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# NCERT Solution - Application of Derivative JEE Notes | EduRev

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

The document NCERT Solution - Application of Derivative JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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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.

1.  when r = 3 cm,

dA/dr = 2Ï€(3) = 6Ï€

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 = 2Ï€(4) = 8Ï€

Hence, the area of the circle is changing at the rate of 8Ï€ 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:

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 = Ï€r2

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

[By chain rule]

It is given that,

dr/dt = 3cm/s

âˆ´ dA/dt = 2Ï€r(3) = 6Ï€r

Thus, when r = 10 cm,

dA/dt = 6Ï€(10) = 60Ï€ 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,

It is given that,

dx/dt = 3cm/s

âˆ´ 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,

[By chain rule]

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

Thus, when r = 8 cm,

dA/dt = 2Ï€ (8) (5) = 80Ï€

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80Ï€ 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= = 2Ï€r.

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

It is given that  dr/dt = 0.7 cm/s

Hence, the rate of increase of the circumference  is 2Ï€ (0.7) = 1.4Ï€ 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)

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

A = x.y

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 = 4/3Ï€r3

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

Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is 1/Ï€ 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 Ï€r2

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

dV/dr = 4Ï€(10)= 400Ï€

Hence, the volume of the balloon is increasing at the rate of 400Ï€ 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]

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

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:

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

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

x-coordinate i.e.,  , we have:

When

When

Hence, the points required on the curve are (4,11) and

Question 12: The radius of an air bubble is increasing at the rate of Â½ 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 = 4/3Ï€r3

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

[By chain rule]

It iis given that  dr/dt = 1/2 cm/s

Therefore , when r = 1 cm,

Hence, the rate at which the volume of the bubble increases is 2Ï€ 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 = 4/3Ï€r3

It is given that:

Diameter

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

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 = 1/3Ï€r2h
It is riven that.

h= 1/6r = r=6h

âˆ´ V = 1/3Ï€(6h)2h = 12Ï€h3

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

It is also given that

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.

âˆ´ Marginal cost (MC)

= 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.

âˆ´ 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) 10Ï€    (B) 12Ï€   (C) 8Ï€    (D) 11Ï€

ANSWER: - The area of a circle (A) with radius (r) is given by,
A = Ï€r2
Therefore, the rate of change of the area urith respect to its radius r is

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

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

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

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

âˆ´ When x = 15,

MR = 6(15) + 36 = 90 + 36 = 126

Hence, the required marginal revenue is Rs 126.

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:

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:

Hence, f is strictly increasing on R.

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

(a) strictly increasing in  (b) strictly decreasing in

(c) neither increasing nor decreasing in (0, Ï€)

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

(a) Since for each   we have

Hence,/is strictly increasing in

(b) Since for each , we have

Hence,f is strictly decreasing in

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

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

(a) strictly increasing
(b) strictly decreasing

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

Now. the point 3/4 divides the real line into two disjoint intervals i.e..

Hence, the given function (f) is strictly decreasing in interval

In interval

Hence, the given function (J) is strictly increasing in interval

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

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

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

Hence, the given function (f) is strictly increasing in intervals (-âˆž, -2) and (3, âˆž) , 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 - 6xâ€” 2x2
(c) -2x3 - 9x2 - 12x+ 1
(d) 6 - 9x -x2
(e) (x + 1)3 (x â€” 3)3

f(x) = x2 +2x-5

âˆ´ f'(x) = 2x+2

Now,

f'(x) = 0 â‡’ x = -1

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

In interval

âˆ´ f is strictly decreasing in interval (-âˆž, -1).
Thus, f is strictly decreasing for x < -1.
In interval

âˆ´ f is strictly increasing in interval (-1, âˆž)
Thus, f is strictly deere asing for x < - 1.

