# Important Questions Class 12 Maths Chapter 5

## Important Questions for CBSE Class 12 Maths Chapter 5 – Continuity and Differentiability

Chapter 5 of Class 12 Maths “Continuity and Differentiability” is an important chapter of the CBSE Class 12 Maths curriculum. You might anticipate a few questions on this subject in the Class 12 Maths exam. If you gain a firm grasp of the subjects presented in Class 12 Maths Chapter 5, you will be able to solve the problems regarding continuity and differentiability in board exams with ease. For this, you must grasp the fundamental strategies for answering the many kinds of questions based on this subject to master the concepts.

We at Extramarks have made an effort to provide Important Questions Class 12 Maths Chapter 5 to give thorough information about the subject and various types of questions that are significant from an exam perspective. This set of Important Questions from Differentiability And Continuity are curated by subject matter experts to aid in better preparations for the Class 12 Maths exams.

You can also find CBSE Class 12 Maths Important Questions Chapter-by-Chapter Important Questions here:

 CBSE Class 12 Maths Important Questions Sr No Chapter No Chapter Name 1 Chapter 1 Relations and Functions 2 Chapter 2 Inverse Trigonometric Functions 3 Chapter 3 Matrices 4 Chapter 4 Determinants 5 Chapter 5 Continuity and Differentiability 6 Chapter 6 Application of Derivatives 7 Chapter 7 Integrals 8 Chapter 8 Application of Integrals 9 Chapter 9 Differential Equations 10 Chapter 10 Vector Algebra 11 Chapter 11 Three Dimensional Geometry 12 Chapter 12 Linear Programming 13 Chapter 13 Probability

## CBSE Class 12 Maths Chapter-5 Important Questions

Very Short Answer Type Questions (1 Mark)

1. For what value of x,f(x) = |2x – 7|x,f(x) = |2x – 7|. is not derivable.

Ans. The input value for which the function cannot be derived must be identified. In other words, the point where the function has a break (there are two different tangents at that point)

f(x)=  2x-772
7-2x < 72

Hence, for the value x= 72x = 72, the function is not derivable.

1. Write the set of points of continuity of g(x) = |x – 1| + |x + 1|.

Ans. Let’s look for the breakpoints of  piecewise function.

g (x) = (x-1) + (x+1)

(1−x)+(x+1)−1⩽x⩽1

(1−x)−(x+1)x<−1

We can observe that the function is a polynomial at all of the breakpoints since polynomials are continuous across their entire domain and hence the points of continuity of g(x) is R.

1. What is the derivative of |x – 3| at x=- 1

Ans: Writing the function in piecewise form, we get

f (x) = x−3 x>3

3−x x⩽3

For x=−1, we have

f(−1)=3−(−1)=4

(x)=ddx(3−x)=−1

f′(−1)=−1

1. What are the points of discontinuity of f(x) = (x – 1) + (x + 1)(x – 7)(x – 6)

Ans. It is clear from the provided function that it is not defined for the values x=6,7. We can argue that the function is continuous for the aforementioned points because it is a polynomial. Therefore, the inputs x=6 and x=7 will be where the points of discontinuity are located.

1. Write the number of points of discontinuity off(x) = [x] in[3,7].

Ans. Step functions are known to be discontinuous at integer positions. The integer inputs at the specified domain are 4,5,6,7. Consequently, the points of the discontinuity number is 4.

### Short Answer Type Questions (4 Marks)

The Chapter 5 Class 12 Maths Important Questions are created and prepared by Extramarks subject matter experts. When creating Important Questions for Continuity and Differentiability Class 12, these experts take into account all the significant themes of the chapter, including Rolle’s theorem, derivative functions, mean value theorem, etc.

Along with the grading scheme, the answers to each question are also presented in a step-by-step format. You will learn how to present their answers in the best way possible through this, helping them to avoid losing any exam points. Access the set of Class 12 Maths Chapter 5 Important Questions for Continuity and Differentiability from Extramarks to get a better grasp of the chapter.

