# NCERT Solutions for Exercise 6.5 Class 12 Maths Chapter 6 - Application of Derivatives

Throughout the NCERT solutions for exercise 6.5 Class 12 Maths chapter 6 the topic maxima and minima is discussed. NCERT solutions for Class 12 Maths chapter 6 exercise 6.5 uses the concept of derivatives to find the maximum and minimum of different functions. Exercise 6.5 Class 12 Maths also give ideas about absolute minimum and maximum. In the NCERT Class 12 Mathematics Book, some real-life examples of finding maximum and minimum values are given. And certain definitions are discussed after the examples in the NCERT book. Such as definitions of maximum and minimum values, extreme point, monotonic functions, local maxima and minima and certain theorems etc. After these the Class 12 Maths chapter 6 exercise 6.5 is given for practice. Students should try to solve NCERT syllabus exercise before looking into the solutions of Class 12th Maths chapter 6 exercise 6.5.

**Also, read**

Application of Derivatives Exercise 6.1

Application of Derivatives Exercise 6.2

Application of Derivatives Exercise 6.3

Application of Derivatives Exercise 6.4

Application of Derivatives Miscellaneous Exercise

## Application of Derivatives Exercise 6.5

**Question:1(i) **Find the maximum and minimum values, if any, of the following functions

given by

( )

** Answer: **

Given function is,

add and subtract 2 in given equation

Now,

for every

Hence, minimum value occurs when

Hence, the minimum value of function occurs at

and the minimum value is

and it is clear that there is no maximum value of

**Question:1(ii) **Find the maximum and minimum values, if any, of the following functions

given by

**Answer:**

Given function is,

add and subtract 2 in given equation

Now, for every

Hence, minimum value occurs when

Hence, the minimum value of function occurs at and the minimum value isand it is clear that there is no maximum value of

** Question:1(iii) ** Find the maximum and minimum values, if any, of the following functions

given by

** Answer: **

Given function is,

for every

Hence, maximum value occurs when

Hence, maximum value of function occurs at x = 1

and the maximum value is

and it is clear that there is no minimum value of

** Question:1(iv) ** Find the maximum and minimum values, if any, of the following functions

given by

** Answer: **

Given function is,

value of varies from

Hence, function neither has a maximum or minimum value

** Question:2(i) ** Find the maximum and minimum values, if any, of the following functions

given by

** Answer: **

Given function is

Hence, minimum value occurs when |x + 2| = 0

x = -2

Hence, minimum value occurs at x = -2

and minimum value is

It is clear that there is no maximum value of the given function

** Question:2(ii) ** Find the maximum and minimum values, if any, of the following functions

given by

** Answer: **

Given function is

Hence, maximum value occurs when -|x + 1| = 0

x = -1

Hence, maximum value occurs at x = -1

and maximum value is

It is clear that there is no minimum value of the given function

** Question:2(iii) ** Find the maximum and minimum values, if any, of the following functions

given by

** Answer: **

Given function is

We know that value of sin 2x varies from

Hence, the maximum value of our function is 6 and the minimum value is 4

** Question:2(iv) ** Find the maximum and minimum values, if any, of the following functions

given by

** Answer: **

Given function is

We know that value of sin 4x varies from

Hence, the maximum value of our function is 4 and the minimum value is 2

** Question:2(v) ** Find the maximum and minimum values, if any, of the following functions

given by

** Answer: **

Given function is

It is given that the value of

So, we can not comment about either maximum or minimum value

Hence, function has neither has a maximum or minimum value

** Question:3(i) ** 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:

** Answer: **

Given function is

So, x = 0 is the only critical point of the given function

So we find it through the 2nd derivative test

Hence, by this, we can say that 0 is a point of minima

and the minimum value is

** Question:3(ii) ** 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:

** Answer: **

Given function is

Hence, the critical points are 1 and - 1

Now, by second derivative test

Hence, 1 is the point of minima and the minimum value is

Hence, -1 is the point of maxima and the maximum value is

** Question:3(iii) ** 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:

** Answer: **

Given function is

Now, we use the second derivative test

Hence, is the point of maxima and the maximum value is which is

** Question:3(iv) ** 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:

** Answer: **

Given function is

Now, we use second derivative test

Hence, is the point of maxima and maximum value is which is

** Question:3(v) ** 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:

** Answer: **

Givrn function is

Hence 1 and 3 are critical points

Now, we use the second derivative test

Hence, x = 1 is a point of maxima and the maximum value is

Hence, x = 1 is a point of minima and the minimum value is

** Question:3(vi) ** 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:

** Answer: **

Given function is

( but as we only take the positive value of x i.e. x = 2)

