  is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

We know that two curves intersect at right angles if the tangents to the curves at the point of intersection i.e.. at  NCERT Solution - Application of Derivative JEE Notes | EduRev are peipendicular to each other.

NCERT Solution - Application of Derivative JEE Notes | EduRev

This implies that we should have the product of the tangents as − 1.

Thus, the given two curves cut at right angles if the product of the slopes of their respective tangents

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the given two curves cut at right angels if 8k= 1.

 

Quetion 62: Find the equations of the tangent and normal to the hyperbola   NCERT Solution - Application of Derivative JEE Notes | EduRev at the point NCERT Solution - Application of Derivative JEE Notes | EduRev

ANSWER.: - Differentiating  NCERT Solution - Application of Derivative JEE Notes | EduRev with respect to x. we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, the slope of the tangent at NCERT Solution - Application of Derivative JEE Notes | EduRev is  NCERT Solution - Application of Derivative JEE Notes | EduRev

Then, the equation of the tangent atNCERT Solution - Application of Derivative JEE Notes | EduRevis given by,

  NCERT Solution - Application of Derivative JEE Notes | EduRevNCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRevNCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Now. the slope of the normal at NCERT Solution - Application of Derivative JEE Notes | EduRev is given by,

   NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the equation of the normal at  NCERT Solution - Application of Derivative JEE Notes | EduRev is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Quetion 63: Find the equation of the tangent to the curve  NCERT Solution - Application of Derivative JEE Notes | EduRev which is parallel to the line 4x - 2y + 5 = 0.

ANSWER : - The equation of the given curve is  NCERT Solution - Application of Derivative JEE Notes | EduRev

The slope of the tangent to the given curve at any point (x,y) is given by,

  NCERT Solution - Application of Derivative JEE Notes | EduRev

The equation of the given line is 4x- 2y+ 5 = 0.

NCERT Solution - Application of Derivative JEE Notes | EduRev (which is of the form y = mx + c)

∴ Slope of the line = 2
Now. the tangent to the given curve is parallel to the line 4x - 2y - 5 = 0 if the slope of the tangent is equal to the slope of the line.

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ Equation of the tangent passing through the point  NCERT Solution - Application of Derivative JEE Notes | EduRev is given by. NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the equation of the required tangent is 48x - 24y = 23

Question 64: The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is 

(A) 3    
(B) 1/3  
(C) -3    
(D) -1/3

ANSWER : - The equation of the given curve is y = 2x2 + 3sin x 

Slope of the tangent to the given curve atx = 0 is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the slope of the normal to the given curve at x = 0 is

NCERT Solution - Application of Derivative JEE Notes | EduRev

The correct answer is D.

Question 65: The line yx  1 is a tangent to the curve y2 = 4x at the point

(A) (1, 2)
 (B) (2, 1)                   
 (C) (1, −2)                 
 (D) (−1, 2)

ANSWER : - The equation of the given curve is.

Differentiating with respect to x, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, the slope of the tangent to the given curve at. any point (x. y) is given by

NCERT Solution - Application of Derivative JEE Notes | EduRev

The given line is y = x +1 (which is of the form v = mx + c)
∴ Slope of the line = 1
The line y = x + 1 is a tangent to the given curve if the slope.

Question 66: Using differentials, find the approximate value of each of the following up to 3 places of decimal

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Consider NCERT Solution - Application of Derivative JEE Notes | EduRev Let x = 25 and Ax = 0.3. Then

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

NCERT Solution - Application of Derivative JEE Notes | EduRevNCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev
Consider  NCERT Solution - Application of Derivative JEE Notes | EduRevLetx = 49 and Δx = 0.5.
Then,
NCERT Solution - Application of Derivative JEE Notes | EduRev

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

NCERT Solution - Application of Derivative JEE Notes | EduRevNCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of  NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
Now, dy is approximately equal to Δy and is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of  NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Consider  NCERT Solution - Application of Derivative JEE Notes | EduRev Letx = 0.008 and Δx = 0.001.

