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Edited By Ramraj Saini | Updated on Dec 03, 2023 08:53 PM IST | #CBSE Class 12th

**NCERT Solutions for Exercise 6.5 Class 12 Maths Chapter 6 Application of Derivatives **are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the **latest syllabus and pattern of CBSE 2023-24. **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.

This Story also Contains

- NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5
- Access NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5
- Application of Derivatives Exercise 6.5
- More About NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5
- Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5
- Key Features Of NCERT Solutions for Exercise 6.5 Class 12 Maths Chapter 6
- NCERT Solutions Subject Wise
- Subject Wise NCERT Exemplar Solutions

**12th class Maths exercise 6.5 **answers** **are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

**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

**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

** 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

** 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

** Answer: **

Given function is

Now, we use the second derivative test

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

** Answer: **

Given function is

Now, we use second derivative test

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

** 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

** 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

** 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

** 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

** 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

** 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

** 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

** 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

** 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

** 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

- 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.

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

- The
**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 preparation. - Not 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.
- Students can access these solutions for free.

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

**Comprehensive Coverage:**The solutions encompass all the topics covered in ex 6.5 class 12, ensuring a thorough understanding of the concepts.**Step-by-Step Solutions:**In this class 12 maths ex 6.5, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.**Accuracy and Clarity:**Solutions for class 12 ex 6.5 are presented accurately and concisely, using simple language to help students grasp the concepts easily.**Conceptual Clarity:**In this 12th class maths exercise 6.5 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.**Inclusive Approach:**Solutions for ex 6.5 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.**Relevance to Curriculum:**The solutions for class 12 maths ex 6.5 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

- NCERT Solutions Class 12 Chemistry
- NCERT Solutions for Class 12 Physics
- NCERT Solutions for Class 12 Biology
- NCERT Solutions for Class 12 Mathematics

1. What are the topics covered in exercise 6.5?

The topic of maxima and minima is covered in Class 12th Maths chapter 6 exercise 6.5.

2. How many examples are solved before Class 12 Maths exercise 6.5 on the topic maxima and minima?

16 questions are explained before exercise 6.5

3. What is the total number of solved examples in the NCERT book till Exercise 6.5 Class 12 Maths?

A sum of 41 questions are solved till the Class 12th Maths chapter 6 exercise 6.5

4. Give some applications of maxima and minima?

The problems of maxima and minima are used in business, science and mathematics etc. Examples are problems related to maximising profit, minimising the distance between the two points etc

5. What is a monotonic function in a given interval?

In the given interval, a monotonic function is either increasing or decreasing.

6. For a function f, if k is a point of local maxima, then what is the local maximum value of the function is?

The local maximum value is f(k)

7. For a function f, if u is a point of local minima, then what is the local minimum value of the function is?

f(u) is the local minimum value of function f

8. Is the topic maxima and minima important for board exams?

Yes, maxima and minima are one of the important topics of the chapter from where 1 question can be expected for CBSE board exam

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6 minHave a question related to CBSE Class 12th ?

You can use them people also used problem

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

**
Thank you and wishing you all the best for your bright future.
**

Hello student,

**
If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:
**

- No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
- You have to appear for the 2025 12th board exams.
- Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
- Aim to register before late October to avoid extra fees.
- Schools might not offer classes for private students, so focus on self-study or coaching.

**
Remember
**
, these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

**
Good luck with your studies!
**

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