NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.5

Class 12 is a crucial year in a student’s academic career. Several colleges and institutions base their admissions on the results of the Class 12 board exams. An essential factor in determining admission to different courses is the marks obtained in the board exams. As a result, students must perform well in the board examination. But getting ready for the board exam is a difficult task. It necessitates a great deal of effort and careful planning. Students must practice a lot of questions, especially for a subject like mathematics. Teachers advise students to practice NCERT questions for a thorough understanding. The CBSE also suggests using the NCERT textbook for its board exams. Students should therefore complete the NCERT exercises before appearing for the board exams.

Chapter 6 of the NCERT book for Mathematics is on the Application of Derivatives. Students should be thorough with the previous chapter on Continuity and Differentiability to be able to understand the concepts in this chapter. This chapter is about the application of the concepts learned in the previous chapter. Exercise 6.5 of this chapter is based on the topic of Maxima and Minima. Students should study the concepts under this exercise thoroughly to be able to solve the questions in this exercise. It is always helpful to solve the exercises from the NCERT book, as these questions are often asked in board examinations. However, some questions in this exercise can be difficult for students to solve. That is why Extramarks provides the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5.

For students in Class 12 to advance in their preparation for the Class 12 board examinations in Mathematics, NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 is considered a very crucial resource. The CBSE suggests using the Mathematics textbook from NCERT for Class 12. The NCERT textbook is also suggested in the syllabus of several other boards. Hence, the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 provided by Extramarks will not only help students studying under the CBSE but also students from other boards.

Derivatives find application in a wide range of subjects other than Mathematics. They are used in Physics extensively, for example, to calculate Velocity, Acceleration, etc. Derivatives help in finding the rate of change of one quantity with respect to another quantity. They are widely applied in Chemistry to find the rate of a chemical reaction. Derivatives can also be used in the equation of a Tangent to a Curve. Hence, students need to be thorough with this chapter to understand the concepts in several other subjects. To help students understand the concepts in this chapter, Extramarks provides a wide range of study materials. The study materials, including the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 contain all the key ideas outlined in the CBSE-recommended syllabus for the board exams. To help students prepare for the questions posed in the CBSE board examinations, qualified teachers have created the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5. Apart from that, students can prepare for a variety of college entrance exams, such as the IIT-JEE and VITEEE, by practising the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5. To fully understand the questions, students should practise the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 at least twice.

Finding the best resource for their learning goals is typically a challenge for students. The appropriate and essential answer to their educational demands is provided by NCERT Solutions. The availability of NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 satisfies the requirements of students in their board exam preparation. The CBSE prescribes the NCERT textbooks in its school curriculum. There are several exam-relevant and crucial-for-practice questions in the textbooks. Students are frequently perplexed by these questions, which causes them to lack confidence in the subject and the topic. It is critical to understand the ideas right from the very first topic. Students who have access to the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 will benefit from having a firm understanding of both fundamental and complex ideas when attempting the questions presented in the NCERT textbooks. By completing it, students can have a strong conceptual knowledge of the Class 12th Maths Exercise 6.5. Additionally, they can analyse themselves and make improvements by comparing their answers with the solutions.

Studying the concepts under the Class 12 Maths Chapter 6 Exercise 6.5 helps students find the maximum and minimum values of a given function within a given interval. Students can also learn to find the maxima and minima of a given function. There are multiple theorems on this topic. These theorems also include the first and second derivative tests, which are quite important from an examination point of view. The students’ duty as they get ready for the board exams is made significantly easier by the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5. The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 provided by Extramarks offer clear answers to the problems posed in this exercise. Experienced subject specialists have created the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 while considering the students’ level of comprehension. The most recent curriculum recommended by the CBSE for Class 12 has been used to create the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5. Students can therefore stop looking elsewhere and not have to worry about the syllabus. Students in Class 12 can go to the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 as a one-stop resource for simple, reliable answers to all the exercises’ questions.

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.5) Exercise 6.5

The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 may not always be available to students. The ability to connect to the internet is crucial, yet it may not always be possible. Students who want to continue learning should have the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 in PDF format on hand. Extramarks offers a comprehensive solution for all the learning demands of students. Although NCERT textbooks are an excellent source of study material, students still have a real concern about finding trustworthy NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5. Extramarks offers accurate and comprehensive NCERT solutions. In addition to this, it provides a comprehensive learning solution that combines the benefits of live classrooms and learning apps to give students a seamless learning environment.

Access NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives

NCERT Solution Class 12 Maths of Chapter 6 All Exercises

Chapter 6 of the Mathematics NCERT textbook for Class 12 deals with the Application of Derivatives. This chapter includes 5 exercises for students to solve the questions based on this chapter. Apart from that, students can also find various exercises in Chapter 6 at the end. Students can also find various examples that can provide them with help in solving the questions in the exercises. Exercise 6.1 is based on the Rate of Change of Quantities. There are various problems in this exercise where students have to find the rate of change of one quantity with respect to another quantity. This exercise contains 18 questions. The last two questions are multiple choice-type questions, while the rest are subjective-type questions. Students can refer to the solutions provided by Extramarks to help themanswer the questions.

Exercise 6.2 includes 19 questions. One of these questions is objective type while the other questions are subjective type. This exercise is based on Increasing and Decreasing Functions. The questions require students to find out if the given functions are increasing, decreasing, or constant in a given interval.

Exercise 6.3 has 27 questions in total. 25 questions are subjective-type questions, and the rest are objective-type questions. This exercise is based on Tangents and Normals. Students are required to learn to find the equation of the tangent line and the normal line to a given curve at a specific point using differentiation. Some questions also require students to find the slope of the tangent line and the normal line to a given curve. Students should practice as many questions as they can because these questions are very important from an examination point of view.

Exercise 6.4 is based on Approximations. Students are required to find the approximate values of certain quantities using differentiation. This is a smaller exercise as compared to other exercises, as it contains nine questions in total. One of these questions is an objective-type question, while the others are subjective-type questions.

Exercise 6.5 has 29 questions in total. Three of these questions are objective-type questions, while the rest are subjective-type questions. This exercise is based on the concepts of Maxima and Minima. Students are required to find the maximum and minimum values of the given functions using the concept of derivatives. Some questions require students to find the points of maximum and minimum values. They are also expected to learn to find the points of local maxima and minima for a given function. They are required to learn to use the first and second derivative tests to solve these problems. Students should practice as many questions as they can. Some questions might be challenging to solve. Students can take the help of the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 provided by Extramarks to solve the difficult questions. They should also refer to the stepwise answers to learn how to write answers in a way that can earn them maximum marks.

Apart from the above-mentioned exercises, the NCERT textbook also contains a miscellaneous exercise in chapter 6. There are 24 questions in this exercise. Six of these questions are multiple-choice types. The rest of the questions are subjective-type questions. The questions in this exercise are based on all the concepts learnt in the chapter. These questions are generally of a higher difficulty level than the questions in other exercises. Students should solve all the questions from other exercises to be able to solve the questions in this exercise.

Students can check out the NCERT Solutions Class 12 provided by Extramarks to find the solutions to the questions in all the exercises.

Class 12 Maths Chapter 6 Exercise 6.5

The Class 12th Math 6.5 exercise deals with the concepts of Maxima and Minima. This exercise includes 29 questions. Three of these questions are of the multiple choice type. Students should try to answer all 29 questions correctly to be thorough with the concepts. Students who have access to the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 can assess their comprehension and identify their strong and weak areas. The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 also offer reviewed and analysed learning while aiding students in the development of a variety of skills, including logical and reasoning skills. The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 also aid students in successfully and efficiently preparing for the board exams. It is considered an essential step from the standpoint of the exam. In this exercise, conceptual comprehension and application are crucial. Students gain a clear understanding from having access to the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 on Extramarks in a variety of ways. The clear and step-by-step solutions provided in the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 help them resolve challenging problems. The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 also helps them in acquiring the skill of problem-solving.

What Is Maxima and Minima?

Maxima and Minima are two of the most important concepts in the Application of Derivatives. Maxima is a point where a given function attains its maximum value. On the other hand, Minima is a point where the function attains the minimum value. Students can use the first and second derivative tests to find the points of Maxima and Minima of a function. The board examinations often include questions based on these concepts. Hence, students must answer all the questions in this exercise. Apart from that, they should also go through the examples in the textbook for a clearer understanding of solving the questions. They can also check the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 provided by Extramarks for help.

What Is The Significance Of Ex 6.5 Class 12 NCERT Solutions?

NCERT exercises are considered the most important source for practising questions for the CBSE board examinations. Hence, students are always advised to solve the questions from the NCERT textbook. Along with that, they also need the solutions to the questions so that they can check their answers. Hence, the NCERT solutions provide students with the support they need while preparing for the board exams. Students can more easily and precisely apply the ideas covered in chapter 3 by using the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5. The accurate and error-free NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 have been created by highly qualified teachers. They create these solutions while taking into account the comprehension level of Class 12 students. To make composing exam answers easier, the questions in this exercise have been broken down into smaller, more manageable steps. Examiners consistently emphasise the need for step-by-step written answers. To achieve perfect scores on exams, students should use the step-by-step solutions provided in the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5. Many students struggle to comprehend and retain the numerous formulas provided in the NCERT textbook, which prevents them from applying them to the exercises’ questions. However, it is simpler to recall the formulas and use them correctly if they practice the stepwise solutions in the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5. This offers the students an advantage in their exams because they know how to approach problems correctly. Class 12 students can fully understand the questions even if they are tackling them for the first time thanks to Extramarks’ step-by-step preparation of the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5. Alternative and simpler solutions to some problems are also provided in Extramarks’ NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5.

Along with the CBSE board exams, the chapter on Applications of Derivatives is a crucial component of competitive exams like JEE Main and JEE Advanced. Since college admissions are based on the results of these exams, a student’s performance in these exams determines their career. As a result of the tremendous strain placed on them, they get overwhelmed and occasionally demotivated. However, using the study guide and the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 provided by Extramarks together allows students to fully grasp the concepts and retain them for a long time. This gives them more self-assurance as they prepare for various exams. Students who study the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 and other solutions can easily pass the JEE, IIT, BITSAT, and other entrance exams. To help students succeed in these challenging exams, the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 clearly explains the solutions. Additionally, the structure of the Class 12 board exams ensures that the student’s ability for problem-solving is put to the ultimate test. Anyone who can pass this difficult test will be able to pass other competitive exams. The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 will help them to improve their fundamentals and play a significant role in their entire preparation period, whether it be for the board examination or any competitive examination.