(b) We have,

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

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

âˆ´ f is strictly increasing for x< -3/2

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

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

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

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

Now,

f'(x) = 0 â‡’ x = -1 and x = -2

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

(-âˆž, -2),(-2, -1) and (-1, âˆž)

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

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

âˆ´ 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

âˆ´ f is strictly increasing for 2 < x < 1

(d) We have.

f(x) = 6-9x = x2

âˆ´  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., (-âˆž, -9/2) and (-9/2,âˆž )

In interval  (-âˆž, -9/2) i.e., for

âˆ´ f is strictly increasing for x < -9/2

In interval  (-9/2,âˆž ) i.e. for

âˆ´ f is strictly decreasing for

(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., (-âˆž, -1) (-1,1), (1,3), and (3, âˆž)

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

âˆ´ f is strictly decreasing in intervals  (-âˆž, -1) and (-1,1)

In intervals

âˆ´ f is strictly increasing in intervals  (1,3), and (3, âˆž)

Question 25: Show that  , is an increasing function of x throughout its domain.

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

When -1 < x < 0. we have:

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.

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

âˆ´ y is strictly decreasing in intervals  (-âˆž , 0) and (1,2)

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

âˆ´ y is strictly increasing in intervals (0,1) and (2, âˆž).
âˆ´ y is strictly increasing for 0 < x < 1 and x > 2.

Question 27: Prove that   is an increasing function of Î¸ in [0, Ï€/2]

In interval   we have cos

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

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

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

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

The given function is f (x) = logx.

âˆ´ 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;âˆž).

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.

The point 1/2 divides the interval (-1. 1) into two disjoint intervals i.e..

Now. in interval

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

However, in interval

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,Ï€/2) ?

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

is strictly decreasing in interval

is strictly decreasing in interval

The point   divides the interval  1 into two disjoint intervals i.e., 0

Now, in interval
f3 is strictly decreasing in interval

However, in interval

âˆ´ f3 is strictly increasing in interval

Hence, fis neither increasing nor decreasing in interval .

âˆ´ f4 is strictly increasing in interval

Therefore, functions cos x and cos 2.x are strictly decreasing in

Hence, the correct answers are A andB.

Question 31: On which of the following intervals is the function f given by  strictly decreasing?

(D) ) None of these

In interval

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

In interval

Thus, function f is strictly increasing in interval

âˆ´ f is strictly increasing in interval .

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

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).

Now, function f will be increasing in

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

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

-a/2 =1

-a/2 =1 â‡’ a = -2

Hence, the required value of a is -2.

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

The points x = 1 andx = -1 divide the real line in three disjoint intervals i.e.,

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

In intervals   , it is observed that:

âˆ´ f is strictly increasing on  (-âˆž, 1) and (1, âˆž)

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

In interval

âˆ´ f  is strictly decreasing in

Question 35: Prove that the function f given by f (x) = log cos x is strictly decreasing on    and strictly increasing on

âˆ´ f is strictly decreasing on

In interval

âˆ´ f is strictly increasing on

Question 36: Prove that the function given by  is increasing in R.

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) (-âˆž,âˆž)
(B) (-2,0)
(C) (2,âˆž)
(D) (0,2)

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

In interv als   is always positive.

âˆ´ f is decreasing on (-âˆž,0) and (2, âˆž)

In interval (0,2),

âˆ´ f is strictly increasing on (0.2).
Hence f is strictly increasing in interval (0,2).

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;

Question 39: Find the slope of the tangent to the curve  , x â‰  2 at x = 10.

ANSWER : - The riven curve is

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

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

The slope of the tangent to a curve at

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

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

The slope of the tangent to a curve at

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

Question 42: Find the slope of the norm al to the curve x = a cos3Î¸, y = a sin3Î¸ at

ANSWER: - It is given thatx = a cos3Î¸, and y = a sin3Î¸.

Tli ere fore, the slope of the tangent at   is given by..

Hence, the slope of the normal at   is given by,

Question 43: Find the slope of the norm al to the curve x = 1 - a sin Î¸, y = b cos2Î¸ at   Î¸ = Ï€/2

ANSWER: - It is given thatx = 1 - a sin Î¸ and y = b cos2Î¸.

Therefore, the slope of the tangent at   is given by.