### A Brief Overview of Class 12 Maths Chapter 5

Chapter 5 of Class 12 Maths covers the crucial ideas of continuity and differentiability. Along with the new class of functions known as exponential and logarithmic functions, you will also learn how to differentiate inverse trigonometric functions. These functions provide strong strategies for differentiation. The chapter also introduces certain geometrically clear circumstances that allow students to grasp basic theorems, like Rolle’s theorem and the Mean Value Theorem.

### Continuous Functions

A continuous function in calculus is a real-values function whose graphs do not have any holes or breaks.

A function f(x) is said to be continuous at x = k, if the following first three conditions satisfy.

f (k) exists.

lim x→k f(x) exists.

lim x→kf(x)=f(k).

Algebra of Composite Function

Continuous functions have continuous sums, differences, products, and quotients. It implies that if f and g are two continuous real-valued functions at real number c, then

• f + g is continuous at x = C.
• f – g is continuous at x = C.
• f . g is continuous at x = C.
• f/g is continuous at x = C.

Differentiable Function

The derivative function is continuous, meaning that its derivative exists at every location inside its domain. This means that there must be a (non-vertical) tangent line at each point in the domain of a graph of a differentiable function.

If the following criterion is met, a function is said to be differentiable at x0.

At x0, the function f is continuous.

The slope of the tangent at x0 has a clearly defined slope.

Derivatives of Inverse Trigonometric Functions

The inverse function theorem can be used to calculate the derivatives of inverse trigonometric functions. For instance, the inverse function for y = f(x) = arcsin x is the sine function x = (y) is equal to sin y. The derivative of the function y = arcsin x is so obtained as follows:

(arc sinx) =f′(x)=1ψ′(y) = 1Sin′(y) 1Cos(y)

= 11−sin2y = 11−sin2(arc sinx) = 11−x2 (−1<x<1)

Using the same method, we can find the derivative of other inverse trigonometric functions like arccos x, arctan x, arccot x, arcsec x, and arccosec x.

Logarithmic Differentiation

Logarithmic differentiation is the method of differentiating functions by first figuring out their logarithms, followed by differentiation.

The logarithmic derivative of the first function y = f is the name for the derivative of the logarithmic functions (x). With the help of these differentiable functions, we are able to calculate the derivatives of power-exponential functions, or functions with the form y = u(x)V(x), where u(x) and v(x) are differentiable functions of x, efficiently.

Derivative of Functions in Parametric Form

When a third variable is linked to each of the two other variables, a relationship between the two variables is established, even if it may not always be clear or implicit. In this instance, we say that a third variable is used to indicate the relationship between the two. To put it another way, a relationship between the variables x and y expressed in the form of x = f(t), y = f(t), is said to be in parametric form with t acting as a parameter.

Through the chain rule, we may calculate the derivative of such functions:

dy/dx = dt/dx.

Or

dy/dx = dy/dt ÷ dx/dt (Whenever dx/dt ≠ 0)

Rolle’s Theorem

Rolle’s theorem states that if f: x,y → R is continuous on

x, y and differentiable on (x, y) such that f(x) = f(y), where x and y are some real numbers, then there exists some c in (x, y) such that f'(c) =0.

Mean Value Theorem

Mean Value Theorem states that if f:x,y → R is continuous on x,y, and differentiable on (x, y). Then there exists some c in (x, y) such that:

f′(c)= f(x)−f(y)y−a

List of the Topics Covered in Class 12 Maths Chapter 5

The list of topics covered in Class 12 Maths Chapter 5 “Continuity and Differentiability” is given below:

5.1: Introduction

5.2: Continuity

5.2.1: Algebra of Continuous Function

5.3: Differentiability

5.3.1: Derivatives of Composite Function

5.3.2: Derivatives of Implicit Function

5.3.3: Derivatives of Inverse Trigonometric Function

5.4: Exponential And Logarithmic Function

5.5: Logarithmic Differentiation

5.6: Derivatives of Function in Parametric Form

5.7: Second Order Derivative

5.8: Mean Value Theorem

Benefits of Important Questions For Class 12 Maths Chapter 5 By Extramarks

The benefits of using Important Questions For Class 12 Maths Chapter 5 are as follows:

• This set of important questions come with answers in a simple format. Therefore, you can accurately understand these answers supplied. This enables you to more efficiently prepare the chapter and improve final exam preparation.
• You will grasp the main points of the chapter as you practise the key questions. This will help you perform well in the upcoming Maths exam.
• By dedicating enough time to answering crucial problems, you develop your time management abilities and accuracy.
• Last but not least, the more you practise, the more proficient you get with the themes of the chapter, which boosts your confidence, thereby making you feel better prepared.