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NCERT books strictly adhere to the CBSE curriculum. The NCERT textbook serves as the source for the majority of board exam questions. To perform well in their board exams, students must diligently practice the exercises. Also, the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 are easily accessible on Extramarks. The platform’s in-house subject-matter experts have put together the solutions and study materials according to the curriculum. The information given to the students has been fact-checked and is supported by facts.

Students preparing for board exams can benefit from the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 since they are presented in straightforward language that every student can understand. The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 have been written in a step-by-step format to assist students in obtaining full marks in their exams. To prevent students from having to look elsewhere for the concepts covered in the syllabus or the distribution of marks, the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 have been created following the most recent syllabus. The solutions follow all the guidelines established by the CBSE. All the problems in the exercise are answered in the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5, which are also totally error-free. Students can attempt such questions fast and correctly in exams by repeatedly practising the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5. Because they are prepared straightforwardly, students can utilise the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 whenever and wherever they like without requiring any outside assistance.

What is the Procedure of Practising with Ex 6.5 Class 12 Maths NCERT Solutions?

Exercise 6.5 is one of the essential exercises for students preparing for the Class 12 board examinations. Students attempting to solve the questions in this exercise should follow simple procedures to be able to easily solve the problems. Students are required to write stepwise answers in Mathematics. Hence, they should know what steps to follow while writing answers to score full marks. Students should first thoroughly read and understand the concepts of Maxima and Minima from the textbook. They can also rely on the study materials provided by Extramarks if they are facing any difficulties understanding the concepts. Once they are thorough with the concepts, they should go through the examples provided in the NCERT book. It will help them get an idea of how to solve the questions.

Class 12 students should attempt to solve the questions in the exercise sequentially. They should first try to solve the questions without looking at the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5. They should try to recall all the processes that could possibly be used while solving the questions. Once they have attempted all the questions, they should refer to the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 to check their answers. If they have written wrong answers, they should follow the solutions to rectify them. Students should try to solve such questions multiple times until they can answer them correctly. They should also refer to the solutions to find answers to the questions they are unable to solve. Once they have learned to correctly solve the questions from the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5, they should try to solve them again. This will help them get acquainted with the questions and speed up their writing. They should solve the questions in the shortest time possible. This will help them solve questions in the board exam faster and save time so that they can complete the examination paper on time.

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Q.1 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

Ans

$\begin{array}{l} (i) The given function is \\ f (x) = {(2 x - 1)}^{2} + 3 \\ It is clear that {(2 x - 1)}^{2} \geq 0 for \forall x \in R \\ Therefore, f (x) = {(2 x - 1)}^{2} + 3 \geq 3 for \forall x \in R \\ The minimum value of f is attained \\ when 2 x - 1 = 0 \Rightarrow x = \frac{1}{2} \\ ∴ Minimum value of f = f (\frac{1}{2}) \\ = {(2 \times \frac{1}{2} - 1)}^{2} + 3 \\ = 3 \\ Hence, function f does not have a maximum value . \\ (ii) The given function is \\ f (x) = 9 x^{2} + 12 x + 2 = {(3 x + 2)}^{2} - 2. \\ It is clear that {(3 x + 2)}^{2} \geq 0 for \forall x \in R \\ Therefore, f (x) = {(3 x + 2)}^{2} - 2 \geq - 2 for \forall x \in R \\ The minimum value of f is attained \\ when 3 x + 2 = 0 \Rightarrow x = - \frac{2}{3} \\ ∴ Minimum value of f = f (- \frac{2}{3}) \\ = {(2 \times - \frac{2}{3} + 2)}^{2} - 2 \\ = - 2 \\ Hence, function f does not have a maximum value . \\ (iii) The given function is f (x) = - {(x - 1)}^{2} + 10 \\ It is clear that (x - 1)^{2} \geq 0 for every x \in R \\ Therefore, f (x) = - {(x - 1)}^{2} + 10 \leq 10 for every x \in R \\ The maximum value of f is attained when \\ (x - 1) = 0 \Rightarrow x = 1 \\ ∴ Maximum value of f = f (1) \\ = - {(1 - 1)}^{2} + 10 \\ = 10 \\ Hence, function f does not have a minimum value . \\ (iv) The given function is g (x) = x^{3} + 1 \\ Here, function g has neither a maximum value nor a minimum value . \end{array}$

Q.2

$\begin{array}{l} Find the maximum and minimum values, if any, of the \\ following functions given by : \\ (i) |x + 2| - 1 (ii) g (x) = - |x + 1| + 3 \\ (iii) h (x) = \sin 2 x + 5 (iv) f (x) = |\sin 4 x + 3| \\ (v) h (x) = x + 1, x \in (- 1, 1) \end{array}$

Ans

$\begin{array}{l} (i) f (x) = | x + 2 | - 1 \\ We know that | x + 2 | \geq 0 \forall x \in R \\ Therefore, f (x) = | x + 2 | - 1 \geq - 1 \forall x \in R \\ The minimum value of f is attained when \\ | x + 2 | = 0 \Rightarrow x = - 2 \\ ∴ Minimum value of f = f (- 2) \\ = | - 2 + 2 | - 1 \\ = - 1 \\ Thus, function f does not have a maximum value . \\ (ii) g (x) = - | x + 1 | + 3 \\ Since, - | x + 1 | \leq 0 \forall x \in R \\ Therefore, g (x) = - | x + 1 | + 3 \leq 3 \forall x \in R \\ The maximum value of g is attained when \\ - | x + 1 | = 0 \Rightarrow x = - 1 \\ Maximum value of g = g (- 1) \\ = - | - 1 + 1 | + 3 \\ = 3 \\ Hence, function g does not have a minimum value . \\ (iii) f (x) = | \sin 4 x + 3 | \\ Since, - 1 \leq \sin 4 x \leq 1 \\ \Rightarrow 2 \leq \sin 4 x + 3 \leq 4 \\ Thus, the maximum and minimum values of f are 4 \\ and 2 respectively . \\ (v) h (x) = x + 1, x \in (- 1, 1) \\ Here, if a point x_{0} is closest to - 1, then we find \\ \frac{x_{0} + 1}{2} < x_{0} + 1 for all x_{0} \in (- 1, 1) \\ Also, if x_{1} is closest to 1, then x_{1} + 1 < \frac{x_{1} + 1}{2} + 1 \forall x \in (- 1, 1) \\ Hence, function h (x) has neither maximum nor \\ minimum value in (- 1, 1) . \end{array}$

Q.3 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:

$\begin{array}{l} (i) f (x) = x^{2} (i i) g (x) = x^{3} - 3 x \\ (i i i) h (x) = s i n x + c o s x, 0 < x < \frac{π}{2} \\ (i v) f (x) = s i n x - c o s x, 0 < x < \frac{π}{2} \\ (v) f (x) = x^{3} - 6 x^{2} + 9 x + 15 \\ (v i) g (x) = \frac{x}{2} + \frac{2}{x}, x > 0 \\ (v i i) g (x) = \frac{1}{x^{2} + 2} (v i i i) f (x) = x \sqrt{1 - x}, x > 0 \end{array}$

