Short Introduction

Continuity is one of the most important concepts of Calculus. It helps us understand whether a function behaves smoothly around a point without any sudden breaks or jumps. In CBSE Class 12 Mathematics, Chapter 5 “Continuity and Differentiability” introduces students to the formal definition of continuity and its applications.

Exercise 5.1 focuses on checking the continuity of polynomial, rational, modulus, piecewise and trigonometric functions at specific points.

This complete solution guide by www.mymockmate.com provides easy-to-understand step-by-step explanations that are highly useful for CBSE Board Exams and competitive examinations.

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Quick Information Box

Particular	Details
Chapter	5
Chapter Name	Continuity and Differentiability
Exercise	5.1
Board	CBSE
Class	12
Difficulty Level	Easy to Moderate
Important For	Board Exams, JEE, CUET

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Concepts Used (Topics Covered)

✔ Definition of Continuity

✔ Left Hand Limit (LHL)

✔ Right Hand Limit (RHL)

✔ Existence of Function Value

✔ Continuous Functions

✔ Polynomial Functions

✔ Rational Functions

✔ Modulus Functions

✔ Piecewise Functions

✔ Trigonometric Functions

✔ Greatest Integer Function

✔ Composite Functions

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Important Formulas

Definition of Continuity

A function f(x) is continuous at x = a if

lim x→a f(x) = f(a)

That is,

LHL = RHL = f(a)

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Polynomial Functions

Every polynomial function is continuous for all real numbers.

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Rational Functions

A rational function is continuous wherever its denominator is not zero.

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Modulus Function

f(x)=|x|

is continuous for every real number.

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Sum, Difference and Product Rule

If f(x) and g(x) are continuous at x=a, then

• f(x)+g(x)
• f(x)-g(x)
• f(x)×g(x)

are also continuous.

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Question 1

1. Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

Solution

Since $f(x)=5x-3$ is a polynomial function, it is continuous for every real number.

Still, we verify the three points.

At $x=0$

$$f(0)=5(0)-3=-3$$$$\lim_{x\to0}f(x) =\lim_{x\to0}(5x-3) =-3$$

Therefore,$$\lim_{x\to0}f(x)=f(0)$$

Hence, ff is continuous at x=0x=0.

At $x=-3$

$$f(-3)=5(-3)-3=-18$$$$\lim_{x\to-3}(5x-3)=-18$$

Thus,$$\lim_{x\to-3}f(x)=f(-3)$$

Hence, ff is continuous at x=−3x=-3.

At $x=5$

$$f(5)=25-3=22$$$$\lim_{x\to5}(5x-3)=22$$

Therefore, ff is continuous at x=5x=5.

Answer

f(x)=5x−3 is continuous at 0,−3, and 5.​

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Question 2

Examine the continuity of the function f(x) = 2×2 – 1 at x = 3.

Solution

First calculate $f(3)$:$$f(3)=2(3)^2-1=18-1=17$$

Now,$$\lim_{x\to3}f(x) =\lim_{x\to3}(2x^2-1) =2(3)^2-1 =17$$

Therefore,$$\lim_{x\to3}f(x)=f(3)=17$$

Hence, f(x) is continuous at x=3.​

Question 3

Examine the following functions for continuity.

(a) $f(x)=x-5$

This is a polynomial function.

For any $a\in\mathbb R$,$$\lim_{x\to a}(x-5)=a-5=f(a)$$

Therefore,$$\boxed{f(x)=x-5\text{ is continuous for all }x\in\mathbb R.}$$

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(b)

$$f(x)=\frac1{x-5},\qquad x\ne5$$

The denominator is zero at $x=5$, so the function is not defined there.

For every $a\ne5$,$$\lim_{x\to a}\frac1{x-5} =\frac1{a-5} =f(a)$$

Hence,$$\boxed{f\text{ is continuous on }\mathbb R-\{5\}.}$$

It is not defined, and hence discontinuous as an extended real-line function, at $x=5$.

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(c)

$$f(x)=\frac{x^2-25}{x+5},\qquad x\ne-5$$

Factor the numerator:$$x^2-25=(x-5)(x+5)$$

Therefore,$$f(x)=\frac{(x-5)(x+5)}{x+5}=x-5,\qquad x\ne-5$$

Thus, for every $x\ne-5$, $f(x)=x-5$, which is continuous.