Hence, the slope of the normal at

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

ANSWER : - The equation of the given curve is

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

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:

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

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

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

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

âˆ´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:

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

ANSWER : - The equation of the given curve is

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

If the slope of the tangent is -1. then we have:

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

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

âˆ´The equation of the tangent through (0, âˆ’1) is given by,

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

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

ANSWER : - The equation of the given curve is

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

If the slope of the tangent is 2, then we have:

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

ANSWER : - The equation of the given curve is

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

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

âˆ´The equation of the tangent through   is given by.

Hence, the equation of the required line is  y = 1/2.

Question 50: Find points on the curve  at which the tangents are

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

ANSWER.: - The equation of the given curve is

On differentiating both sides with respect to x, we have:

(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.

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

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 âˆ’ 6x3  13x2 âˆ’ 10x  5 at (0, 5)
(ii) yx4 âˆ’ 6x3  13x2 âˆ’ 10x  5 at (1, 3)
(iii) yx3 at (1, 1)
(iv) yx2 at (0, 0)
(v) x= cos t, y = sin t at
t = Ï€/4

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

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

y-5 = -10(x-0)

â‡’ y-5 = -10x
â‡’10x+y =5

The slope of the normal at (0, 5) is

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

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

On differentiating with respect to x, we get:

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

y-3 =2(x-1)

â‡’y-3 = 2x-2

â‡’ y = 2x+1
The slope of the normal at (1,3) is

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

(iii) The equation of the curve is yx3.

On differentiating with respect to x, we get:

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)

â‡’ y = 3x-2

The slope of the normal at (1. 1) is

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

â‡’3y-3 = -x+1

â‡’ x+3y-4 =0

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

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)

â‡’ y=0

The slope of the normal at (0, 0) is  ,  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.

âˆ´ The slope of the tangent at

Thus, the equation of the tangent to the given curve at.

The slope of the normal at

Therefore, the equation of the normal to the riven curve at

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

(a) parallel to the line 2x âˆ’ y  9 = 0
(b) perpendicular to the line 5y âˆ’ 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 â‡’ y = 2x+9
This is of the form y = mx + c.
âˆ´ 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

â‡’2x = 4

â‡’ x=2

Now, x =2

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

y-7 = 2(x-2)

â‡’ 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.

This is of the form y = mx + c.

âˆ´ Slope of the line = 3

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

Thus, the equation of the tangent passing through   is given by.

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

âˆ´3x2x3

â‡’ x2 (x âˆ’ 3) = 0

â‡’ 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 âˆ’ 2x5, find all the points at which the tangents passes through the origin.

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

âˆ´ 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

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

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

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

On differentiating with respect to x, we have:

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

Hence, the points at which the tangents are parallel to the x-axis are (1, 2) and (1, âˆ’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:

The slope of a tangent to the curve at

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

âˆ´ Slope of normal at

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

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,

âˆ´ Slope of the normal to the given curve at any point

The equation of the given line is x - 14y + 4 = 0.
(which is of the form y = mx + c)

âˆ´ Slone of the given line =

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.

Therefore, there are two normals to the given curve with slope   and passing through the points (2, 18) and (-2.-6).

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

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

Hence, the equations of the normals to the given curve (which are parallel to the given line) are

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:

.-.The slope of the tangent at
Then, the equation of the tangent at   is given by..

Now, the slope of the normal at   is given by,

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

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:

Thus, the point of intersection of the given curves is

Differentiating x = y2 with respect to x, we have:

Therefore, the slope of the tangent to the curve

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

Slope of the tangent to the curve xy = kat   is given by.

We know that two curves intersect at right angles if the tangents to the curves at the point of intersection i.e.. at   are peipendicular to each other.

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

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

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    at the point

ANSWER.: - Differentiating   with respect to x. we have:

Therefore, the slope of the tangent at  is

Then, the equation of the tangent atis given by,

Now. the slope of the normal at  is given by,

Hence, the equation of the normal at   is given by.