Ans

$\begin{array}{l} (i) f (x) = x^{2} . .. (i) \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 2 x . .. (ii) \\ For maxima or minima, \\ f ‘ (x) = 0 \\ \Rightarrow 2 x = 0 \Rightarrow x = 0 \\ Thus, x = 0 is the only critical point which could possibly \\ be the point of local maxima or local minima of f . \\ differentiating equation (ii) w . r . t . x, we get \\ f ” (x) = 2, which is positive . \\ Therefore, by second derivative test, x = 0 is a point of local \\ minima and local minimum value of f at x = 0 is f (0) = 0. \\ (ii) g (x) = x^{3} - 3 x . .. (i) \\ Differentiating w . r . t . x, we get \\ g ‘ (x) = 3 x^{2} - 3 . .. (ii) \\ For maxima or minima, we have \\ g ‘ (x) = 0 \Rightarrow 3 x^{2} - 3 = 0 \\ \Rightarrow x = \pm 1 \\ Differentiating equation (ii) w . r . t . x, we get \\ g ‘ ‘ (x) = 6 x \\ Putting x = 1, we get \\ g ‘ ‘ (1) = 6 (1) = 6 > 0 \end{array}$ $\begin{array}{l} S o, g (x) is minimum at x = 1 and minimum value is \\ g (1) = {(1)}^{3} - 3 (1) = - 2 \\ P u t t i n g x = - 1, we get \\ g ‘ ‘ (- 1) = 6 (- 1) = - 6 < 0 \\ S o, g (x) is maximum at x = - 1 and maximum value is \\ g (- 1) = {(- 1)}^{3} - 3 (- 1) = 2. \\ (i i i) h (x) = sinx + cosx, 0 < x < \frac{π}{2} … (i) \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ h’ (x) = cosx - sinx … (i i) \\ F o r maxima or minima, we have \\ h’ (x) = 0 \Rightarrow cosx - sinx = 0 \\ \Rightarrow \cos x = \sin x \\ \Rightarrow \frac{\cos x}{\sin x} = 1 \\ \Rightarrow \cot x = \cot 45 ° \Rightarrow x = 45 ° \\ D i f f e r e n t i a t i n g equation (i i) w . r . t . x, we get \\ h” (x) = - sinx - cosx \\ putting x = 45 °, w e get \\ h” (45 °) = - sin 45 ° - cos 45 ° \\ = - (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) < 0 \\ T h u s, h (x) is maximum at x = 45°, then maximum value is \\ h (45 °) = sin 45 ° + cos 45 ° \\ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \\ = \frac{2}{\sqrt{2}} = \sqrt{2} . \\ (i v) f (x) = \sin x - \cos x, 0 < x < 2 π \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ f’ (x) = c o x + \sin x \\ F o r maxima or minima, \\ f’ (x) = 0 \Rightarrow c o x + \sin x = 0 \\ \Rightarrow \sin x = - \cos x \\ \Rightarrow \tan x = - 1 \Rightarrow x = \frac{3 π}{4}, \frac{7 π}{4} \in (0, 2 π) \\ a n d f ‘ ‘ (x) = - \sin x + \cos x \\ \Rightarrow f ‘ ‘ (\frac{3 π}{4}) = - \sin \frac{3 π}{4} + \cos \frac{3 π}{4} \\ = - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \\ = - \sqrt{2} < 0 \\ Therefore, by second derivative test,x = \frac{3 π}{4} is a point of local \\ maxima and the local maximum value of f at x = \frac{3 π}{4} i s \\ f (\frac{3 π}{4}) = \sin \frac{3 π}{4} - \cos \frac{3 π}{4} \end{array}$ $\begin{array}{l} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \\ a n d \\ f ‘ ‘ (\frac{7 π}{4}) = - \sin \frac{7 π}{4} + \cos \frac{7 π}{4} \\ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \\ = \sqrt{2} > 0 \\ Therefore, by second derivative test,x = \frac{7 π}{4} is a point of local \\ minima and the local minimum value of f at x = \frac{7 π}{4} i s \\ f (\frac{7 π}{4}) = \sin \frac{7 π}{4} - \cos \frac{7 π}{4} \\ = - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = - \sqrt{2} \\ (v) f (x) = x^{3} - {6x}^{2} + 9x + 15 \\ Differentiating w .r .t . x, we get \\ f’ (x) = 3 x^{2} - 12x + 9 \\ For maxima or minima, we have \\ f’ (x) = 0 \Rightarrow 3 (x^{2} - 4x + 3) = 0 \\ \Rightarrow (x - 3) (x - 1) = 0 \\ \Rightarrow x = 3, 1 \\ N o w, \\ f ‘ ‘ (x) = 6 x - 12 \\ Putting x=3 in f” (x), we get \\ f ‘ ‘ (3) = 6 (3) - 12 = 6>0 \\ So, f (x) i s l o c a l minimum at x = 3, then l o c a l \\ minimum value = f (3) \\ = {(3)}^{3} - 6 {(3)}^{2} + 9 (3) +15 \\ = 15 \\ Putting x = 1 in f” (x), we get \\ f ‘ ‘ (1) = 6 (1) - 12 = - 6<0 \\ So, f (x) i s l o c a l maximum at x = 1, then l o c a l \\ maximum value = f (1) \\ = {(1)}^{3} - 6 {(1)}^{2} + 9 (1) +15 \\ = 19 \\ (v i) g (x) = \frac{x}{2} + \frac{2}{x} \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ g’ (x) = \frac{1}{2} - \frac{2}{x^{2}} a n d g ‘ ‘ (x) = \frac{4}{x^{3}} \\ F o r \max i m a or minima, we have \\ g’ (x) = 0 \Rightarrow \frac{1}{2} - \frac{2}{x^{2}} = 0 \\ \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2 \\ S i n c e, x>0 so, x = 2 \\ then, g” (2) = \frac{4}{2^{3}} = \frac{1}{2} > 0 \\ Therefore, by second derivative test, x = 2 is a point of local \\ minima and the local minimum value of g at x = 2 is \end{array}$ $\begin{array}{l} g(2) = \frac{2}{2} + \frac{2}{2} = 2 \\ (v i i) g (x) = \frac{1}{x^{2} + 2} \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ g’ (x) = - \frac{1}{{(x^{2} + 2)}^{2}} \frac{d}{d x} (x^{2} + 2) \\ = - \frac{2 x}{{(x^{2} + 2)}^{2}} \\ F o r maxima or minima, we have \\ g’ (x) = 0 \Rightarrow - \frac{2 x}{{(x^{2} + 2)}^{2}} = 0 \\ \Rightarrow x = 0 \\ 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 the \\ 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 g (0) = \frac{1}{0 + 2} = \frac{1}{2} \\ (v i i i) f (x) = x \sqrt{1 - x}, x > 0 \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ f ‘ (x) = \frac{d}{d x} (x \sqrt{1 - x}) \\ = x \frac{d}{d x} \sqrt{1 - x} + \sqrt{1 - x} \frac{d}{d x} x \\ = x (- \frac{1}{2 \sqrt{1 - x}}) + \sqrt{1 - x} \\ = \frac{- x + 2 - 2 x}{2 \sqrt{1 - x}} \\ = \frac{2 - 3 x}{2 \sqrt{1 - x}} \\ F o r maximum or minimum, we have \\ f ‘ (x) = 0 \Rightarrow \frac{2 - 3 x}{2 \sqrt{1 - x}} = 0 \\ \Rightarrow 2 - 3 x = 0 \Rightarrow x = \frac{2}{3} \\ f ‘ ‘ (x) = \frac{d}{d x} (\frac{2 - 3 x}{2 \sqrt{1 - x}}) \\ = \frac{2 \sqrt{1 - x} \frac{d}{d x} (2 - 3 x) - (2 - 3 x) \frac{d}{d x} 2 \sqrt{1 - x}}{{(2 \sqrt{1 - x})}^{2}} \\ = \frac{2 \sqrt{1 - x} (0 - 3) - (2 - 3 x) (- \frac{2}{2 \sqrt{1 - x}})}{{(2 \sqrt{1 - x})}^{2}} \\ = \frac{- 6 \sqrt{1 - x} + \frac{(2 - 3 x)}{\sqrt{1 - x}}}{4 (1 - x)} \\ = \frac{\frac{- 6 (1 - x) + 2 - 3 x}{\sqrt{1 - x}}}{4 (1 - x)} \\ = \frac{3 x - 4}{4 {(1 - x)}^{\frac{3}{2}}} \\ P u t t i n g x = \frac{2}{3} in f” (x), we get \\ f” (\frac{2}{3}) = \frac{3 (\frac{2}{3}) - 4}{4 {(1 - \frac{2}{3})}^{\frac{3}{2}}} \\ = \frac{2 - 4}{4 {(\frac{1}{3})}^{\frac{3}{2}}} \\ = \frac{- 2}{(\frac{4}{3 \sqrt{3}})} \\ = \frac{- 3 \sqrt{3}}{2} < 0 \\ S o, f (x) is maximum at x = \frac{2}{3} and maximum value is \\ f (\frac{2}{3}) = \frac{2}{3} \sqrt{1 - \frac{2}{3}} \\ = \frac{2}{3} \times \frac{1}{\sqrt{3}} = \frac{2 \sqrt{3}}{9} \end{array}$

Q.4 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

Ans

$\begin{array}{l} (i) We have, \\ f (x) = e^{x} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = e^{x} \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow e^{x} = 0 \\ Which is impossible, because the exponential function can never \\ assume 0 for any value of x . \\ Therefore, there is no number c for that f ‘ (c) = 0 . \\ Thus, function f is neither maxima nor minima . \\ (ii) We have, \\ g (x) = \log x \\ Differentiating w . r . t . x, we get \\ g ‘ (x) = \frac{1}{x} \\ For maxima or minima, \\ g ‘ (x) = 0 \Rightarrow \frac{1}{x} = 0 \\ \Rightarrow x = \frac{1}{0} \\ Thus, there is no point for maxima or minima . \\ Hence, function g (x) does not have maxima or minima . \\ (iii) We have, \\ h (x) = x^{3} + x^{2} + x + 1 \\ Differentiating w . r . t . x, we get \\ h ‘ (x) = 3 x^{2} + 2 x + 1 \\ For maxima or minima, we have \\ h ‘ (x) = 0 \Rightarrow 3 x^{2} + 2 x + 1 = 0 \\ \Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ = \frac{- 2 \pm \sqrt{2^{2} - 4 (3) (1)}}{2 (3)} \\ = \frac{- 2 \pm \sqrt{- 8}}{6} \\ = \frac{- 1 \pm i \sqrt{2}}{3} = \notin R \\ Hence, function h does not have maxima or minima . \end{array}$

Q.5

$\begin{array}{l} Find the absolute maximum value and the absolute minimum \\ value of the following functions in the given intervals : \\ (i) f (x) = x^{3}, x \in [- 2, 2] (ii) f (x) = sinx + cosx, x \in [, π] \\ (iii) f (x) = 4 x - \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}] (iv) f (x) = {(x - 1)}^{2} + 3, x \in [- 3, 1] \end{array}$

Ans

$\begin{array}{l} (i) The given function is f (x) = x^{3} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 3 x^{2} \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow 3 x^{2} = 0 \Rightarrow x = 0 \\ Then, we evaluate the value of f at critical point x = 0 and at \\ end points of the interval [- 2, 2] . \\ f (0) = 0 \\ f (- 2) = {(- 2)}^{3} = - 8 \\ f (2) = {(2)}^{3} = 8 \\ Hence, we can conclude that the absolute maximum value of f \\ on [- 2, 2] is 8 occurring at x = 2 . Also, the absolute minimum \\ value of f on [- 2, 2] is - 8 occurring at x = - 2. \\ (ii) The given function is f (x) = \sin x + \cos x . \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = \cos x - sinx \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow \cos x - sinx = 0 \\ \Rightarrow tanx = 1 \Rightarrow x = \frac{π}{4} \\ Then, we evaluate the value of f at critical point x = \frac{π}{4} \\ and at the end points of the interval [0, π] . \\ f (\frac{π}{4}) = \sin \frac{π}{4} + \cos \frac{π}{4} \\ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \\ = \frac{2}{\sqrt{2}} \\ = \sqrt{2} \\ f (0) = \sin 0 + \cos 0 \\ = 0 + 1 \\ = 1 \\ f (π) = sinπ + cosπ \\ = 0 - 1 \\ = - 1 \\ Hence, we can conclude that the absolute maximum value of \\ f on [0, π] is \sqrt{2} occurring at x = \frac{π}{4} and the absolute minimum \\ value of f on [0, π] is - 1 occurring at x = π . \\ (iii) The given function is f (x) = 4 x - \frac{1}{2} x^{2} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 4 - x \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow 4 - x = 0 \\ \Rightarrow x = 4 \\ Then, we evaluate the value of f at critical point x = 4 and at \\ the end points of the interval [- 2, \frac{9}{2}] . \\ f (4) = 4 (4) - \frac{1}{2} {(4)}^{2} = 16 - 8 = 8 \\ f (- 2) = 4 (- 2) - \frac{1}{2} {(- 2)}^{2} = - 8 - 2 = - 8 \\ f (\frac{9}{2}) = 4 (\frac{9}{2}) - \frac{1}{2} {(\frac{9}{2})}^{2} = 18 - \frac{81}{8} = \frac{63}{4} \\ Hence, we can conclude that the absolute maximum value \\ of f on [- 2, \frac{9}{2}] is 8 occurring at x = 4 and the absolute minimum \\ value of f on [- 2, \frac{9}{2}] is - 10 occurring at x = - 2. \\ (iv) The given function is f (x) = {(x - 1)}^{2} + 3 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 2 (x - 1) \\ For maximum and minimum, we have \\ f ‘ (x) = 0 \Rightarrow 2 (x - 1) = 0 \\ \Rightarrow x = 1 \\ Then, we evaluate the value of f at critical point x = 1 and at \\ the end points of the interval [- 3, 1] . \\ f (1) = {(1 - 1)}^{2} + 3 = 3 \\ f (- 3) = {(- 3 - 1)}^{2} + 3 = 19 \\ Hence, we can conclude that the absolute maximum value of f \\ on [- 3, 1] is 19 occurring at x = - 3 and the minimum value of \\ f on [- 3, 1] is 3 occurring at x = 1. \end{array}$