Hence,$$\boxed{f\text{ is continuous for every }x\ne-5.}$$

At $x=-5$, the original function is not defined.

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(d)

$$f(x)=|x-5|$$

The modulus function is continuous, and $x-5$ is continuous.

Therefore, ∣x−5∣ is continuous for all real x.​

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Question 4

Prove that the function f(x) = xⁿ is continuous at x = n, where n is a positive integer.

Solution

We have$$f(n)=n^n$$

Also,$$\lim_{x\to n}f(x) =\lim_{x\to n}x^n =n^n$$

Thus,$$\lim_{x\to n}f(x)=f(n)$$

Therefore,$$\boxed{f(x)=x^n\text{ is continuous at }x=n.}$$

In fact, $x^n$ is continuous for every real $x$.

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At $x=0$

$$f(0)=0$$

Since the first rule applies near $0$,$$\lim_{x\to0}f(x)=\lim_{x\to0}x=0$$

Thus,$$\lim_{x\to0}f(x)=f(0)$$

So, continuous at x=0x=0.

At $x=1$

$$f(1)=1$$

Left-hand limit:$$\lim_{x\to1^-}f(x)=\lim_{x\to1^-}x=1$$

Right-hand limit:$$\lim_{x\to1^+}f(x)=5$$

Since$$1\ne5$$

the function is not continuous at x=1x=1.

At $x=2$

For $x>1$,$$f(x)=5$$

Therefore,$$\lim_{x\to2}f(x)=5=f(2)$$

Hence, continuous at x=2x=2.

Answer

Continuous at x=0,2;discontinuous at x=1.​

Find all points of discontinuity of f, where f is defined by

Question 6

$$f(x)= \begin{cases} 2x+3,&x\le2\\ 2x-3,&x>2 \end{cases}$$

The possible point of discontinuity is $x=2$. $$f(2)=2(2)+3=7$$

LHL:$$\lim_{x\to2^-}f(x)=2(2)+3=7$$

RHL:$$\lim_{x\to2^+}f(x)=2(2)-3=1$$

Since$$7\ne1$$

the function is discontinuous at $x=2$.

Answer

$$\boxed{x=2\text{ is the only point of discontinuity.}}$$

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Question 7

$$f(x)= \begin{cases} |x|+3,&x\le-3\\ -2x,&-3<x<3\\ 6x+2,&x\ge3 \end{cases}$$

Possible discontinuities occur at $x=-3$ and $x=3$.

At $x=-3$

$$f(-3)=|-3|+3=6$$

LHL:$$\lim_{x\to-3^-}f(x)=|-3|+3=6$$

RHL:$$\lim_{x\to-3^+}f(x)=-2(-3)=6$$

Thus,$$\text{LHL}=\text{RHL}=f(-3)=6$$

So $f$ is continuous at $-3$.

At $x=3$

$$f(3)=6(3)+2=20$$

LHL:$$\lim_{x\to3^-}f(x)=-2(3)=-6$$

RHL:$$\lim_{x\to3^+}f(x)=6(3)+2=20$$

Since$$-6\ne20$$

$f$ is discontinuous at $x=3$.

Answer

$$\boxed{x=3\text{ is the only point of discontinuity.}}$$

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Question 8

$$f(x)= \begin{cases} \dfrac{|x|}{x},&x\ne0\\ 0,&x=0 \end{cases}$$

At $x=0$:

For $x<0$,$$\frac{|x|}{x}=\frac{-x}{x}=-1$$

Therefore,$$\lim_{x\to0^-}f(x)=-1$$

For $x>0$,$$\frac{|x|}{x}=1$$

Therefore,$$\lim_{x\to0^+}f(x)=1$$

Since$$-1\ne1$$

the limit does not exist.

Hence,$$\boxed{f\text{ is discontinuous at }x=0.}$$

It is continuous for $x\ne0$.