Hence, 2 is the only critical point

Now, we use the second derivative test

Hence, 2 is the point of minima and the minimum value is

** Question:3(vii) ** 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:

** Answer: **

Gien function is

Hence., x = 0 is only critical point

Now, we use the second derivative test

Hence, 0 is the point of local maxima and the maximum value is

** Question:3(viii) ** 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:

** Answer: **

Given function is

Hence, is the only critical point

Now, we use the second derivative test

Hence, it is the point of minima and the minimum value is

** Question:4(i) ** Prove that the following functions do not have maxima or minima:

** Answer: **

Given function is

But exponential can never be 0

Hence, the function does not have either maxima or minima

** Question:4(ii) ** Prove that the following functions do not have maxima or minima:

** Answer: **

Given function is

Since log x deifne for positive x i.e.

Hence, by this, we can say that for any value of x

Therefore, there is no such that

Hence, the function does not have either maxima or minima

** Question:4(iii) ** Prove that the following functions do not have maxima or minima:

** Answer: **

Given function is

But, it is clear that there is no such that

Hence, the function does not have either maxima or minima

** Question:5(i) ** Find the absolute maximum value and the absolute minimum value of the following

functions in the given intervals:

** Answer: **

Given function is

Hence, 0 is the critical point of the function

Now, we need to see the value of the function at x = 0 and as

we also need to check the value at end points of given range i.e. x = 2 and x = -2

Hence, maximum value of function occurs at x = 2 and value is 8

and minimum value of function occurs at x = -2 and value is -8

** Question:5(ii) ** Find the absolute maximum value and the absolute minimum value of the following

functions in the given intervals:

** Answer: **

Given function is

as

Hence, is the critical point of the function

Now, we need to check the value of function at and at the end points of given range i.e.

Hence, the absolute maximum value of function occurs at and value is

and absolute minimum value of function occurs at and value is -1

** Question:5(iii) ** Find the absolute maximum value and the absolute minimum value of the following

functions in the given intervals:

** Answer: **

Given function is

Hence, x = 4 is the critical point of function

Now, we need to check the value of function at x = 4 and at the end points of given range i.e. at x = -2 and x = 9/2

Hence, absolute maximum value of function occures at x = 4 and value is 8

and absolute minimum value of function occures at x = -2 and value is -10

** Question:5(iv) ** Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

** Answer: **

Given function is

Hence, x = 1 is the critical point of function

Now, we need to check the value of function at x = 1 and at the end points of given range i.e. at x = -3 and x = 1

Hence, absolute maximum value of function occurs at x = -3 and value is 19

and absolute minimum value of function occurs at x = 1 and value is 3

** Question:6 ** . Find the maximum profit that a company can make, if the profit function is

given by

** Answer: **

Profit of the company is given by the function

x = -2 is the only critical point of the function

Now, by second derivative test

At x = -2

Hence, maxima of function occurs at x = -2 and maximum value is

Hence, the maximum profit the company can make is 113 units

** Question:7 ** . Find both the maximum value and the minimum value of

on the interval [0, 3].

** Answer: **

Given function is

Now, by hit and trial let first assume x = 2

Hence, x = 2 is one value

Now,

which is not possible

Hence, x = 2 is the only critical value of function

Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 0 and x = 3

Hence, maximum value of function occurs at x = 0 and vale is 25

and minimum value of function occurs at x = 2 and value is -39

** Question:8 ** . At what points in the interval does the function attain its maximum value?

** Answer: **

Given function is

So, values of x are

These are the critical points of the function

Now, we need to find the value of the function at and at the end points of given range i.e. at x = 0 and

Hence, at function attains its maximum value i.e. in 1 in the given range of

** Question:9 ** What is the maximum value of the function ?

** Answer: **

** ** Given function is

Hence, is the critical point of the function

Now, we need to check the value of the function at

Value is same for all cases so let assume that n = 0

Now

Hence, the maximum value of the function is

** Question:10. ** Find the maximum value of in the interval [1, 3]. Find the

the maximum value of the same function in [–3, –1].