Conclusion

Continuity and Differentiability Class 12 important questions is a great way to prepare the chapter concepts. The most recent NCERT textbook is taken into consideration when creating the crucial questions.By practising these important questions, you can quickly clear up any doubts they may have about any topic in the chapter. Utilise these questions to prepare for the upcoming Class 12 board exams.

Q1.

$\text{If}\mathrm{y}={\mathrm{log}}_{\mathrm{cosx}}\mathrm{sinx},\text{then}\frac{\mathrm{dy}}{\mathrm{dx}}\text{is equal to}$

Opt.

1. $\frac{\mathrm{cotx}.\mathrm{logcosx}?\mathrm{tanx}.\mathrm{logxsinx}}{{\left(\mathrm{logcosx}\right)}^{2}}$
2. $\frac{?\mathrm{tanxlogsinx}?\mathrm{logcosxcotx}}{{\left(\mathrm{logsinx}\right)}^{2}}$
3. $\frac{\mathrm{tanxlogsinx}?\mathrm{logcosxcotx}}{{\left(\mathrm{logsinx}\right)}^{2}}$
4. $\frac{\mathrm{tanxlogsinx}+\mathrm{logcosxcotx}}{{\left(\mathrm{logsinx}\right)}^{2}}$

Ans.

$\frac{\mathrm{cotx}.\mathrm{logcosx}\mathrm{tanx}.\mathrm{logxsinx}}{{\left(\mathrm{logcosx}\right)}^{2}}$

Q2.

$\text{I}\mathrm{f}\left(\mathrm{x}\right)=\left(2\mathrm{x}?1\right){\mathrm{tan}}^{1}\left({\mathrm{e}}^{2\mathrm{x}}\right),\text{then}\mathrm{f}‘\left(0\right)\text{is equal to}$

Opt.

1. $\frac{\mathrm{?}}{2}$
2. $\frac{\mathrm{?}}{2}+1$
3. $\frac{\mathrm{?}}{2}–1$
4. $\frac{\mathrm{?}}{6}+5$

Ans.

Q3.

$\mathrm{Find}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{if}\mathrm{y}=\mathrm{sin}\left({\mathrm{tan}}^{–1}{\mathrm{e}}^{–\mathrm{x}}\right)\mathrm{.}$

Opt.

$\begin{array}{l}\mathrm{Here}\mathrm{y}=\mathrm{sin}\left({\mathrm{tan}}^{1}{\mathrm{e}}^{1}\right)\\ \mathrm{differentiating}\mathrm{y}\mathrm{w}.\mathrm{r}.\mathrm{t}\mathrm{x}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{sin}\left({\mathrm{tan}}^{1}{\mathrm{e}}^{1}\right)\right]\end{array}$

Ans.

$\begin{array}{l}\mathrm{Here}\mathrm{y}=\mathrm{sin}\left({\mathrm{tan}}^{1}{\mathrm{e}}^{1}\right)\\ \mathrm{differentiating}\mathrm{y}\mathrm{w}.\mathrm{r}.\mathrm{t}\mathrm{x}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{}=\mathrm{}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{}\left[\mathrm{sin}\left({\mathrm{tan}}^{1}{\mathrm{e}}^{1}\right)\right]\\ =\mathrm{}\mathrm{cos}\left({\mathrm{tan}}^{1}{\mathrm{e}}^{1}\right)\mathrm{}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{}\left({\mathrm{tan}}^{1}{\mathrm{e}}^{\mathrm{x}}\right)\\ =\mathrm{}\mathrm{cos}\left({\mathrm{tan}}^{1}{\mathrm{e}}^{1}\right)\mathrm{}.\frac{1}{1+{\left({\mathrm{e}}^{\mathrm{x}}\right)}^{2}}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\mathrm{x}}\right)\\ =\mathrm{}\mathrm{cos}\left({\mathrm{tan}}^{1}{\mathrm{e}}^{\mathrm{x}}\right)\mathrm{}.\frac{1}{1+{\left({\mathrm{e}}^{\mathrm{x}}\right)}^{2}}.{\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\right)\\ =\mathrm{}\mathrm{cos}\left({\mathrm{tan}}^{1}{\mathrm{e}}^{\mathrm{x}}\right)\mathrm{}.\frac{1}{1+{\left({\mathrm{e}}^{\mathrm{x}}\right)}^{2}}.{\mathrm{e}}^{\mathrm{x}}.\left(1\right)\\ =\mathrm{}\frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{cos}\left({\mathrm{tan}}^{1}{\mathrm{e}}^{\mathrm{x}}\right)}{1+{\mathrm{e}}^{2\mathrm{x}}}\mathrm{}\end{array}$