Q.6 Find the maximum profit that a company can make, if the profit function is given by

p(x) = 41 − 24x − 18x²

Ans

$\begin{array}{l} We have, \\ p (x) = 41 - 24 x - 18 x^{2} \\ Differentiating w . r . t . x, we get \\ p ‘ (x) = - 24 - 36 x \\ and \\ p ” (x) = - 36 \\ For maxima or minima, we have \\ p ‘ (x) = 0 \Rightarrow - 24 - 36 x = 0 \\ \Rightarrow x = \frac{24}{- 36} = - \frac{2}{3} \\ So, p ” (- \frac{2}{3}) = - 36 < 0 \\ By second derivative test, x = - \frac{2}{3}, is the point of local \\ maxima of p . \\ ∴ maximum profit = 41 - 24 (- \frac{2}{3}) - 18 {(- \frac{2}{3})}^{2} \\ = 41 + 16 - 8 \\ = 49 \\ Hence, the maximum profit that the company can make is \\ 49 units . \end{array}$

Q.7

$\begin{array}{l} Find both the maximum value and the minimum value of \\ {3x}^{4} - {8x}^{3} {+12x}^{2} - 48x+25 on the interval [0,3] \end{array}$ $\begin{array}{l} . \end{array}$

Ans

$\begin{array}{l} Let f (x) = 3 x^{4} - 8 x^{3} + 12 x^{2} - 48 x + 25 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 12 x^{3} - 24 x^{2} + 24 x - 48 \\ f ‘ ‘ (x) = 36 x^{2} - 48 x + 24 \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow 12 x^{3} - 24 x^{2} + 24 x - 48 = 0 \\ \Rightarrow (x - 2) (x^{2} + 2) = 0 \\ \Rightarrow x = 2, \sqrt{- 2} \\ \sqrt{- 2} is not real, so x = 2. \\ Then, we evaluate the value of f at critical point x = 2 and at \\ the end points of the interval [0, 3] . \\ f (0) = 3 {(0)}^{4} - 8 {(0)}^{3} + 12 {(0)}^{2} - 48 (0) + 25 = 25 \\ f (2) = 3 {(2)}^{4} - 8 {(2)}^{3} + 12 {(2)}^{2} - 48 (2) + 25 = - 39 \\ f (3) = 3 {(3)}^{4} - 8 {(3)}^{3} + 12 {(3)}^{2} - 48 (3) + 25 = 16 \\ So, the functin f (x) is minimum at x =2 and minimum value \\ is - 39 . The function is maximum at x =0 and maximum value \\ is 25 . \end{array}$

Q.8 At what point in the interval [0, 2π], does the function sin 2x attain its maximum value?

Ans

$\begin{array}{l} Let f (x) = \sin 2 x \\ Differentiating w . r . t . x, we get \\ f^{‘} (x) = 2 \cos 2 x \\ For maxima or minima, we have \\ f^{‘} (x) = 0 \Rightarrow 2 \cos 2 x = 0 \\ \Rightarrow 2 x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2} \\ \Rightarrow x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4} \\ Then, we evaluate the values of f at critical points \\ x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4} and at the end point s o f the interval [0, 2 π] . \\ f (\frac{π}{4}) = \sin 2 (\frac{π}{4}) = \sin (\frac{π}{2}) = 1, \\ f (\frac{3 π}{4}) = \sin 2 (\frac{3 π}{4}) = \sin (\frac{3 π}{2}) = - 1, \\ f (\frac{5 π}{4}) = \sin 2 (\frac{5 π}{4}) = \sin (\frac{5 π}{2}) = 1, \\ f (\frac{7 π}{4}) = \sin 2 (\frac{7 π}{4}) = \sin (\frac{7 π}{2}) = - 1, \\ f (0) = \sin 2 (0) = \sin (0) = 0, \\ f (2 π) = \sin 2 (2 π) = \sin 4 π = 0, \\ Hence, we can conclude that the ab solute maximum value of f \\ on [0, 2 π] is o ccurring at x = \frac{π}{4} and at x = \frac{5 π}{4} . \end{array}$

Q.9 What is the maximum value of the function sin x + cos x?

Ans

$\begin{array}{l} Let f (x) = sinx + cosx \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = cosx - sinx \\ f ‘ ‘ (x) = - sinx - cosx \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow cosx - sinx = 0 \\ \Rightarrow tanx = 1 \\ \Rightarrow x = \frac{π}{4}, \frac{5 π}{4}, . .. \\ So, \\ f ‘ ‘ (\frac{π}{4}) = - \sin \frac{π}{4} - \cos \frac{π}{4} \\ = - (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) \\ = - \sqrt{2} < 0 \\ By second derivative test, f will be the maximum at x = \frac{π}{4}, \\ So maximum value = f (\frac{π}{4}) \\ = \sin \frac{π}{4} + \cos \frac{π}{4} \\ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \end{array}$

Q.10 Find the maximum value of 2x³ – 24x + 107 in the interval [1, 3]. Find the maximum
value of the same function in [- 3, -1].

Ans

$\begin{array}{l} Let f (x) = 2 x^{3} - 24 x + 107 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 6 x^{2} - 24 \\ f ‘ ‘ (x) = 12 x \\ For maxima or minima, we get \\ f ‘ (x) = 0 \Rightarrow 6 x^{2} - 24 = 0 \\ \Rightarrow 6 (x^{2} - 4) = 0 \\ \Rightarrow x = \pm 2 \\ We first consider the interval [1, 3] . \\ Then, we evaluate the value of f at the critical point \\ x = 2 \in [1, 3] and at the end points of the interval [1, 3] . \\ f (2) = 2 (8) - 24 (2) + 107 = 16 - 48 + 107 = 75 \\ f (1) = 2 (1) - 24 (1) + 107 = 2 - 24 + 107 = 85 \\ f (3) = 2 (27) - 24 (3) + 107 = 54 - 72 + 107 = 89 \\ Hence, the absolute maximum value of f (x) in the interval \\ [1, 3] is 89 occurring at x = 3. \\ Next, we consider the interval [- 3, - 1] . \\ Now, we find the value of f (x) at the critical point \\ x = - 2 \in [- 3, - 1] and at the end points of \\ the interval [- 3, - 1] . \\ f (- 3) = 2 (- 27) - 24 (- 3) + 107 \\ = - 54 + 72 + 107 \\ = 125 \\ f (- 1) = 2 (- 1) - 24 (- 1) + 107 \\ = - 2 + 24 + 107 \\ = 129 \\ f (- 2) = 2 (- 8) - 24 (- 2) + 107 \\ = - 16 + 48 + 107 \\ = 139 \\ Hence, the absolute maximum value of f (x) in the interval \\ [- 3, - 1] is 139 occurring at x = - 2. \end{array}$

Q.11 It is given that x = 1, the function x⁴ – 62 x² + ax + 9 attains its maximum value, on the interval [0, 2].

Find the value of a.

Ans

$\begin{array}{l} Let f (x) = x^{4} - 62 x^{2} + ax + 9 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 4 x^{3} - 124 x + a \\ Since, the given that function f attains its maximum value on the \\ interval [0, 2] at x = 1. \\ ∴ f ‘ (1) = 0 \\ \Rightarrow 4 {(1)}^{3} - 124 (1) + a = 0 \\ 4 - 124 + a = 0 \\ - 120 + a = 0 \Rightarrow a = 120 \\ Hence, the value of a is 120 . \end{array}$

Q.12 Find the maximum and minimum values of x + sin2x on [0, 2π].