Then.

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

NCERT Solution - Application of Derivative JEE Notes | EduRev
Hence, the aooroximate value of  NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
Now, dy is approximately equal to Δy and is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRevNCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

 

NCERT Solution - Application of Derivative JEE Notes | EduRev
Hence, the approximate value of  NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Then,

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of   NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev
Now, dy is approximately equal to Δy and is given by,
NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

NCERT Solution - Application of Derivative JEE Notes | EduRev

 Hence, the approximate value of   NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Then,

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
Thus, the approximate value of  NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

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

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of  NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

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

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of      NCERT Solution - Application of Derivative JEE Notes | EduRev

slope of the line is equal to the slope of the tangent. Also, the line must intersect the curve. Thus, we must have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

Question 67: Find the approximate value of f (2.01), where f(x) = 4x2+ 5x + 2

ANSWER: - Let x = 2 and Δx = 0.01. Then, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate value of f (2.01) is 28.21.

Question 68: Find the approximate value of f (5.001), where f (x) = x3 - 7x2 + 15.

ANSWER : - Let x = 5 and Δx = 0.001. Then, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 69: Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%.

ANSWER : - The volume of a cube (V) of side x is given by Vx3.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate change in the volume of the cube is 0.03x3 m3.

Question 70: Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%

ANSWER : - The surface area of a cube (S) of side x is given by S = 6x2.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate change in the surface area of the cube is 0.12x2 m2.

Question 71: If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume.

ANSWER : - Let r be the radius of the sphere and Δr be the error in measuring the radius.

Then,r = 7 m and Δr = 0.02 m

Now, the volume V of the sphere is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate error in calculating the volume is 3.92 π m3.
Question 72: If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating in surface area.
ANSWER : - Let r be the radius of the sphere and Δr be the error in measuring the radius.
Then, r = 9 m and Δr = 0.03 m

Now, the surface area of the sphere (S) is given by.
NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 73: If f (x) = 3x2 15x  5, then the approximate value of (3.02) is

A. 47.66                    
 B. 57.66                    
 C. 67.66                     
 D. 77.66

ANSWER : - Let x = 3 and Δx = 0.02. Then, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 74: The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is

A. 0.06 x3 m3                        
 B. 0.6 x3 m3              
 C. 0.09 x3 m3                         
 D. 0.9 x3 m3

ANSWER : - The volume of a cube (V) of side x is given by Vx3.

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the approximate change in the volume of the cube is 0.09x3 m3

The correct answer is C.
NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 75: Find the maximum and minimum values,, if any, of the following functions given by

(i) f(x) =(2x- 1)2+ 3    
(ii) f(x) = 9x2 + 12x+2
(iii) f(x) = -(x - 1)2+ 10    
(iv) g(x) = x3 + 1

ANSWER: - (i) The given function is f(x) = (2x- 1)2 + 3.
It can be observed that (2x- 1)2 > 0 for every x ∈ R.
There fore ,f(x)= (2x - 1)2 + 3 > 3 for every x ∈ R.
The minimum value of/is attained when 2x - 1 = 0.

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ Minimum value of  NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, function f does not have a maximum value.
(ii) The given function is f(x) = 9x2 + 12x + 2 = (3x+2)2- 2.
It can be observed that (3x+2)2 > 0 for every x ∈ R.

There fore f = (2x - 1)2 + 3 > 3 for every x ∈ R.
The minimum value of fis attained when 2x - 1 = 0.

  NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ Minimum value of  NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, function f does not have a maximum value.
(ii) The given function is f(x) = 9x2 + 12x + 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.

Hence, function f does not have a maximum value.

(iii) The given function is f(x) = − (x − 1)2  10.

It can be observed that (x − 1)2 ≥ 0 for every x ∈ R.

