NCERT Solutions for Class 12 Maths Chapter 6

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Chapter 6 – APPLICATION OF DERIVATIVES 

NCERT Solutions for Class 12 Maths Chapter 6

6.1 Introduction

In Ch 6 Maths Class 12, the introduction part will be a recap of the last chapter where you got acquainted with concepts of derivative of a composite function, implicit functions, logarithmic functions, inverse trigonometric functions and exponential functions. 

In this chapter, you will learn how the application of derivatives works in different disciplines like social science, engineering, science, etc. This chapter will go through how to apply the application of derivatives to find out the rate of change of quantities.

Other Key Concepts Discussed in This Chapter Are

• Equations of a tangent and normal to a curve at a point.

• Find out turning points on the graph which will help you determine the points where maximum and minimum values of a function occur.

• Find intervals where a function increases and decreases.

• Approximations and errors.

The chapter will also give you ample practice with a variety of exercises of different kinds like short and long answers, objective types questions, etc. These activities would further strengthen all the important topics learned in this chapter and help you tremendously in your preparation for your board’s exams.

6.2 Rate of Change of Quantities

In NCERT Solutions for Class 12 Chapter 6 Applications of Derivatives, you would brush up all the learnings from the previous class. You would be reminded that to represent the rate of change of quantity with respect to another, we use the notation dy/dx, this means that y changes when x is changed.

You would further expand this concept to find out if changes in x and y are dependent on a 3rd variable (say z)  i.e. x= f(z) and y=g(z), Chain rule can be applied as shown below:

dy/dx = (dy/dz)/(dx/dz), given dx/dz <> =0

Exercise 6.1 Solutions: 18 Questions (10 Long Questions, 6 Short Questions, 2 MCQs)

6.3 Increasing and Decreasing Functions

In this section of AOD Class 12 NCERT solutions, you would understand what is meant by increasing and decreasing function and how to determine whether a function is increasing or decreasing in a given range by using differentiation. You would define that a function is increasing in a range if the first derivative is positive in that range. Similarly, you would learn that a function is decreasing in a range if the derivative is negative in that range. 

You will learn that in the case of trigonometric functions, one has to determine the quadrant in which they lie in order to know if they are increasing or decreasing. So, in this case, you would understand that for increasing value of x, y is increasing then it is an increasing function, and if y decreases with an increase in the value of x, then it is a decreasing function. For example, sin x increases in the first quadrant while cos x is a decreasing function in the same quadrant. 

You Should Write Down The Above Laws in The Formula As Depicted Below

• Increasing Functions –> x1 < x2 in interval i implies f(x1) < f(x2) where x1 and x2 є i

• Decreasing Functions -> x1 < x2 in interval i implies f(x1) > f(x2) where x1 and x2 є i

Exercise 6.2 Solutions: 19 Questions (10 Long questions, 7 Short questions, 2 MCQ)

6.4 Tangents and Normals

In this part of NCERT Solutions for Class 12 Maths Application of Derivatives, you would build your knowledge of the equation of straight lines learned in previous chapters. You would rekindle that the equation of a straight line on a curve y = f(x) with a finite slope s, passing through a given point (a, b) is given by:

y - b = s (x – a)

You would learn that a tangent to a point (a,b) on a curve is given by:

[dy/dx](a,b) and is denoted by f’(a) hence the equation of a tangent to the curve is given by : y – b = f’(a) (x – a). 

You would also get to know that normal is perpendicular to the tangent whose equation is given by y – b = (-1/ f’(a)) (x – a)

With Application of Derivatives Class 12 Solutions, you would get a good idea about what methods to adopt to calculate the expressions utilizing the slope formula.

Exercise 6.3 Solutions: 27 Questions (25 Short Questions, 2 MCQ)

6.5 Approximations

This part of Ch 6 Class 12 Maths NCERT Solutions will introduce you to the concept of using differentiation in order to find approximate values of certain quantities. You would get to know how approximations are useful when an exact numerical number is not easy to obtain. You would understand how to derive the approximate value of a quantity that changes in small measures by a change in another variable. So if y = f(x) and dx is a small increment in x which makes an increment of dy in the value of y, you would learn that dy = f (x + dx) – f(x) or dy = f’(x) dx.

The small change is generally denoted by Δ and approximation by ≈. So you will learn how to use these symbols to express the above equation as dy = (dy/dx) Δx.

From the learnings in this chapter, you will be able to figure out that the derivative of the dependent variable (y) is not equal to the increase in the variable while the derivative of the independent variable (x) is equal to the increase in the variable.

Exercise 6.4 Solutions: 9 Questions (7 Short Questions, 2 MCQ)

6.6 Maxima and Minima

This unit of NCERT Solutions for Class 12 Maths Chapter 6 describes the minimum and maximum values of a function in the form of derivatives. You will learn about turning points on the graph of a function where the graph reaches its highest point and the lowest point locally in a domain. You would understand how these points can be used to sketch the graph of a function. 

You will also be able to find the absolute maximum and absolute minimum of a function which are useful in solving many applied problems. We would learn various definition around maxima and minima which are used for solving problems related to it like:

• In a range r, function f is said to have maximum value if there exists a point p in r where f(p) >= f(x) for all x є r. The value f(p) defines the maximum value of f in r, and the point p is termed as the point of the maximum value in r.

• In a range r, function f is said to have minimum value if there exists a point p in r where f(p) <= f(x) for all x є r. The value f(p) defines the maximum value of f in r, and the point p is termed as the point of the minimum value in r.

• In a range r, function f is said to have extreme value if there exists a point p in r where f(p) is either a maximum value or a minimum value of f in r and the point p is termed as the extreme point in r.

You would also learn how local maxima and local minima are represented and many theorems like the first derivative test, second derivative test around this concept.

Exercise 6.5 Solutions: 29 Questions (15 Long Questions, 11 Short Questions, 3 MCQ)

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