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Question 9

$$f(x)= \begin{cases} \dfrac{x}{|x|},&x<0\\ -1,&x\ge0 \end{cases}$$

At $x=0$,$$f(0)=-1$$

LHL:$$\lim_{x\to0^-}\frac{x}{|x|} =-1$$

RHL:$$\lim_{x\to0^+}(-1)=-1$$

Thus,$$\text{LHL}=\text{RHL}=f(0)=-1$$

Therefore,$$\boxed{f\text{ is continuous at }x=0.}$$

Each part is continuous on its respective open interval, so$$\boxed{f\text{ is continuous for all real }x.}$$

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Question 10

$$f(x)= \begin{cases} x+1,&x\ge1\\ x^2+1,&x<1 \end{cases}$$

At $x=1$,$$f(1)=1+1=2$$

LHL:$$\lim_{x\to1^-}(x^2+1)=2$$

RHL:$$\lim_{x\to1^+}(x+1)=2$$

Therefore,$$\text{LHL}=\text{RHL}=f(1)=2$$

Hence,$$\boxed{f\text{ is continuous for all real }x.}$$

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Question 11

$$f(x)= \begin{cases} x^3-3,&x\le2\\ x^2+1,&x>2 \end{cases}$$

At $x=2$,$$f(2)=2^3-3=5$$

LHL:$$\lim_{x\to2^-}(x^3-3)=5$$

RHL:$$\lim_{x\to2^+}(x^2+1)=5$$

Thus,$$\text{LHL}=\text{RHL}=f(2)=5$$

Therefore,$$\boxed{f\text{ is continuous for all real }x.}$$

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Question 12

$$f(x)= \begin{cases} x^{10}-1,&x\le1\\ x^2,&x>1 \end{cases}$$

At $x=1$,$$f(1)=1^{10}-1=0$$

LHL:$$\lim_{x\to1^-}(x^{10}-1)=0$$

RHL:$$\lim_{x\to1^+}x^2=1$$

Since$$0\ne1$$

the function is discontinuous at $x=1$.

Answer

x=1 is the only point of discontinuity.​

At $x=1$,$$f(1)=1+5=6$$

LHL:$$\lim_{x\to1^-}(x+5)=6$$

RHL:$$\lim_{x\to1^+}(x-5)=-4$$

Since$$6\ne-4$$

the function is discontinuous at $x=1$.

Answer

$$\boxed{\text{The function is not continuous.}}$$

The exercise asks this continuity check explicitly.

Discuss the continuity of the function f, where f is defined by

Question 14

$$f(x)= \begin{cases} 3,&0\le x\le1\\ 4,&1<x<3\\ 5,&3\le x\le10 \end{cases}$$

Possible points are $x=1$ and $x=3$.

At $x=1$

$$f(1)=3$$

LHL $=3$, while RHL $=4$.

Therefore, discontinuous at $x=1$.

At $x=3$

$$f(3)=5$$

LHL $=4$, while RHL $=5$.

Therefore, discontinuous at $x=3$.

Answer

$$\boxed{\text{Points of discontinuity: }x=1,3.}$$

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Question 15

$$f(x)= \begin{cases} 2x,&x<0\\ 0,&0\le x\le1\\ 4x,&x>1 \end{cases}$$

Possible points: $0,1$.

At $x=0$

$$f(0)=0$$

LHL:$$\lim_{x\to0^-}2x=0$$

RHL:$$\lim_{x\to0^+}0=0$$

Thus, continuous at $0$.

At $x=1$

$$f(1)=0$$

LHL:$$\lim_{x\to1^-}0=0$$

RHL:$$\lim_{x\to1^+}4x=4$$

Since $0\ne4$,$$\boxed{x=1\text{ is a point of discontinuity.}}$$

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Question 16

$$f(x)= \begin{cases} -2,&x\le-1\\ 2x,&-1<x\le1\\ 2,&x>1 \end{cases}$$

At $x=-1$

$$f(-1)=-2$$

LHL:$$-2$$

RHL:$$2(-1)=-2$$

Hence continuous at $-1$.

At $x=1$

$$f(1)=2$$

LHL:$$2(1)=2$$

RHL:$$2$$

Hence continuous at $1$.

Therefore, f is continuous for all real x.​

For continuity at $x=3$,$$\text{LHL}=\text{RHL}=f(3)$$

LHL:$$3a+1$$

RHL:$$3b+3$$

Therefore,$$3a+1=3b+3$$$$3a-3b=2$$

Hence,$$\boxed{a-b=\frac23}$$

or a=b+2/3​​

At $x=0$

$$f(0)=\lambda(0-0)=0$$

LHL:$$\lim_{x\to0^-}\lambda(x^2-2x)=0$$

RHL:$$\lim_{x\to0^+}(4x+\lambda)=\lambda$$

For continuity,$$0=\lambda$$

Thus,$$\boxed{\lambda=0}$$

At $x=1$

Since $1>0$, near $x=1$,$$f(x)=4x+\lambda$$

which is a linear function and therefore continuous.