Therefore, f(x) = − (x − 1)2  10 ≤ 10 for every x ∈ R.

The maximum value of f is attained when (x − 1) = 0.

(x − 1) = 0 ⇒ x = 0

∴Maximum value of ff(1) = − (1 − 1)2  10 = 10

Hence, function f does not have a minimum value.

(iv) The given function is g(x) = x3  1.

Hence, function g neither has a maximum value nor a minimum value.

Question 76: Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = |x + 2| − 1                                      
 (ii) g(x) = − |x + 1| + 3
 (iii) h(x) = sin(2x) 5                                   
 (iv) f(x) = |sin 4x + 3|
 (v) h(x) =  +1, x 
 (−1, 1)

ANSWER : NCERT Solution - Application of Derivative JEE Notes | EduRev

We know that NCERT Solution - Application of Derivative JEE Notes | EduRev for every x ∈ R.
Therefore, NCERT Solution - Application of Derivative JEE Notes | EduRev for every x ∈ R.

The minimum value of/is attained when NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ Minimum value of NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence: function f does not have a maximum value.

(ii) NCERT Solution - Application of Derivative JEE Notes | EduRev
We know that  NCERT Solution - Application of Derivative JEE Notes | EduRev for every x ∈ R.
Therefore,  NCERT Solution - Application of Derivative JEE Notes | EduRev for every x ∈ R.

The maximum value of g is attained when  NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ Maximum value of g = g(-1) NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, function g does not have a minimum value.
(iii)h(x) = sin2x+ 5

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the maximum and minimum values of h are 6 and 4 respectively.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the maximum and minimum values of f are 4 and 2 respectively.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Here, if a point x0 is closest to  NCERT Solution - Application of Derivative JEE Notes | EduRev

Also, if x1 is closest to 1. then  NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, function h(x) has neither maximum nor minimum value in (-1. 1).

Quetion 77: Find the local maxima and local minima,, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Thus.x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.

We have f''(0) = 2 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) = 0.

NCERT Solution - Application of Derivative JEE Notes | EduRev

By second derivative test. x= 1 is a point of local minima and local minimum value ofg atr = 1 is g(l) = l3 - 3 = 1 - 3 = -2.

However. x = -1 is a point of local maxima and local maximum value of g at

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, by second derivative test.  NCERT Solution - Application of Derivative JEE Notes | EduRev is a point: of local maxima and the local maximum value

NCERT Solution - Application of Derivative JEE Notes | EduRev

 Therefore, by second derivative test,  NCERT Solution - Application of Derivative JEE Notes | EduRev is a point of local maxima and the local maximum value of  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev 
However,  NCERT Solution - Application of Derivative JEE Notes | EduRev is a point of local minima and the local minimum value of f at NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

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

NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, by second derivative test. x = 2 is a point of local minima and the local minimum value of g at NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Now, for values close to x = 0 and to the left of 0, g (x) > 0. Also, for values close to x = 0 and to thi right of 0, g'(x) < 0

Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, by second derivative test.  NCERT Solution - Application of Derivative JEE Notes | EduRev  is a point of local maxima and the local maximum value off at  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 78: Prove that the following functions do not have maxima or minima:

(i) f(x) = ex   
(ii) g(x) = logx    
(iii) h(x) = x3 + x2 + x+ 1

ANSWER: -

NCERT Solution - Application of Derivative JEE Notes | EduRev

Now., if  NCERT Solution - Application of Derivative JEE Notes | EduRev . But. the exponential function can never assume 0 for any value of x.

Therefore, there does not exist  NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, function f does not have maxima or minima.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, there does not exist ∈ R such that .

Hence, function g does not have maxima or minima.

iii. We have,

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRevNCERT Solution - Application of Derivative JEE Notes | EduRev

Therefore, there does not exist c∈ R such that h'(c)=0.
Hence, function h does not have maxima or minima.
Question 79: Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

ANSWER: - (i) The given function is f (x) = x3.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Then, we evaluate the value of/ at critical point x = 0 and at endpoints of the interval [—2,2].