Ans

$\begin{array}{l} Let f (x) = x + \sin 2 x . \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = 1 + 2 \cos 2 x \\ f ‘ ‘ (x) = - 4 \sin 2 x \\ Now, for maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow 1 + 2 \cos 2 x = 0 \\ \Rightarrow \cos 2 x = - \frac{1}{2} = - \cos \frac{π}{3} \\ = \cos (π - \frac{π}{3}) \\ \Rightarrow \cos 2 x = \cos \frac{2 π}{3} \\ \Rightarrow 2 x = 2 nπ \pm \frac{2 π}{3}, n \in Z \\ \Rightarrow x = nπ \pm \frac{π}{3}, n \in Z \\ \Rightarrow x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3} \in (0, 2 π) \\ Then, we evaluate the value of f at critical points \\ x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3} and at the end points of the interval [0, 2 π]. \\ f (\frac{π}{3}) = \frac{π}{3} + \sin (\frac{2 π}{3}) \\ = \frac{π}{3} + \sin (π - \frac{π}{3}) \\ = \frac{π}{3} + \sin (\frac{π}{3}) \\ = \frac{π}{3} + \frac{\sqrt{3}}{2} \\ f (\frac{π}{3}) = \frac{π}{3} + \sin (\frac{2 π}{3}) = \frac{2 π}{3} + \frac{\sqrt{3}}{2} \\ f (\frac{2 π}{3}) = \frac{2 π}{3} + \sin (\frac{4 π}{3}) = \frac{2 π}{3} - \frac{\sqrt{3}}{2} \\ f (\frac{4 π}{3}) = \frac{4 π}{3} + \sin (\frac{8 π}{3}) = \frac{4 π}{3} + \frac{\sqrt{3}}{2} \\ f (\frac{5 π}{3}) = \frac{5 π}{3} + \sin (\frac{10 π}{3}) = \frac{5 π}{3} - \frac{\sqrt{3}}{2} \\ f (0) = 0 + \sin 0 = 0 \\ f (2 π) = 2 π + \sin 2 π = 0 \\ Hence, we can conclude that the absolute maximum value of \\ f (x) in the interval [0, 2 π] is 2 π occurring at x = 2 π and the \\ absolute minimum value of f (x) in the interval [0, 2 π] \\ is 0 occurring at x = 0 . \end{array}$

Q.13 Find two numbers whose sum is 24 and whose product is as large as possible.

Ans

$\begin{array}{l} Let two numbers be x and (24 - x) andP (x) denotes the product \\ of two numbers, then \\ P (x) = x (24 - x) = 24 x - x^{2} \\ Differentiating w . r . t . x, we get \\ \frac{dP (x)}{dx} = 24 - 2 x \\ P ‘ ‘ (x) = - 2 \\ For maxima or minima, we have \\ P ‘ (x) = 0 \Rightarrow 24 - 2 x = 0 \\ \Rightarrow x = 12 \\ So, P ‘ ‘ (12) = - 2 < 0 \\ ∴ By second derivative test, x = 12 is the point of local maxima \\ of P . Hence, the product of the numbers is the maximum when \\ the numbers are 12 and 24 - 12 = 12. \end{array}$

Q.14 Find two positive numbers x and y such that x + 6 = 60 and xy³ is maximum.

Ans

$\begin{array}{l} The two numbers are x and y such that x + y = 60 . \\ ∴ y = 60 - x \\ L e t f (x) = x y^{3} \Rightarrow f (x) = x {(60 - x)}^{3} \\ D i f f e r e n t i a t i n g f (x) w .r .t . x, we get \\ f’ (x) = \frac{d}{d x} {x {(60 - x)}^{3}} \\ = x \frac{d}{d x} {(60 - x)}^{3} + {(60 - x)}^{3} \frac{d}{d x} x \\ = x .3 {(60 - x)}^{2} (0 - 1) + {(60 - x)}^{3} \times 1 \\ = {(60 - x)}^{3} - 3 x {(60 - x)}^{2} \\ = {(60 - x)}^{2} (60 - x - 3 x) \\ = {(60 - x)}^{2} (60 - 4 x) \\ f ‘ ‘ (x) = {(60 - x)}^{2} \frac{d}{d x} (60 - 4 x) + (60 - 4 x) \frac{d}{d x} {(60 - x)}^{2} \\ = {(60 - x)}^{2} (0 - 4) + (60 - 4 x) .2 (60 - x) (0 - 1) \\ = - 4 {(60 - x)}^{2} - 2 (60 - 4 x) (60 - x) \\ = - 2 (60 - x) (120 - 2 x + 60 - 4 x) \\ = - 2 (60 - x) (180 - 6 x) \\ = - 12 (60 - x) (30 - x) \\ F o r maxima or minima, we have \\ f’ (x) = 0 \Rightarrow {(60 - x)}^{2} (60 - 4 x) = 0 \\ \Rightarrow x = 60, \frac{60}{4} = 60, 16 \\ W h e n, x = 60 \\ f” (60) = - 12 (60 - 60) (30 - 60) = 0 \\ T h u s, at x = 60, f (x) is neither maximum nor minimum . \\ W h e n, x = 15 \\ f” (60) = - 12 (60 - 15) (30 - 15) < 0 \\ S o, f (x) is maximum when x = 15 and y = 60 - 15 = 45 \\ Hence, the required numbers are 15 and 45 . \end{array}$

Q.15 Find two positive numbers x and y such that their sum is 35 and the product x²y⁵ is a maximum.

Ans

$\begin{array}{l} Let one number be x . Then, the other number is y = (35 - x) . \\ Let P(x) = x^{2} y^{5} . Then, we have: \\ P (x) = x^{2} {(35 - x)}^{5} \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ P’ (x) = x^{2} \frac{d}{d x} {(35 - x)}^{5} + {(35 - x)}^{5} \frac{d}{d x} x^{2} \\ = x^{2} \times 5 {(35 - x)}^{4} \frac{d}{d x} (35 - x) + {(35 - x)}^{5} \times 2 x \\ = 5 x^{2} {(35 - x)}^{4} (0 - 1) + 2 x {(35 - x)}^{5} \\ = x {(35 - x)}^{4} (- 5 x + 70 - 2 x) \\ = {(35 - x)}^{4} (70 x - 7 x^{2}) \\ P ‘ ‘ (x) = {(35 - x)}^{4} \frac{d}{d x} (70 x - 7 x^{2}) + \{\frac{d}{d x} {(35 - x)}^{4}\} (70 x - 7 x^{2}) \\ P ‘ ‘ (x) = {(35 - x)}^{4} (70 - 14 x) + \{4 {(35 - x)}^{3} \frac{d}{d x} (35 - x)\} (70 x - 7 x^{2}) \\ P ‘ ‘ (x) = {(35 - x)}^{4} (70 - 14 x) + 4 {(35 - x)}^{3} (0 - 1) (70 x - 7 x^{2}) \\ = 7 {(35 - x)}^{3} \{(35 - x) (10 - 2 x) - 4 (10 x - x^{2})\} \\ = 7 {(35 - x)}^{3} (6 x^{2} - 120 x + 350) \\ F o r maximum or minimum, we have \end{array}$ $\begin{array}{l} P’ (x) = 0 \Rightarrow {(35 - x)}^{4} (70 x - 7 x^{2}) = 0 \\ \Rightarrow 7 x {(35 - x)}^{4} (10 - x) = 0 \Rightarrow x = 0, 10, 35 \\ A t x = 0, \\ P ‘ ‘ (0) = 7 {(35 - 0)}^{3} (6 \times 0^{2} - 120 \times 0 + 350) \\ = 7 {(35)}^{3} (350) > 0 \\ S o, function is minimum at x = 0 . \\ A t x = 35, \\ y = 35 - 35 = 0 \\ S o, P (x) = 35 \times 0 = 0 \\ T h u s, x=35 will not be possible value of x . \\ A t x = 10, \\ P ‘ ‘ (10) = 7 {(35 - 10)}^{3} (6 \times 10^{2} - 120 \times 10 + 350) \\ = 7 {(25)}^{3} (600 - 1200 + 350) < 0 \\ S o, function is maximum at x = 10 . \\ Therefore, the two positive numbers are \\ x = 10 a n d y = 35 - 10 = 25 \\ H e n c e, the required numbers are 10 and 25 . \end{array}$

Q.16 Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Ans

$\begin{array}{l} Let one number be x . Then, the other number is y = (16 - x) . \\ Let S (x) = x^{3} + y^{3} . Then, we have : \\ S (x) = x^{3} + {(16 - x)}^{3} \\ Differentiating w . r . t . x, we get \\ S ‘ (x) = \frac{d}{dx} x^{3} + \frac{d}{dx} {(16 - x)}^{3} \\ = 3 x^{2} + 3 {(16 - x)}^{2} \frac{d}{dx} (16 - x) \\ = 3 x^{2} + 3 {(16 - x)}^{2} (0 - 1) \\ = 3 x^{2} - 3 {(16 - x)}^{2} \\ S ” (x) = \frac{d}{dx} 3 x^{2} - \frac{d}{dx} 3 {(16 - x)}^{2} \\ = 6 x - 6 (16 - x) \frac{d}{dx} (16 - x) \\ = 6 x - 6 (16 - x) (0 - 1) \\ = 6 x + 6 (16 - x) = 96 \\ For maxima or minima, \\ f ‘ (x) = 0 \Rightarrow 3 x^{2} - 3 {(16 - x)}^{2} = 0 \\ \Rightarrow 3 x^{2} - 3 (256 - 32 x + x^{2}) = 0 \\ \Rightarrow 3 (x^{2} - 256 + 32 x - x^{2}) = 0 \\ \Rightarrow 3 (32 x - 256) = 0 \Rightarrow x = \frac{256}{32} = 8 \\ Then, \\ S ‘ ‘ (8) = 96 > 0 \\ By second derivative test, x = 8 is the point of local minima of S . \\ Hence, the sum of the cubes of the numbers is the minimum when \\ the numbers are 8 and 16 - 8 = 8. \end{array}$

Q.17 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?

Ans

$\begin{array}{l} Let the side of the square to be cut off be x cm . Then, the \\ length and the breadth of the box will be (18 - 2x) cm each \\ and the height of the box is x cm . \end{array}$

$\begin{array}{l} \end{array}$ $\begin{array}{l} Therefore, the volume V(x) of the box is given by, \\ V (x) = x(18 - {2x)}^{2} \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ V’ (x) = x \frac{d}{d x} {(18 - 2x)}^{2} + {(18 - 2x)}^{2} \frac{d}{d x} x \end{array}$ $\begin{array}{l} = x . 2 (18 - 2 x) \frac{d}{dx} (18 - 2 x) + {(18 - 2 x)}^{2} \times 1 \\ = 2 x (18 - 2 x) (0 - 2) + {(18 - 2 x)}^{2} \\ = (18 - 2 x) (- 4 x + 18 - 2 x) \\ = 2 (9 - x) \times 6 (3 - x) \\ = 12 (9 - x) (3 - x) \\ V ” (x) = 12 {(9 - x) \frac{d}{dx} (3 - x) + (3 - x) \frac{d}{dx} (9 - x)} \\ = 12 {(9 - x) (0 - 1) + (3 - x) (0 - 1)} \\ = 12 (- 9 + x - 3 + x) \\ = 24 (x - 6) \\ For maxima or minima, we have \\ v ‘ (x) = 0 \Rightarrow 12 (9 - x) (3 - x) = 0 \\ \Rightarrow x = 3, 9 \\ If x = 9, then the length and the breadth will become 0 . \\ \Rightarrow x \neq 9 . \\ When x = 3, \\ V ” (3) = 24 (3 - 6) = - 72 < 0 \\ ∴ By second derivative test, x = 3 is the point of maxima of V . \\ Hence, if we remove a square of side 3 cm from each corner \\ of the square tin and make a box from the remaining sheet, \\ then the volume of the box obtained is the largest possible . \end{array}$

Q.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’ ?