Quetion 63: Find the equation of the tangent to the curve   which is parallel to the line 4x - 2y + 5 = 0.

ANSWER : - The equation of the given curve is

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

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

(which is of the form y = mx + c)

âˆ´ 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.

âˆ´ Equation of the tangent passing through the point   is given by.

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,

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

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, âˆ’2)
(D) (âˆ’1, 2)

ANSWER : - The equation of the given curve is.

Differentiating with respect to x, we have:

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

The given line is y = x +1 (which is of the form v = mx + c)
âˆ´ 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

Consider  Let x = 25 and Ax = 0.3. Then

Now, dy is approximately equal to Î”y and is given by,

Hence, the approximate value of

Consider  Letx = 49 and Î”x = 0.5.
Then,

Now, dy is approximately equal to Î”y and is given by.

Hence, the approximate value of

Now, dy is approximately equal to Î”y and is given by.

Hence, the approximate value of

Consider   Letx = 0.008 and Î”x = 0.001.

Then.

Now, dy is approximately equal to Î”y and is given by,

Hence, the aooroximate value of

Now, dy is approximately equal to Î”y and is given by,

Hence, the approximate value of

Now, dy is approximately equal to Î”y and is given by,

Hence, the approximate value of

Then,

Now, dy is approximately equal to Î”y and is given by,

Hence, the approximate value of

Now, dy is approximately equal to Î”y and is given by,

Hence, the approximate value of

Now, dy is approximately equal to Î”y and is given by.

Hence, the approximate value of

Now, dy is approximately equal to Î”y and is given by.

Hence, the approximate value of

Then,

Now, dy is approximately equal to Î”y and is given by,

Thus, the approximate value of

Now. dy is approximately equal to Î”y and is given by.

Hence, the approximate value of

Now, dy is approximately equal to Î”y and is given by.

Hence, the approximate value of

Now, dy is approximately equal to Î”y and is given by,

Hence, the approximate value of

Hence, the approximate value of

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

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 Î”x = 0.01. Then, we have:

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 Î”x = 0.001. Then, we have:

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.

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.

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 Î”r be the error in measuring the radius.

Then,r = 7 m and Î”r = 0.02 m

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

Hence, the approximate error in calculating the volume is 3.92 Ï€ 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 Î”r be the error in measuring the radius.
Then, r = 9 m and Î”r = 0.03 m

Now, the surface area of the sphere (S) is given by.

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 Î”x = 0.02. Then, we have:

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.

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

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

âˆ´ Minimum value of

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 âˆˆ R.

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

âˆ´ Minimum value of

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 âˆˆ R.

Therefore, f(x) = (3x + 2)2 âˆ’ 2 â‰¥ âˆ’2 for every x âˆˆ 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) = âˆ’ (x âˆ’ 1)2  10.

It can be observed that (x âˆ’ 1)2 â‰¥ 0 for every x âˆˆ R.

Therefore, f(x) = âˆ’ (x âˆ’ 1)2  10 â‰¤ 10 for every x âˆˆ R.

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

(x âˆ’ 1) = 0 â‡’ x = 0

âˆ´Maximum value of ff(1) = âˆ’ (1 âˆ’ 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| âˆ’ 1
(ii) g(x) = âˆ’ |x + 1| + 3
(iii) h(x) = sin(2x) 5
(iv) f(x) = |sin 4x + 3|
(v) h(x) =  +1, x
âˆˆ (âˆ’1, 1)

We know that  for every x âˆˆ R.
Therefore,  for every x âˆˆ R.

The minimum value of/is attained when

âˆ´ Minimum value of

Hence: function f does not have a maximum value.

(ii)
We know that   for every x âˆˆ R.
Therefore,   for every x âˆˆ R.

The maximum value of g is attained when

âˆ´ Maximum value of g = g(-1)

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

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

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

Here, if a point x0 is closest to

Also, if x1 is closest to 1. then

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:

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.

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

Therefore, by second derivative test.   is a point: of local maxima and the local maximum value

Therefore, by second derivative test,   is a point of local maxima and the local maximum value of

However,   is a point of local minima and the local minimum value of f at

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.