** Answer: **

Given function is

we neglect the value x =- 2 because

Hence, x = 2 is the only critical value of function

Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 1 and x = 3

Hence, maximum value of function occurs at x = 3 and vale is 89 when

Now, when

we neglect the value x = 2

Hence, x = -2 is the only critical value of function

Now, we need to check the value at x = -2 and at the end points of given range i.e. x = -1 and x = -3

Hence, the maximum value of function occurs at x = -2 and vale is 139 when

** Question:11. ** It is given that at x = 1, the function attains its maximum value, on the interval [0, 2]. Find the value of a.

** Answer: **

Given function is

Function attains maximum value at x = 1 then x must one of the critical point of the given function that means

Now,

Hence, the value of a is 120

** Question:12 ** . Find the maximum and minimum values of

** Answer: **

Given function is

So, values of x are

These are the critical points of the function

Now, we need to find the value of the function at and at the end points of given range i.e. at x = 0 and

Hence, at function attains its maximum value and value is in the given range of

and at x= 0 function attains its minimum value and value is 0

** Question:13 ** . Find two numbers whose sum is 24 and whose product is as large as possible.

** Answer: **

Let x and y are two numbers

It is given that

x + y = 24 , y = 24 - x

and product of xy is maximum

let

Hence, x = 12 is the only critical value

Now,

at x= 12

Hence, x = 12 is the point of maxima

Noe, y = 24 - x

= 24 - 12 = 12

Hence, the value of x and y are 12 and 12 respectively

** Question:14 ** Find two positive numbers x and y such that x + y = 60 and is maximum.

** Answer: **

It is given that

x + y = 60 , x = 60 -y

and is maximum

let

Now,

Now,

hence, 0 is neither point of minima or maxima

Hence, y = 45 is point of maxima

x = 60 - y

= 60 - 45 = 15

Hence, values of x and y are 15 and 45 respectively

** Question:15 ** Find two positive numbers x and y such that their sum is 35 and the product is a maximum.

** Answer: **

It is given that

x + y = 35 , x = 35 - y

and is maximum

Therefore,

Now,

Now,

Hence, y = 35 is the point of minima

Hence, y= 0 is neither point of maxima or minima

Hence, y = 25 is the point of maxima

x = 35 - y

= 35 - 25 = 10

Hence, the value of x and y are 10 and 25 respectively

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

** Answer: **

let x an d y are positive two numbers

It is given that

x + y = 16 , y = 16 - x

and is minimum

Now,

Hence, x = 8 is the only critical point

Now,

Hence, x = 8 is the point of minima

y = 16 - x

= 16 - 8 = 8

Hence, values of x and y are 8 and 8 respectively

** Question:17 ** . A square piece of tin of side 18 cm is to be 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: **

It is given that the side of the square is 18 cm

Let assume that the length of the side of the square to be cut off is x cm

So, by this, we can say that the breath of cube is (18-2x) cm and height is x cm

Then,

Volume of cube =

But the value of x can not be 9 because then the value of breath become 0 so we neglect value x = 9

Hence, x = 3 is the critical point

Now,

Hence, x = 3 is the point of maxima

Hence, the length of the side of the square to be cut off is 3 cm so that the volume of the box is the maximum possible

** Question:18 ** A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum ?

** Answer: **

It is given that the sides of the rectangle are 45 cm and 24 cm

Let assume the side of the square to be cut off is x cm

Then,

Volume of cube

But x cannot be equal to 18 because then side (24 - 2x) become negative which is not possible so we neglect value x= 18

Hence, x = 5 is the critical value

Now,

Hence, x = 5 is the point of maxima

Hence, the side of the square to be cut off is 5 cm so that the volume of the box is maximum