Ans

$\begin{array}{l} Let the side of the square to be cut off be x cm . Then, the \\ height of the box is x, the length is (45 - 2x), and the breadth \\ is (24 - 2x) . \end{array}$

$\begin{array}{l} Therefore, the volume V (x) of the box is given by, \\ V (x) = x (45 - 2 x) (24 - 2 x) \\ = x (1080 - 90 x - 48 x + 4 x^{2}) \\ = 1080 x - 138 x^{2} + 4 x^{3} \\ Differentiating w . r . t . x, we get \\ V ‘ (x) = \frac{d}{dx} (1080 x - 138 x^{2} + 4 x^{3}) \end{array}$ $\begin{array}{l} = 1080 - 276 x + 12 x^{2} \\ = 12 (x^{2} - 23 x + 90) \\ V ‘ ‘ (x) = 12 (2 x - 23) \\ F o r maxima or minima, we have \\ V’ (x) = 0 \Rightarrow 12 (x^{2} - 23 x + 90) = 0 \\ \Rightarrow (x - 18) (x - 5) = 0 \Rightarrow x = 5, 18 \\ It is not possible to cut off a square of side 18 cm from each \\ corner of the rectangular sheet . Thus, x cannot be equal to 18 . \\ When x = 5, \\ V ‘ ‘ (5) = 12 (2 \times 5 - 23) \\ = 12 (10 - 23) = - 156 < 0 \\ ∴ By second derivative test, x = 5 is the point of maxima . \\ Hence, the side of the square to be cut off to make the volume \\ of the box maximum is 5 cm . \end{array}$

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

Ans

$\begin{array}{l} Let a rectangle of length l and breadth b be inscribed in the \\ given circle of radius a .Then, the diagonal passes through the \\ centre and is of length 2a cm . \end{array}$

$\begin{array}{l} Now, by applying the Pythagoras theorem, we have: \\ {(2 a)}^{2} = l^{2} + b^{2} \\ \Rightarrow b^{2} = 4 a^{2} - l^{2} \\ \Rightarrow b = \sqrt{4 a^{2} - l^{2}} \\ ∴ Area of the rectangle, \\ A = l b \\ = l \sqrt{4 a^{2} - l^{2}} \\ D i f f e r e n t i a t i n g w .r .t . l, we get \\ \frac{d A}{d l} = \frac{d}{d l} (l \sqrt{4 a^{2} - l^{2}}) \\ = \sqrt{4 a^{2} - l^{2}} \frac{d}{d l} l + l \frac{d}{d l} \sqrt{4 a^{2} - l^{2}} \\ = \sqrt{4 a^{2} - l^{2}} + l \frac{1}{2 \sqrt{4 a^{2} - l^{2}}} (- 2 l) \\ = \sqrt{4 a^{2} - l^{2}} - \frac{l^{2}}{\sqrt{4 a^{2} - l^{2}}} \\ = \frac{4 a^{2} - l^{2} - l^{2}}{\sqrt{4 a^{2} - l^{2}}} \\ = \frac{4 a^{2} - 2 l^{2}}{\sqrt{4 a^{2} - l^{2}}} \\ \frac{d^{2} A}{d l^{2}} = \frac{d}{d l} (\frac{4 a^{2} - 2 l^{2}}{\sqrt{4 a^{2} - l^{2}}}) \\ = \frac{\sqrt{4 a^{2} - l^{2}} \frac{d}{d l} (4 a^{2} - 2 l^{2}) - (4 a^{2} - 2 l^{2}) \frac{d}{d l} \sqrt{4 a^{2} - l^{2}}}{{(\sqrt{4 a^{2} - l^{2}})}^{2}} \\ = \frac{\sqrt{4 a^{2} - l^{2}} (0 - 4 l) - (4 a^{2} - 2 l^{2}) \times \frac{1}{2 \sqrt{4 a^{2} - l^{2}}} (- 2 l)}{4 a^{2} - l^{2}} \\ = \frac{- 4 l \sqrt{4 a^{2} - l^{2}} + \frac{4 a^{2} l - 2 l^{3}}{\sqrt{4 a^{2} - l^{2}}}}{(4 a^{2} - l^{2})} \\ = \frac{- 4 l (4 a^{2} - l^{2}) + 4 a^{2} l - 2 l^{3}}{{(4 a^{2} - l^{2})}^{\frac{3}{2}}} \\ = \frac{- 16 a^{2} l + 4 l^{3} + 4 a^{2} l - 2 l^{3}}{{(4 a^{2} - l^{2})}^{\frac{3}{2}}} \\ = \frac{2 l^{3} - 12 a^{2} l}{{(4 a^{2} - l^{2})}^{\frac{3}{2}}} \\ F o r maxima or minima, we have \\ \frac{d A}{d l} = 0 \Rightarrow \frac{4 a^{2} - 2 l^{2}}{\sqrt{4 a^{2} - l^{2}}} = 0 \\ \Rightarrow 4 a^{2} - 2 l^{2} = 0 \Rightarrow l^{2} = \frac{4 a^{2}}{2} = 2 a^{2} \\ \Rightarrow l = \sqrt{2} a a n d b = \sqrt{4 a^{2} - 2 a^{2}} = \sqrt{2} a \\ W h e n l= \sqrt{2} a, \\ \frac{d^{2} A}{d l^{2}} = \frac{2 \sqrt{2} a (2 a^{2} - 6 a^{2})}{{(4 a^{2} - 2 a^{2})}^{\frac{3}{2}}} = \frac{2 \sqrt{2} a (- 4 a^{2})}{{(2 a^{2})}^{\frac{3}{2}}} < 0 \\ ∴ By the second derivative test, when l= \sqrt{2} a, then the area of \\ the rectangle is the maximum . \\ Since l = b = \sqrt{2} a, 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 . \end{array}$

Q.20 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.

Ans

$\begin{array}{l} Let r and h be the radius and height of the cylinder respectively . \\ Then, the surface area (S) of the cylinder is given by, \\ S = 2 {πr}^{2} + 2 πrh \Rightarrow h = \frac{S - 2 {πr}^{2}}{2 πr} = \frac{S}{2 π} (\frac{1}{r}) - r \\ Let V be the volume of the cylinder . Then, \\ V = 2 {πr}^{2} h \\ = 2 {πr}^{2} {\frac{S}{2 π} (\frac{1}{r}) - r} \\ = \frac{Sr}{2} - {πr}^{3} \\ Differentiating w . r . t . r, we get \\ \frac{dV}{dr} = \frac{S}{2} - 3 {πr}^{2} and \frac{d^{2} V}{{dr}^{2}} = - 6 πr \\ For maxima or minima, we have \\ \frac{dV}{dr} = 0 \Rightarrow \frac{S}{2} - 3 {πr}^{2} = 0 \Rightarrow r^{2} = \frac{S}{6 π} \\ \Rightarrow r = \pm \sqrt{\frac{S}{6 π}} = \sqrt{\frac{S}{6 π}}, - \sqrt{\frac{S}{6 π}} (Neglect) \\ When r = \sqrt{\frac{S}{6 π}} \\ \frac{d^{2} V}{{dr}^{2}} = - 6 π (\sqrt{\frac{S}{6 π}}) < 0 \\ ∴ By second derivative test, the volume is the maximum when \\ r = \sqrt{\frac{S}{6 π} \Rightarrow} S = 6 {πr}^{2} . \\ ∴ h = \frac{6 {πr}^{2} - 2 {πr}^{2}}{2 πr} \\ = \frac{4 {πr}^{2}}{2 πr} = 2 r \\ Hence, the volume is the maximum when the height is twice \\ the radius i . e ., when the height is equal to the diameter . \end{array}$

Q.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?

Ans

$\begin{array}{l} Let r and h be the radius and height of the cylinder respectively . \\ Then, volume (V) of the cylinder is given by, \\ V = {πr}^{2} h = 100 \\ \Rightarrow h = \frac{100}{{πr}^{2}} \\ Surface area (S) of the cylinder is given by, \\ S = 2 {πr}^{2} + 2 πrh = 2 {πr}^{2} + 2 πr (\frac{100}{{πr}^{2}}) \\ = 2 {πr}^{2} + (\frac{200}{r}) \\ Differentiating w . r . t . x, we get \\ \frac{dS}{dr} = \frac{d}{dr} (2 {πr}^{2}) + \frac{d}{dr} (\frac{200}{r}) \\ = 4 πr - \frac{200}{r^{2}} \\ \frac{d^{2} S}{{dr}^{2}} = \frac{d}{dr} 4 πr - \frac{d}{dr} \frac{200}{r^{2}} \\ = 4 π + \frac{400}{r^{3}} \\ For maxima or minima, we have \\ \frac{dS}{dr} = 0 \Rightarrow 4 πr - \frac{200}{r^{2}} = 0 \\ 4 πr = \frac{200}{r^{2}} \Rightarrow r^{3} = \frac{200}{4 π} = \frac{50}{π} \\ When r = {(\frac{50}{π})}^{\frac{1}{3}} \\ So, \frac{d^{2} S}{{dr}^{2}} > 0 \\ ∴ By second derivative test, the surface area is the minimum \\ when the radius of the cylinder is {(\frac{50}{π})}^{\frac{1}{3}} cm . \\ When r = {(\frac{50}{π})}^{\frac{1}{3}} cm, h = \frac{100}{π {(\frac{50}{π})}^{\frac{2}{3}}} = 2 {(\frac{50}{π})}^{\frac{1}{3}} cm \\ Hence, the required dimensions of the can which has the \\ minimum surface area is given by radius = {(\frac{50}{π})}^{\frac{1}{3}} and \\ height = 2 {(\frac{50}{π})}^{\frac{1}{3}} cm . \end{array}$

Q.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?