Hence,$$\boxed{f\text{ is continuous at }x=1\text{ for every }\lambda.}$$

For the value required for continuity at $x=0$, $\lambda=0$.

19. Show that the function defined by g(x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

Let $n$ be any integer.

At $x=n$,$$[n]=n$$

Therefore,$$g(n)=n-n=0$$

For $x\to n^-$,$$[x]=n-1$$

Hence,$$\lim_{x\to n^-}g(x) =\lim_{x\to n^-}[x-(n-1)] =1$$

For $x\to n^+$,$$[x]=n$$

Therefore,$$\lim_{x\to n^+}g(x)=0$$

Thus,$$\text{LHL}=1,\qquad \text{RHL}=0$$

Since LHL $\ne$ RHL, g(x) is discontinuous at every integral point.

20. Is the function defined by f(x) = x² – sin x + 5 continuous at x = π?

The functions $x^2$, $\sin x$, and the constant function $5$ are continuous everywhere.

Therefore, their sum/difference is also continuous.

Hence,$$\boxed{f(x)=x^2-\sin x+5\text{ is continuous at }x=\pi.}$$

21. Discuss the continuity of the following functions:
    (a) f(x) = sin x + cos x
    (b) f(x) = sin x – cos x
    (c) f(x) = sin x . cos x

(a) $f(x)=\sin x+\cos x$

Both $\sin x$ and $\cos x$ are continuous for all real $x$.

Therefore,$$\boxed{\sin x+\cos x\text{ is continuous on }\mathbb R.}$$

(b) $f(x)=\sin x-\cos x$

Difference of two continuous functions is continuous.$$\boxed{\sin x-\cos x\text{ is continuous on }\mathbb R.}$$

(c) $f(x)=\sin x\cdot\cos x$

Product of continuous functions is continuous.$$\boxed{\sin x\cos x\text{ is continuous on }\mathbb R.}$$

The three functions are specified in the exercise statement.

22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Discuss the continuity of cosine, cosecant, secant and cotangent functions.

1. Cosine function

$$f(x)=\cos x$$

Cosine is continuous for every real number.$$\boxed{\cos x\text{ is continuous on }\mathbb R}$$

2. Cosecant function

$$\csc x=\frac1{\sin x}$$

It is continuous wherever $\sin x\ne0$.

Since$$\sin x=0\quad\text{at }x=n\pi$$

we have$$\boxed{\csc x\text{ is continuous for }x\ne n\pi,\ n\in\mathbb Z.}$$

3. Secant function

$$\sec x=\frac1{\cos x}$$

It is continuous wherever $\cos x\ne0$.$$\cos x=0$$

at$$x=\frac{(2n+1)\pi}{2}$$

Therefore,$$\boxed{\sec x\text{ is continuous for }x\ne\frac{(2n+1)\pi}{2}.}$$

4. Cotangent function

$$\cot x=\frac{\cos x}{\sin x}$$

It is continuous wherever $\sin x\ne0$.

Therefore, cotx is continuous for x Not = nπ.

LHL:$$\lim_{x\to0^-}\frac{\sin x}{x}=1$$

RHL:$$\lim_{x\to0^+}(x+1)=1$$

Also,$$f(0)=0+1=1$$

Therefore,$$\text{LHL}=\text{RHL}=f(0)=1$$

Hence, the function is continuous at $0$.

Answer

There are no points of discontinuity.​

At $x=0$,$$-1\le\sin\frac1x\le1$$

Multiplying by $x^2\ge0$,$$-x^2\le x^2\sin\frac1x\le x^2$$

As $x\to0$,$$-x^2\to0,\qquad x^2\to0$$

Hence, by the squeeze theorem,$$\lim_{x\to0}x^2\sin\frac1x=0$$

Since$$f(0)=0$$

we get$$\lim_{x\to0}f(x)=f(0)$$

Therefore,$$\boxed{f\text{ is continuous at }x=0.}$$

It is also continuous for $x\ne0$.

