NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 Application of Derivatives - PDF Download

NCERT solutions for class 12 maths chapter 6 exercise 6.3 Application of Derivatives using derivatives explains how derivatives can be used to figure out the maximum or minimum value of different functions. Class 12 maths chapter 6 exercise 6.3 NCERT solutions provide the theoretical portion of this exercise which is very helpful in understanding how to draw the graph of a function. In addition, students will also be able to identify the absolute maximum and absolute minimum of a function which is essential for solving many practical problems.

The academic team of mathematics at eSaral have developed these solutions that will help you to have clarity of all concepts and theorems included in exercise 6.3 to prepare for board exams. Ex 6.3 class 12 maths chapter 6 consists of 29 questions based on finding the value of maximum and minimum functions. Every question in ex 6.3 is explained by eSaral’s experts with a step by step approach for students to score good marks in exams.

Ex 6.3 class 12 maths solutions are also available in free PDF format. You can easily download these PDFs from the official website of eSaral and practice this exercise anytime anywhere.

Topics Covered in Exercise 6.3 Class 12 Mathematics Questions

Ex 6.3 class 12 maths ch 6 describes maxima and minima, maximum and minimum values of a function in a closed interval and some important theorems. You can check below the detailed explanation of these topics and theorems by subject experts of eSaral.

1. Maxima and Minima 

Let f be a function defined on an interval I. Then

(a) f is said to have a maximum value in I, if there exists a point c in I such that f(c ) > f(x) , for all x ∈ I.

The number f(c) is called the maximum value of f in I and the point c is called a point of maximum value of f in I.

(b) f is said to have a minimum value in I, if there exists a point c in I such that f (c) < f (x), for all x ∈ I.

The number f (c), in this case, is called the minimum value of f in I and the point c, in this case, is called a point of minimum value of f in I. 

(c) f is said to have an extreme value in I if there exists a point c in I such that f (c) is either a maximum value or a minimum value of f in I.

The number f (c), in this case, is called an extreme value of f in I and the point c is called an extreme point.

Every monotonic function assumes its maximum/minimum value at the end points of the domain of definition of the function.

Every continuous function on a closed interval has a maximum and a minimum value. 

Turning Points - The function changes its nature from decreasing to increasing or vice-versa. These points may be called turning points of the given function.

Local Maximum and Local Minimum Values 

Let f be a real valued function and let c be an interior point in the domain of f. Then 

(a) c is called a point of local maxima if there is an h > 0 such that

                 f (c) ≥ f (x), for all x in (c – h, c + h), x ≠ c

The value f(c) is called the local maximum value of f.

(b) c is called a point of local minima if there is an h > 0 such that

                  f (c) ≤ f (x), for all x in (c – h, c + h)

The value f(c) is called the local minimum value of f .

Theorem 2: Let f be a function defined on an open interval I. Suppose c ∈ I be any point. If f has a local maxima or a local minima at x = c, then either f ′(c) = 0 or f is not differentiable at c.

Theorem 3: (First Derivative Test) Let f be a function defined on an open interval I. Let f be continuous at a critical point c in I. Then

(i) If f ′(x) changes sign from positive to negative as x increases through c, i.e., if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima. 

(ii) If f ′(x) changes sign from negative to positive as x increases through c, i.e., if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima.

(iii) If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflection.

Theorem 4 (Second Derivative Test): Let f be a function defined on an interval I and c ∈ I. Let f be twice differentiable at c. Then

(i) x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0

           The value f (c) is local maximum value of f.

(ii) x = c is a point of local minima if f c ′( ) 0 = and f ″(c) > 0

           In this case, f (c) is local minimum value of f .

(iii) The test fails if f ′(c) = 0 and f ″(c) = 0.

In this case, we go back to the first derivative test and find whether c is a point of local maxima, local minima or a point of inflexion. 

• Maximum and Minimum Values of a Function in a Closed Interval

Suppose we have a function f defined by

               f(x) = x + 2, x ∈ (0, 1)

Note that the function is constant at (0, 1), and there is no maximum or minimum value for this function. In addition, it can be observed that the function does not have either a local maximum or local minimum value. However, if we extend the domain of f to the closed interval [0, 1], then f still may not have a local maximum (minimum) values but it certainly does have maximum value 3 = f(1) and minimum value 2 = f(0). The maximum value 3 of f at x = 1 is called absolute maximum value (global maximum or greatest value) of f on the interval [0, 1]. Similarly, the minimum value 2 of f at x = 0 is called the absolute minimum value (global minimum or least value) of f on [0, 1].

Theorem 5:  Let f be a continuous function on an interval I = [a, b]. Then f has the absolute maximum value and f attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in I.

Theorem 6: Let f be a differentiable function on a closed interval I and let c be any interior point of I. Then

(i) f ′(c) = 0 if f attains its absolute maximum value at c.

(ii) f ′(c) = 0 if f attains its absolute minimum value at c.

Based on the above, we have a working rule to find absolute max and/or minimum values of the function in the closed interval (a, b).

Step 1: Find all critical points of f in the interval, i.e., find points x where either f′(x)=0 or f is not differentiable.

Step 2: Take the end points of the interval.

Step 3: At all these points (listed in Step 1 and 2), calculate the values of f.

Step 4: Identify the maximum and minimum values of f out of the values calculated in Step 3. This maximum value will be the absolute maximum (greatest) value of f and the minimum value will be the absolute minimum (least) value of f .

Tips for Solving Exercise 6.3 Class 12 Chapter 6 Application of Derivatives

Our expert teachers of eSaral have provided some useful tips to solve ex 6.3 class 12 maths chapter 6.

1. In order to comprehend all the fundamental concepts, students must follow the simple and precise NCERT solutions for class 12 maths chapter 6 ex 6.3.

2. Explaining the derivative questions and examples in these solutions also provides information about the function.

3. Through solving the sums and going over each concept in the class 12 maths chapter 6 ex 6.3 NCERT solutions , students will become acquainted with the format and types of questions that are asked in exams.
   

Importance of Solving Ex 6.3 Class 12 Maths Chapter 6 Application of Derivatives

There are numerous benefits of solving ex 6.3 class 12 maths chapter 6 ex 6.3 which you can check below.

1. NCERT solutions class 12 Maths chapter 6 exercise 6.3 explain each and every definition, formula, theorem and method in an easy and simple way to deliver deep knowledge of the concepts to solve questions of ex 6.3.

2. These solutions are solved by the experienced teachers of eSaral for you to prepare for board exams.

3. NCERT solutions PDF are also provided for ex 6.3 that provides you accurat answers to all the questions.

4. NCERT solutions will help you to revise the exercise questions and examples so that you can be well-versed with the stepwise solutions of ex 6.3 questions.

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