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, we can. conclude that the absolute maximum value off on [-2. 2] is 8 occurring at x = 2. Also, the absolute minimum value off on [-2,2] is -8 occurring at x = -2.
(ii) The given function is f(x) = sin x + cosx

NCERT Solution - Application of Derivative JEE Notes | EduRev

Then, we evaluate the value of/at critical point  NCERT Solution - Application of Derivative JEE Notes | EduRev and at the end points of the interval [0, π] .

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, we can conclude that the absolute maximum value of  NCERT Solution - Application of Derivative JEE Notes | EduRev occurring at  NCERT Solution - Application of Derivative JEE Notes | EduRev and the absolute minimum value of NCERT Solution - Application of Derivative JEE Notes | EduRev occurring at x = π

(iii) The given function is  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Then, we evaluate the value of/at critical point x = 4 and at the end points of the interval  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, we can conclude that the absolute maximum value off on  NCERT Solution - Application of Derivative JEE Notes | EduRev is 8 occurring at x = 4 and the absolute minimum value of f or NCERT Solution - Application of Derivative JEE Notes | EduRev is -10 occurring at x = -2.

(iv) The riven function is 

NCERT Solution - Application of Derivative JEE Notes | EduRev

Then, we evaluate the value off at critical point x = 1 and at the endpoints of the interval [3,1].

NCERT Solution - Application of Derivative JEE Notes | EduRev  

Hence, we can conclude that the absolute maximum value of/on [-3,1] is 19 occurring at x = -3 and the minimum value of f on [-3. 1] is 3 occurring at x = 1.

Question 80: Find the maximum profit that a company can make, if the profit function is given by

p(x) = 41 - 24x - 18x2

ANSWER: - The profit function is riven as p(x) = 41 - 24x - 18x2

NCERT Solution - Application of Derivative JEE Notes | EduRev

By second derivative test.  NCERT Solution - Application of Derivative JEE Notes | EduRev is the point of local maxima of p.

∴ Maximum profit NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the maximum profit that the company can make is 49 units. 

Question 81: Find both the maximum value and the minimum value of NCERT Solution - Application of Derivative JEE Notes | EduRev on the interval [0, 3]

ANSWER:  NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Now,  NCERT Solution - Application of Derivative JEE Notes | EduRev for which there are no real roots.

Therefore, we consider only x= 2 ∈[0,3].
Now, w:e evaluate the value off at critical pointx = 2 and at. the end points ofthe interval [0,3].

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, we can conclude that the absolute maximum value of f on [0, 3] is 25 occurring at x = 0 and the absolute minimum value of f at [0. 3] is - 39 occurring at x = 2.

Question 82: At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?

ANSWER : -Let f(x) = sin 2x.

  NCERT Solution - Application of Derivative JEE Notes | EduRev

Then, we evaluate the values of/at critical points  NCERT Solution - Application of Derivative JEE Notes | EduRev and at the end points of the

NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, we can conclude that the absolute maximum value of f on [0, 2π] is occurring at  NCERT Solution - Application of Derivative JEE Notes | EduRev and  NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 83: What is the maximum value of the function sinx + cos x? 

ANSWER : - Let f (x) = sin x- cos x.

NCERT Solution - Application of Derivative JEE Notes | EduRev

We first consider the interval [1, 3].
Then, we evaluate the value of f at the critical point x = 2 ∈ [1, 3] and at the end points of the interval [1, 3].

f(2) = 2(8) − 24(2) 107 = 16 − 48 107 = 75

f(1) = 2(1) − 24(1) 107 = 2 − 24 107 = 85

f(3) = 2(27) − 24(3) 107 = 54 − 72 107 = 89

Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3.

Next, we consider the interval [−3, −1].