Ans

$\begin{array}{l} Letcirclepiece length = x m \\ Thensquarepiecelength = (28 - x) m \\ For circle, \\ Perimeterofcircle = 2 πr = x \Rightarrow r = \frac{x}{2 π} \\ Andfor square, \\ Perimeterof square = 28 - x = 4 a \Rightarrow a = \frac{28 - x}{4} \\ Nowthesumoftheareas A ofcircleandsquare \\ A = π r^{2} + a^{2} \\ = π {(\frac{x}{2 π})}^{2} + {(\frac{28 - x}{4})}^{2} \\ = \frac{x^{2}}{4 π} + {(\frac{28 - x}{4})}^{2} \\ \frac{dA}{dx} = \frac{x}{2 π} - \frac{28 - x}{8} \\ \frac{d^{2} A}{d x^{2}} = \frac{1}{2 π} + \frac{1}{8} \\ For max ima or minima, we have \\ \frac{dA}{dx} = 0 \Rightarrow \frac{x}{2 π} - \frac{28 - x}{8} = 0 or x = \frac{28 π}{4 + π} \\ \frac{d^{2} A}{d x^{2}} > 0 at x = \frac{28 π}{4 + π} \\ ∴ Aisminimum at circle piece = x \\ = \frac{28 π}{4 + π} \\ Squarepiece = 28 - x = \frac{112}{4 + π} \end{array}$

Q.23

$\begin{array}{l} Prove that the volume of the largest cone that can be \\ inscribed in a sphere of radius R is \frac{8}{27} of the volume of \\ the sphere . \end{array}$

Ans

$\begin{array}{l} DAB is the cone inscribed in a sphere of radius R . \\ Volume of cone V = \frac{1}{3} {πr}^{2} h \\ Where r is the base radius of cone and h its height \\ Height of cone h = OC + OD = R + OC = R + x \\ Radius of Cone = r where r^{2} = R^{2} - {OD}^{2} = R^{2} - x^{2} \\ ∴ V = \frac{1}{3} π (R^{2} - x^{2}) (R + x) = \frac{1}{3} π [R^{3} + R^{2} x - x^{2} R - x^{3}] \\ \frac{dV}{dx} = \frac{1}{3} π [R^{2} - 2 xR - 3 x^{2}] \\ = \frac{1}{3} π [R^{2} - 3 xR - xR - 3 x^{2}] \\ = \frac{1}{3} π (R - 3 x) (R + x) \\ \frac{d^{2} V}{{dx}^{2}} = \frac{1}{3} π {(R - 3 x) \frac{d}{dx} (R + x) + (R + x) \frac{d}{dx} (R - 3 x)} \\ = \frac{1}{3} π (R - 3 x - 3 R - 3 x) \\ = \frac{1}{3} π (- 2 R - 6 x) \\ = - \frac{2}{3} π (R + 3 x) \\ \frac{dV}{dx} = 0 \Rightarrow x = \frac{R}{3} or - R \\ \frac{d^{2} V}{{dx}^{2}} < 0 at x = \frac{R}{3} \\ Hence x = \frac{R}{3} gives maximum volume . \\ V = \frac{1}{3} π (R^{2} - \frac{R^{2}}{9}) (R + \frac{R}{3}) \\ = \frac{1}{3} π \times \frac{8 R^{2}}{9} \times \frac{4 R}{3} \\ = \frac{8}{27} [\frac{4}{3} {πR}^{3}] \\ Hence the volume of cone = \frac{8}{27} (Volume of sphere) is the \\ largest cone that could be inscribed in a given sphere . \end{array}$

Q.24

$\begin{array}{l} Show that the right circular cone of least curved surface \\ and given volume has an altitude equal to \sqrt{2} time the \\ radius of the base . \end{array}$

Ans

$\begin{array}{l} Here, Volume of the cone i s ‘ V ’ . \\ ∴ V = \frac{1}{3} π r^{2} h \Rightarrow r^{2} = \frac{3 V}{π h} \\ and surface area be ‘ S ’ of cone . \\ ∴ S = π r l = π r \sqrt{h^{2} + r^{2}} \\ Where h = height of the cone \\ r = radius of the cone \\ l = slant height of the cone \\ S^{2} = π^{2} r^{2} (h^{2} + r^{2}) \\ Let S_{1} = S^{2} then \\ S_{1} = π^{2} r^{2} (h^{2} + r^{2}) \\ S_{1} = \frac{3 π V}{h} (h^{2} + \frac{3 V}{π h}) = 3 π V h + \frac{9 V^{2}}{h^{2}} [∵ r^{2} = \frac{3 V}{π h}] \\ \frac{d S_{1}}{d h} = 3 π V + 9 V^{2} (\frac{- 2}{h^{3}}) \\ \frac{d^{2} S_{1}}{d h^{2}} = \frac{54 V^{2}}{h^{4}} \\ \frac{d S_{1}}{d h} = 0 for maxima/minima \\ 3 π V + 9 V^{2} (\frac{- 2}{h^{3}}) = 0 \\ \Rightarrow 3 π V = 9 V^{2} (\frac{2}{h^{3}}) \\ \Rightarrow h^{3} = \frac{6 V}{π} \\ \frac{d^{2} S_{1}}{d h^{2}} > 0 at h^{3} = \frac{6 V}{π^{}} \\ Therefore curved surface area is minimum at \frac{π h^{3}}{6} = V . \\ T h u s, \frac{π h^{3}}{6} = \frac{1}{3} π r^{2} h \Rightarrow h^{2} = 2 r^{2} \\ \Rightarrow h = \sqrt{2} r \\ Hence for least curved surface the altitude is \sqrt{2} times radius . \end{array}$

Q.25

$\begin{array}{l} Show that the semi-vertical angle of the cone of the \\ maximum volume and of given slant height is {tan}^{- 1} \sqrt{2} . \end{array}$

Ans

$\begin{array}{l} Let θ be the semi-vertical angle of the cone . \\ It is clear that θ \in [0, \frac{π}{2}] . \\ Let r, h, and l be the radius, height, and the slant height of \\ the cone respectively . The slant height of the cone is given \\ as constant . \end{array}$

$\begin{array}{l} Now, r = l \sin θ and h = l \cos θ \\ The volume (V) of the cone is given by, \\ V = \frac{1}{3} {πr}^{2} h \\ = \frac{1}{3} π {(l \sin θ)}^{2} (l \cos θ) \\ = \frac{1}{3} {πl}^{3} \sin^{2} θcosθ \\ Differentiating w . r . t . θ, we get \\ \frac{dV}{dθ} = \frac{1}{3} {πl}^{3} (\sin^{2} θ \frac{d}{dθ} cosθ + cosθ \frac{d}{dθ} \sin^{2} θ) \\ = \frac{1}{3} {πl}^{3} (- \sin^{3} θ + 2 {sinθcos}^{2} θ) \\ & \frac{d^{2} V}{{dθ}^{2}} = \frac{1}{3} {πl}^{3} (- 3 \sin^{2} θcosθ + 3 \cos^{3} θ - 2 \sin^{2} θcosθ) \\ For maxima or minima, we have \\ \frac{dV}{dθ} = 0 \Rightarrow \frac{1}{3} {πl}^{3} (- \sin^{3} θ + 2 {sinθcos}^{2} θ) = 0 \\ \Rightarrow \sin^{3} θ = 2 {sinθcos}^{2} θ \\ \Rightarrow \tan^{2} θ = 2 \Rightarrow tanθ = \sqrt{2} \\ \Rightarrow θ = \tan^{- 1} \sqrt{2} \\ When θ = \tan^{- 1} \sqrt{2} \Rightarrow \sin^{2} θ = 2 \cos^{2} θ \\ \frac{d^{2} V}{{dθ}^{2}} = \frac{1}{3} {πl}^{3} (- 3 \times 2 \cos^{2} θ . cosθ + 3 \cos^{3} θ - 2 \times 2 \cos^{2} θ . cosθ) \\ = \frac{1}{3} {πl}^{3} (- 6 \cos^{3} θ + 3 \cos^{3} θ - 4 \cos^{3} θ) \\ = \frac{1}{3} {πl}^{3} (- 4 \cos^{3} θ) < 0 \\ ∴ By second derivative test, the volume (V) is the maximum \\ when θ = \tan^{- 1} \sqrt{2} . \end{array}$

Q.26

$\begin{array}{l} Show that semi-vertical angle of right circular cone of \\ {given surface area and maximum volume is sin}^{-1} (\frac{1}{3}) . \end{array}$

Ans

$\begin{array}{l} Here in ΔAOC, l^{2} = h^{2} + r^{2} \\ The total surface area of the cone is \\ S = {πr}^{2} + πrl \Rightarrow \frac{S - {πr}^{2}}{πr} = l \\ or l = \frac{S}{πr} - r \\ Volume of the cone is given by \\ V = \frac{1}{3} {πr}^{2} h = \frac{1}{3} {πr}^{2} \sqrt{l^{2} - r^{2}} \\ Let \\ V_{1} = V^{2} = \frac{1}{9} π^{2} r^{4} (l^{2} - r^{2}) \\ \Rightarrow V_{1} = \frac{1}{9} π^{2} r^{4} ({[\frac{S}{πr} - r]}^{2} - r^{2}) \\ \Rightarrow V_{1} = \frac{1}{9} π^{2} r^{4} (\frac{S^{2}}{π^{2} r^{2}} - \frac{2 S}{π}) = \frac{1}{9} (S^{2} r^{2} - 2 {Sπr}^{4}) \\ Differentiating w . r . t . r, we get \\ V_{1} ‘ = \frac{1}{9} (2 S^{2} r - 8 {Sπr}^{3}) \\ V_{1} ‘ ‘ = \frac{1}{9} (2 S^{2} - 24 {Sπr}^{2}) \\ For maxima or minima, \\ V_{1} ‘ = 0 \\ \Rightarrow \frac{1}{9} (2 S^{2} r - 8 {Sπr}^{3}) = 0 \\ \Rightarrow 2 S^{2} r = 8 {Sπr}^{3} \\ \Rightarrow \frac{S}{4 π} = r^{2} \\ V_{1} ‘ ‘ = \frac{1}{9} (2 S^{2} - 24 {Sπr}^{2}) < 0 at r^{2} = \frac{S}{4 π} \\ \Rightarrow V is maximum at r^{2} = \frac{S}{4 π} \\ ∵ 4 {πr}^{2} = S \\ \Rightarrow 4 {πr}^{2} = {πr}^{2} + πrl \\ \Rightarrow 3 {πr}^{2} = πrl \\ or \frac{r}{l} = \frac{1}{3} \\ i . e \sin ∠ AOC = \frac{1}{3} [∵ InΔAOC, \sin (∠ OAC) = \frac{r}{l}] \\ \Rightarrow semi - vertical angle is \sin^{- 1} (\frac{1}{3}) . \end{array}$