Q4.

$\text{If y =}{\left({\mathrm{tan}}^{1}\mathrm{x}\right)}^{2},\mathrm{?}\mathrm{show}\mathrm{that}{\left({\mathrm{x}}^{2}+1\right)}^{2}{\mathrm{y}}_{2}+2\mathrm{x}\left({\mathrm{x}}^{2}+1\right){\mathrm{y}}_{1}\mathrm{?}=\mathrm{?}2$

Opt.

$\begin{array}{l}\mathrm{Here},\mathrm{y}={\left({\mathrm{tan}}^{1}\mathrm{x}\right)}^{2}\\ \mathrm{differentiating}\mathrm{y}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x}\\ {\mathrm{y}}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\left({\mathrm{tan}}^{1}\mathrm{x}\right)}^{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Here},\mathrm{y}={\left({\mathrm{tan}}^{1}\mathrm{x}\right)}^{2}\\ \mathrm{differentiating}\mathrm{y}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x}\\ {\mathrm{y}}_{1}=\mathrm{}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{}{\left({\mathrm{tan}}^{1}\mathrm{x}\right)}^{2}\\ =\mathrm{}2\left({\mathrm{tan}}^{1}\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{tan}}^{1}\mathrm{x}\right)\\ =\mathrm{}\frac{2{\mathrm{tan}}^{1}\mathrm{x}}{{\mathrm{x}}^{2}+1}\\ \left({\mathrm{x}}^{2}+1\right){\mathrm{y}}_{1}=\mathrm{}2{\mathrm{tan}}^{1}\mathrm{x}\\ \mathrm{again}\mathrm{differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left[\left({\mathrm{x}}^{2}+1\right){\mathrm{y}}_{1}\right]\mathrm{}=\mathrm{}\frac{\mathrm{d}}{\mathrm{dx}}\left(2{\mathrm{tan}}^{1}\mathrm{x}\right)\\ \mathrm{}\left({\mathrm{x}}^{2}+1\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{}{\mathrm{y}}_{1}+{\mathrm{y}}_{1}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{2}+1\right)\mathrm{}=\mathrm{}\frac{\mathrm{d}}{\mathrm{dx}}\left(2{\mathrm{tan}}^{1}\mathrm{x}\right)\\ \left({\mathrm{x}}^{2}+1\right){\mathrm{y}}_{2}+{\mathrm{y}}_{1}\left(2\mathrm{x}\right)\mathrm{}=\mathrm{}\frac{2}{{\mathrm{x}}^{2}+1}\left[\mathrm{as}\mathrm{}{\mathrm{y}}_{2}\mathrm{}=\mathrm{}\frac{{\mathrm{d}}^{2\mathrm{y}}}{{\mathrm{dx}}^{2}}\right]\\ \mathrm{}{\left({\mathrm{x}}^{2}+1\right)}^{2}{\mathrm{y}}_{2}+{\mathrm{y}}_{1}\left({\mathrm{x}}^{2}+1\right)\left(2\mathrm{x}\right)\mathrm{}=2\\ \mathrm{}{\left({\mathrm{x}}^{2}+1\right)}^{2}{\mathrm{y}}_{2}+2\mathrm{x}\left({\mathrm{x}}^{2}+1\right){\mathrm{y}}_{1}\mathrm{}=\mathrm{}2.\end{array}$