** Question:19 ** Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

** Answer: **

Let assume that length and breadth of rectangle inscribed in a circle is l and b respectively

and the radius of the circle is r

Now, by Pythagoras theorem

a = 2r

Now, area of reactangle(A) = l b

Now,

Hence, is the point of maxima

Since, l = b we can say that the given rectangle is a square

Hence, of all the rectangles inscribed in a given fixed circle, the square has the maximum area

** Question:20 ** . Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

** Answer: **

Let r be the radius of the base of cylinder and h be the height of the cylinder

we know that the surface area of the cylinder

Volume of cylinder

Hence, is the critical point

Now,

Hence, is the point of maxima

Hence, the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter(D = 2r) of the base

** Question:21 ** 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 be the radius of base and h be the height of the cylinder

The volume of the cube (V) =

It is given that the volume of cylinder = 100

Surface area of cube(A) =

Hence, is the critical point

Hence, is the point of minima

Hence, and are the dimensions of the can which has the minimum surface area

** Question:22 ** 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: **

Area of the square (A) =

Area of the circle(S) =

Given the length of wire = 28 m

Let the length of one of the piece is x m

Then the length of the other piece is (28 - x) m

Now,

and

Area of the combined circle and square = A + S

Now,

Hence, is the point of minima

Other length is = 28 - x

=

Hence, two lengths are and

** Question:23 ** 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: ** Volume of cone (V) =

Volume of sphere with radius r =

By pythagoras theorem in we ca say that

V =

Now,

Hence, point is the point of maxima

Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is

Volume =

Hence proved

** Question:24 ** 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: **

Volume of cone(V)

curved surface area(A) =

Now , we can clearly varify that

when

Hence, is the point of minima

Hence proved that the right circular cone of least curved surface and given volume has an altitude equal to time the radius of the base

** Question:25 ** Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is

** Answer: **

Let a be the semi-vertical angle of cone

Let r , h , l are the radius , height , slent height of cone

Now,

we know that

Volume of cone (V) =

Now,

Now,

Now, at

Therefore, is the point of maxima

Hence proved

** Question:26 ** Show that semi-vertical angle of the right circular cone of given surface area and maximum volume is

** Answer: **

Let r, l, and h are the radius, slant height and height of cone respectively

Now,

Now,

we know that

The surface area of the cone (A) =

Now,

Volume of cone(V) =

On differentiate it w.r.t to a and after that

we will get

Now, at

Hence, we can say that is the point if maxima

Hence proved

** Question:27 ** The point on the curve which is nearest to the point (0, 5) is

** Answer: **

Given curve is

Let the points on curve be

Distance between two points is given by

Hence, x = 0 is the point of maxima

Hence, the point is the point of minima

Hence, the point is the point on the curve which is nearest to the point (0, 5)

Hence, the correct answer is (A)

** Question:28 ** For all real values of x, the minimum value of

is

(A) 0 (B) 1 (C) 3 (D) 1/3

** Answer: **

Given function is

Hence, x = 1 and x = -1 are the critical points

Now,

Hence, x = 1 is the point of minima and the minimum value is

Hence, x = -1 is the point of maxima

Hence, the minimum value of

is

Hence, (D) is the correct answer

** Question:29 ** The maximum value of

** Answer: **

Given function is

Hence, x = 1/2 is the critical point s0 we need to check the value at x = 1/2 and at the end points of given range i.e. at x = 1 and x = 0

Hence, by this we can say that maximum value of given function is 1 at x = 0 and x = 1

option c is correct

**More About NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5**

There are 29 questions in exercise 6.5 Class 12 Maths.

Question numbers 27 to 29 of Class 12th Maths chapter 6 exercise 6.5 are multiple-choice questions.

- There will be 4 choices given for multiple-choice questions in the NCERT and have to select the correct answer.

**Also Read| **Application of Derivatives Class 12 Notes

**Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5**

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**exercise 6.5 Class 12 Maths**is solved by expert Mathematics faculties and students can rely on Class 12th Maths chapter 6 exercise 6.5 for exam preparationNot only for board exams but also for competitive exams like JEE main the NCERT solutions for Class 12 Maths chapter 6 exercise 6.5 will be helpful

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### Also see-

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