Therefore, by second derivative test. x = 2 is a point of local minima and the local minimum value of g at

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

Therefore, by second derivative test.    is a point of local maxima and the local maximum value off at

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

Now., if   . But. the exponential function can never assume 0 for any value of x.

Therefore, there does not exist

Hence, function f does not have maxima or minima.

Therefore, there does not exist âˆˆ R such that .

Hence, function g does not have maxima or minima.

iii. We have,

Therefore, there does not exist câˆˆ 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:

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

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

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

Then, we evaluate the value of/at critical point   and at the end points of the interval [0, Ï€] .

Hence, we can conclude that the absolute maximum value of   occurring at   and the absolute minimum value of  occurring at x = Ï€

(iii) The given function is

Then, we evaluate the value of/at critical point x = 4 and at the end points of the interval

Hence, we can conclude that the absolute maximum value off on   is 8 occurring at x = 4 and the absolute minimum value of f or  is -10 occurring at x = -2.

(iv) The riven function is

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

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

By second derivative test.   is the point of local maxima of p.

âˆ´ Maximum profit

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

Question 81: Find both the maximum value and the minimum value of  on the interval [0, 3]

Now,   for which there are no real roots.

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

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, 2Ï€], does the function sin 2x attain its maximum value?

ANSWER : -Let f(x) = sin 2x.

Then, we evaluate the values of/at critical points   and at the end points of the

Hence, we can conclude that the absolute maximum value of f on [0, 2Ï€] is occurring at   and

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

ANSWER : - Let f (x) = sin x- cos x.

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

f(2) = 2(8) âˆ’ 24(2) 107 = 16 âˆ’ 48 107 = 75

f(1) = 2(1) âˆ’ 24(1) 107 = 2 âˆ’ 24 107 = 85

f(3) = 2(27) âˆ’ 24(3) 107 = 54 âˆ’ 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 [âˆ’3, âˆ’1].

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

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

f(âˆ’1) = 2(âˆ’1) âˆ’ 24 (âˆ’1) 107 = âˆ’2 24 107 = 129

f(âˆ’2) = 2(âˆ’8) âˆ’ 24 (âˆ’2) 107 = âˆ’16 48 107 = 139

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

Question 85: It is given that at x = 1, the function x4âˆ’ 62x2   ax  9 attains its maximum value, on the interval [0, 2]. Find the value of a.

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

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

Now,   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,  will be negative when

âˆ´ By second derivative test./will be the maximum at  and the maximum value of f  is

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.

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

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

Then, we evaluate the value of f at critical points    and at the end points of the interval [0, 2Ï€].

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

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

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 âˆ’ 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.

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

When x = 35,   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:

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

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

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 âˆ’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 âˆ’ 2x) cm each and the height of the box is x cm.

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

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 âˆ’ 2x, and the breadth is 24 âˆ’ 2x.

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

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

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

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

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:

âˆ´ By the second derivative test, when, then the area of the rectangle is the maximum.

Since   , 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

Let V be the volume of the cylinder. Then.

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,

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

Now,  it is observed that when

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

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 âˆ’ 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,

âˆ´ By second derivative test, the area (A) is the minimum when

Hence, the combined area is the minimum when the length of the wire in making the square is  cm while the length of the wire in making the circle is

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,

Height of the cone is given by,

[ABC is a right triangle]

âˆ´ By second derivative test the volume of the cone is die maximum when

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:

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

Thus, it. can be easily verified that when

Question 99: Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is

ANSWER : - Let Î¸ be the semi-vertical angle of the cone.
It is clear that

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 Î¸ and h = 1 cos Î¸
The volume (V) of the cone is given by,

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

(A)
(B)
(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

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

âˆ´ By second derivative test. d(x) is the minimum at

When

Hence, the point on the curve x2 = 2y which is nearest to the point (0, 5) is

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## Mathematics (Maths) Class 12

209 videos|222 docs|124 tests

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