Q.27

$\begin{array}{l} The point on the curve x^{2} = 2 y which is nearest to the \\ point (0, 5) is \\ (A) (2 \sqrt{2}, 4) (B) (2 \sqrt{2}, 0) (C) (0,0) (D) (2,2) \end{array}$

Ans

$\begin{array}{l} The given curve is x^{2} = 2 y . \\ Let P (x, y) be a point on x^{2} = 2 y and A (0, 5) be the given point . \\ Then, {AP}^{2} = {(x - 0)}^{2} + {(y - 5)}^{2} \\ = x^{2} + {(\frac{x^{2}}{2} - 5)}^{2} [∵ x^{2} = 2 y \Rightarrow y = \frac{x^{2}}{2}] \\ Let Z = {AP}^{2} . Then, Z is maximum or minimum according as AP \\ is maximum or minimum . \\ Now, Z = x^{2} + {(\frac{x^{2}}{2} - 5)}^{2} \\ Differentiating w . r . t . x, we get \\ \frac{dZ}{dx} = \frac{d}{dx} {x^{2} + {(\frac{x^{2}}{2} - 5)}^{2}} \\ = 2 x + 2 (\frac{x^{2}}{2} - 5) \frac{d}{dx} (\frac{x^{2}}{2} - 5) \\ = 2 x + 2 (\frac{x^{2}}{2} - 5) (x - 0) \\ = 2 x + x^{3} - 10 x \\ = x^{3} - 8 x \\ \frac{d^{2} Z}{{dx}^{2}} = 3 x^{2} - 8 \\ For maxima or minima, we have \\ \frac{dZ}{dx} = 0 \Rightarrow x^{3} - 8 x = 0 \\ \Rightarrow x (x^{2} - 8) = 0 \\ \Rightarrow x = 0, \pm 2 \sqrt{2} \\ When x = 0, \\ \frac{d^{2} Z}{{dx}^{2}} = 3 {(0)}^{2} - 8 = - 8 < 0 \\ So, Z is maximum at x = 0 i . e ., point P is not nearest to curve . \\ When x = \pm 2 \sqrt{2} \\ \frac{d^{2} Z}{{dx}^{2}} = 3 {(\pm 2 \sqrt{2})}^{2} - 8 \\ = 24 - 8 = 16 > 0 \\ So, Z is minimum at x = \pm 2 \sqrt{2} i . e ., point P is nearest to curve . \\ Putting x = \pm 2 \sqrt{2} in x^{2} = 2 y, we get \\ y = \frac{x^{2}}{2} = \frac{{(\pm 2 \sqrt{2})}^{2}}{2} = \frac{8}{2} = 4 \\ Hence, the point (\pm 2 \sqrt{2}, 4) on x^{2} = 2 y is the nearest to the \\ point (0, 5) . \\ Thus, the option A is correct . \end{array}$

Q.28

$\begin{array}{l} For all real values of x, the minimum value of \frac{1 - x + x^{2}}{1 + x + x^{2}} is \\ (A) 0 (B) 1 (C) 3 (D) \frac{1}{3} \end{array}$

Ans

$\begin{array}{l} Let f (x) = \frac{1 - x + x^{2}}{1 + x + x^{2}} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = \frac{d}{dx} (\frac{1 - x + x^{2}}{1 + x + x^{2}}) \\ = \frac{(1 + x + x^{2}) \frac{d}{dx} (1 - x + x^{2}) - (1 - x + x^{2}) \frac{d}{dx} (1 + x + x^{2})}{{(1 + x + x^{2})}^{2}} \\ = \frac{(1 + x + x^{2}) (0 - 1 + 2 x) - (1 - x + x^{2}) (0 + 1 + 2 x)}{{(1 + x + x^{2})}^{2}} \\ = \frac{- 1 - x - x^{2} + 2 x + 2 x^{2} + 2 x^{3} - (1 - x + x^{2} + 2 x - 2 x^{2} + 2 x^{3})}{{(1 + x + x^{2})}^{2}} \\ = \frac{- 1 - x - x^{2} + 2 x + 2 x^{2} + 2 x^{3} - 1 + x - x^{2} - 2 x + 2 x^{2} - 2 x^{3}}{{(1 + x + x^{2})}^{2}} \\ = \frac{- 2 + 2 x^{2}}{{(1 + x + x^{2})}^{2}} \\ = \frac{2 (x^{2} - 1)}{{(1 + x + x^{2})}^{2}} \\ f ‘ ‘ (x) = \frac{{(1 + x + x^{2})}^{2} \frac{d}{dx} 2 (x^{2} - 1) - 2 (x^{2} - 1) \frac{d}{dx} {(1 + x + x^{2})}^{2}}{{(1 + x + x^{2})}^{4}} \\ = \frac{{(1 + x + x^{2})}^{2} 2 (2 x - 0) - 2 (x^{2} - 1) 2 (1 + x + x^{2}) \frac{d}{dx} (1 + x + x^{2})}{{(1 + x + x^{2})}^{4}} \\ = \frac{4 x {(1 + x + x^{2})}^{2} - 4 (x^{2} - 1) (1 + x + x^{2}) (0 + 1 + 2 x)}{{(1 + x + x^{2})}^{4}} \\ = \frac{4 (1 + x + x^{2}) {(1 + x + x^{2}) x - (x^{2} - 1) (1 + 2 x)}}{{(1 + x + x^{2})}^{4}} \\ = \frac{4 {x + x^{2} + x^{3} - x^{2} + 1 - 2 x^{3} + 2 x}}{{(1 + x + x^{2})}^{3}} \\ = \frac{4 (- x^{3} + 3 x + 1)}{{(1 + x + x^{2})}^{3}} \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow \frac{2 (x^{2} - 1)}{{(1 + x + x^{2})}^{2}} = 0 \\ \Rightarrow x^{2} - 1 = 0 \Rightarrow x = \pm 1 \\ When x = 1, \\ f ‘ ‘ (1) = \frac{4 (- 1^{3} + 3 \times 1 + 1)}{{(1 + 1 + 1^{2})}^{3}} \\ = \frac{4 (3)}{27} = \frac{4}{9} > 0 \\ f ‘ ‘ (- 1) = \frac{4 {- {(- 1)}^{3} + 3 (- 1) + 1}}{{1 + (- 1) + {(- 1)}^{2}}^{3}} \\ = \frac{4 (- 1)}{1} = - 4 < 0 \\ ∴ By second derivative test, f is the minimum at x = 1 and \\ the minimum value is given by \\ f (1) = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3} \\ Thus, the correct answer is D . \end{array}$

Q.29

$\begin{array}{l} The maximum value of {[x (x-1) +1]}^{\frac{1}{3}}, 0£x£1 is \\ (A) {(\frac{1}{3})}^{\frac{1}{3}} (B) \frac{1}{2} (C) 1 (D) 0 \end{array}$

Ans

$\begin{array}{l} Let f (x) = {[x (x - 1) + 1]}^{\frac{1}{3}}, 0 \leq x \leq 1 \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = \frac{d}{dx} {[x (x - 1) + 1]}^{\frac{1}{3}} \\ = \frac{1}{3} {[x (x - 1) + 1]}^{- \frac{2}{3}} \frac{d}{dx} [x (x - 1) + 1] \\ = \frac{1}{3} {[x (x - 1) + 1]}^{- \frac{2}{3}} (\frac{d}{dx} x^{2} - \frac{d}{dx} x + \frac{d}{dx} 1) \\ = \frac{1}{3} {[x (x - 1) + 1]}^{- \frac{2}{3}} (2 x - 1) \\ = \frac{(2 x - 1)}{3 {[x (x - 1) + 1]}^{\frac{2}{3}}} \\ For maxima or minima, we have \\ f ‘ (x) = 0 \Rightarrow \frac{(2 x - 1)}{3 {[x (x - 1) + 1]}^{\frac{2}{3}}} = 0 \\ \Rightarrow 2 x - 1 = 0 \Rightarrow x = \frac{1}{2} \\ Then, we evaluate the value of f at critical point x = \frac{1}{2} \\ and at the end points of the interval [0, 1] . \\ f (0) = {[(0) (0 - 1) + 1]}^{\frac{1}{3}} = 0 \\ f (\frac{1}{2}) = {[(\frac{1}{2}) (\frac{1}{2} - 1) + 1]}^{\frac{1}{3}} = {(\frac{3}{4})}^{\frac{1}{3}} \\ f (1) = {[(1) (1 - 1) + 1]}^{\frac{1}{3}} = 1 \\ Hence, we can conclude that the maximum value of f in the \\ interval [0, 1] is 1 . \\ The correct answer is C . \end{array}$

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1. Where can students find the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5?

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2. What are the topics covered in the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5?

The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 covers the topics of maxima and minima of a function. Students should refer to the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 to find the points of Maxima and Minima as well as the maximum and minimum values of a given function.

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The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 provides solutions to all the 29 questions in the exercise. Students can find all the steps necessary to solve the questions in the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5 provided by Extramarks to get maximum marks in the examination.

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NCERT Solutions Class 12 Maths Chapter 6 Exercise 6.5

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.5) Exercise 6.5

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