NCERT Solution - Application of Derivative (Part -2) JEE Notes | EduRev

Question 58: Find the equation of the normal at the point (am², am³) for the curve ay² = x³.

ANSWER : - The equation of the given curve is ay² = x³.

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

∴ 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 y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

ANSWER : - The equation of the given curve is y = x³ + 2x + 6.

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

∴ 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)

∴ 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 (—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 x = y² 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 x = y² in xy = k, 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 − 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 8k² = 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)

∴ 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.

∴ 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

The correct answer is D.

Question 65: The line y = x  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:

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)
∴ 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 Δy and is given by, 

Hence, the approximate value of  

Consider  Letx = 49 and Δx = 0.5.
Then,

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

Hence, the approximate value of  

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

Hence, the approximate value of  

Consider   Letx = 0.008 and Δx = 0.001.

Then.

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

Hence, the aooroximate value of  


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

Hence, the approximate value of  

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

Hence, the approximate value of  

Then,

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

Hence, the approximate value of  

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

Hence, the approximate value of

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

 Hence, the approximate value of  

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

Hence, the approximate value of  

Then,

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

Thus, the approximate value of  

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

Hence, the approximate value of  

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

Hence, the approximate value of  

Now, dy is approximately equal to Δy 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 Δx = 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 Δx = 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 V = x³.

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 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,

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.

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

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

ANSWER : - Let x = 3 and Δx = 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 V = x³.

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

The correct answer is C.

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.

∴ 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 ∈ R.

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

∴ 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 ∈ 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 f = f(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) = x  +1, x ∈ (−1, 1)

ANSWER : 

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

The minimum value of/is attained when

∴ Minimum value of

Hence: function f does not have a maximum value.

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

The maximum value of g is attained when  

∴ 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) = e^x   
(ii) g(x) = logx    
(iii) h(x) = x³ + x² + x+ 1

ANSWER: -

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 c ∈ R such that .

Hence, function g does not have maxima or minima.

iii. We have,

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:

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 [—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, π] .

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

(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.

∴ 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]

ANSWER: 

Now,   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].

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.

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, 2π] 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 ∈ [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,   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

∴ 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, π]. 

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, 2π].

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

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.

By second derivative test, x = 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 x and y 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).

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 − 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,

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

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,

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:

∴ 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  

∴ 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 − 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,

∴ 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]

∴ 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 r 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 = πrI (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 θ 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 θ and h = 1 cos θ
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.

∴ 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  

The correct answer is A.