Evaluate the value of f at the critical point x = −2 ∈ [−3, −1] & at the end points of the interval [1, 3].

f(−3) = 2 (−27) − 24(−3) 107 = −54 72 107 = 125

f(−1) = 2(−1) − 24 (−1) 107 = −2 24 107 = 129

f(−2) = 2(−8) − 24 (−2) 107 = −16 48 107 = 139

Hence, the absolute maximum value of f(x) in the interval [−3, −1] is 139 occurring at x = −2.

Question 85: 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.

ANSWER : - Let f(x) = x4 − 62x2   ax  9.

It is given that function f attains its maximum value on the interval [0, 2] at x = 1.

Now,  NCERT Solution - Application of Derivative JEE Notes | EduRev will be negative when (sin x + cos x) is positive i.e,, when sin x ana cos x are both positive. Also, we knowr that sin x and cos x both are positive in the first quadrant. Then, NCERT Solution - Application of Derivative JEE Notes | EduRev will be negative whenNCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ By second derivative test./will be the maximum at NCERT Solution - Application of Derivative JEE Notes | EduRev and the maximum value of f  is

NCERT Solution - Application of Derivative JEE Notes | EduRev

Quetion 84: Find the maximum value of 2x3 - 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].

ANSWER : - Let f(x) = 2x3 - 24x + 107.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 86: Find the maximum and minimum values of x + sin 2x on [0, π]. 

ANSWER: - Let f(x) = x + sin 2x.

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Then, we evaluate the value of f at critical points   NCERT Solution - Application of Derivative JEE Notes | EduRev and at the end points of the interval [0, 2π].

NCERT Solution - Application of Derivative JEE Notes | EduRev

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

Question 87: Find two numbers whose sum is 24 and whose product is as large as possible.

ANSWER : - Let one number be x. Then, the other number is (24 − x).

Let P(x) denote the product of the two numbers. Thus, we have

NCERT Solution - Application of Derivative JEE Notes | EduRev

By second derivative test, x = 12 is the point of local maxima of P. Hence, the product of the numbers is the maximum when the numbers are 12 and 24 − 12 = 12.

Question 88: Find two positive numbers x and y such that x  + y = 60 and xy3 is maximum

ANSWER: - The two numbers are x andy such that x + y = 60.

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

By second derivative test, = 15 is a point of local maxima of f. Thus, function xy3 is maximum when x = 15 and y = 60 − 15 = 45.

Hence, the required numbers are 15 and 45.

Question 89: Find two positive numbers and such that their sum is 35 and the product x2y5 is a maximum

ANSWER : - Let one number be x. Then, the other number is y = (35 − x).

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
When x = 35,  NCERT Solution - Application of Derivative JEE Notes | EduRev and y= 35 - 35 = 0. This will make the product x2y5 equal to 0.

x = 0 andx =35 cannot be the possible values of x.
When x = 10, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev

By second derivative test, P(x) will be the maximum when x = 10 and y = 35 − 10 = 25.

Hence, the required numbers are 10 and 25.

Question 90: Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

ANSWER : - Let one number be x. Then, the other number is (16 − x).

Let the sum of the cubes of these numbers be denoted by S(x). Then,

NCERT Solution - Application of Derivative JEE Notes | EduRev

By second derivative test, x = 8 is the point of local minima of S.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 −8 =8.

Question 91: A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

ANSWER : - Let the side of the square to be cut off be x cm. Then, the length and the breadth of the box will be (18 − 2x) cm each and the height of the box is x cm.

Therefore, the volume V(x) of the box is given by

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

If x = 9. then the length and the breadth will become 0

 

By second derivative test, x = 3 is the point of maxima of V.

Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible

Question 92: 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 the maximum possible?

ANSWER : - Let the side of the square to be cut off be x cm. Then, the height of the box is x, the length is 45 − 2x, and the breadth is 24 − 2x.

Therefore, the volume V(x) of the box is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

If x = 9. then the length and the breadth will become 0

NCERT Solution - Application of Derivative JEE Notes | EduRev

By second derivative test, x = 5 is the point of maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm.

Question 93: Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

ANSWER : - Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.

Then, the diagonal passes through the centre and is of length 2a cm.

Now, by applying the Pythagoras theorem, we have:

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ By the second derivative test, whenNCERT Solution - Application of Derivative JEE Notes | EduRev, then the area of the rectangle is the maximum.

Since  NCERT Solution - Application of Derivative JEE Notes | EduRev , the rectangle is a square.

Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.

Question 94: Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.

ANSWER : - Let r and h be the radius and height of the cylinder respectively.

Then, the surface area (S) of the cylinder is given by

NCERT Solution - Application of Derivative JEE Notes | EduRev

Let V be the volume of the cylinder. Then.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 95: Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

ANSWER : - Let r and h be the radius and height of the cylinder respectively.

Then, volume (V) of the cylinder is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev

Surface area (S) of the cylinder is given by.

NCERT Solution - Application of Derivative JEE Notes | EduRev

Now,  it is observed that when  NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ By second derivative test, the surface area is the minimum when the radius of the cylinder is

NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 96: A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

ANSWER : - Let a piece of length l be cut from the given wire to make a square.

Then, the other piece of wire to be made into a circle is of length (28 − l) m.

Now, side of square = l/4.

Let r be the radius of the circle. Then, 

The combined areas of the square and the circle (A) is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ By second derivative test, the area (A) is the minimum when  NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the combined area is the minimum when the length of the wire in making the square is  NCERT Solution - Application of Derivative JEE Notes | EduRevcm while the length of the wire in making the circle is NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 97: Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

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

Let V be the volume of the cone.

Then, NCERT Solution - Application of Derivative JEE Notes | EduRev

Height of the cone is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev [ABC is a right triangle]

NCERT Solution - Application of Derivative JEE Notes | EduRev

 

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ By second derivative test the volume of the cone is die maximum when NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 98: Show that the right circular cone of least curved surface and given volume has an altitude equal to time the radius of the base.

ANSWER : - Let and h be the radius and the height (altitude) of the cone respectively.

Then, the volume (V) of the cone is given as:

NCERT Solution - Application of Derivative JEE Notes | EduRev

The surface area (S) of the cone is given by.
S = πrI (where l is the slant height)

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRevNCERT Solution - Application of Derivative JEE Notes | EduRev

Thus, it. can be easily verified that when  NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 99: Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is NCERT Solution - Application of Derivative JEE Notes | EduRev

ANSWER : - Let θ be the semi-vertical angle of the cone.
It is clear that  NCERT Solution - Application of Derivative JEE Notes | EduRev

Let r, h. and 1 be the radius, height, and the slant height of the cone respectively.

The slant height of the cone is given as constant.

Now, r = l sin θ and h = 1 cos θ
The volume (V) of the cone is given by,

NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

Question 100: The point on the curve x2 = 2y which is nearest to the point (0, 5) is

(A) NCERT Solution - Application of Derivative JEE Notes | EduRev
(B)  NCERT Solution - Application of Derivative JEE Notes | EduRev
(C) (0, 0)    
(D) (2, 2)

ANSWER : - The given curve is x2 = 2y.

For each value of x. the position of the point will be NCERT Solution - Application of Derivative JEE Notes | EduRev

The distance d(x) between the points NCERT Solution - Application of Derivative JEE Notes | EduRevand (0,5) is given by.
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev
NCERT Solution - Application of Derivative JEE Notes | EduRev

∴ By second derivative test. d(x) is the minimum at NCERT Solution - Application of Derivative JEE Notes | EduRev

When NCERT Solution - Application of Derivative JEE Notes | EduRev

Hence, the point on the curve x2 = 2y which is nearest to the point (0, 5) is  NCERT Solution - Application of Derivative JEE Notes | EduRev

The correct answer is A.

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