Thus, f is continuous on R.​

At $x=0$,$$f(0)=-1$$

Also,$$\lim_{x\to0}(\sin x-\cos x) =\sin0-\cos0 =0-1=-1$$

Therefore,$$\lim_{x\to0}f(x)=f(0)=-1$$

Hence,$$\boxed{f\text{ is continuous at }x=0.}$$

Since $\sin x-\cos x$ is continuous elsewhere, f is continuous for all real x.

Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

Question 26

$$f(x)= \begin{cases} \dfrac{k\cos x}{\pi-2x},&x\ne\frac\pi2\\ 3,&x=\frac\pi2 \end{cases}$$

Find $k$ for continuity at $x=\frac\pi2$.

We require$$\lim_{x\to\pi/2}\frac{k\cos x}{\pi-2x}=3$$

Put $x=\frac\pi2+h$.

Then$$\cos\left(\frac\pi2+h\right)=-\sin h$$

and$$\pi-2\left(\frac\pi2+h\right)=-2h$$

Therefore,$$\lim_{h\to0}\frac{k(-\sin h)}{-2h} = \frac{k}{2}\lim_{h\to0}\frac{\sin h}{h} =\frac{k}{2}$$

For continuity,$$\frac{k}{2}=3$$

Hence,$$\boxed{k=6}$$

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Question 27

$$f(x)= \begin{cases} kx^2,&x\le2\\ 3,&x>2 \end{cases}$$

For continuity at $x=2$,$$f(2)=4k$$

LHL:$$4k$$

RHL:$$3$$

Thus,$$4k=3$$$$\boxed{k=\frac34}$$

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Question 28

$$f(x)= \begin{cases} kx+1,&x\le\pi\\ \cos x,&x>\pi \end{cases}$$

At $x=\pi$,$$f(\pi)=k\pi+1$$

LHL:$$k\pi+1$$

RHL:$$\cos\pi=-1$$

For continuity,$$k\pi+1=-1$$$$k\pi=-2$$

Hence,$$\boxed{k=-\frac2\pi}$$

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Question 29

$$f(x)= \begin{cases} kx+1,&x\le5\\ 3x-5,&x>5 \end{cases}$$

At $x=5$,$$f(5)=5k+1$$

RHL:$$3(5)-5=10$$

For continuity,$$5k+1=10$$$$5k=9$$

Therefore, k=9/5​​

For continuity at $x=2$,$$2a+b=5 \tag{1}$$

For continuity at $x=10$,$$10a+b=21 \tag{2}$$

Subtract (1) from (2):$$8a=16$$$$a=2$$

Substitute into (1):$$2(2)+b=5$$$$b=1$$

Hence,$$\boxed{a=2,\qquad b=1}$$

The original exercise asks for these values so that the entire piecewise function is continuous.

31. Show that the function defined by f(x) = cos (x²) is a continuous function.

We know:$$x^2$$

is continuous for every real $x$, and$$\cos x$$

is continuous for every real $x$.

The composition of continuous functions is continuous.

Therefore, cos(x2) is continuous for all x∈R.​

32. Show that the function defined by f(x) = |cos x| is a continuous function.

We know that$$\cos x$$

is continuous for every real $x$.

Also, the modulus function$$|x|$$

is continuous.

Therefore, the composition$$|\cos x|$$

is continuous.

Hence, ∣cosx∣ is continuous for all real x.​

33. Examine that sin |x| is a continuous function.

The function$$|x|$$

is continuous for every real $x$.

The function$$\sin x$$

is also continuous for every real $x$.

Therefore, their composition$$\sin|x|$$

is continuous.

Hence, sin∣x∣ is continuous for all x∈R.​

34. Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|.

The possible points are where the expressions inside the modulus become zero:$$x=0,\qquad x=-1$$

We examine these points.

For $x<-1$

$$|x|=-x,\qquad |x+1|=-(x+1)$$

Thus,$$f(x)=-x+(x+1)=1$$

For $-1\le x<0$

$$|x|=-x,\qquad |x+1|=x+1$$

Therefore,$$f(x)=-x-(x+1)=-2x-1$$

For $x\ge0$

$$|x|=x,\qquad |x+1|=x+1$$

Hence,$$f(x)=x-(x+1)=-1$$

Now check the critical points.

At $x=-1$

$$\lim_{x\to-1^-}f(x)=1$$

and$$\lim_{x\to-1^+}f(x) =-2(-1)-1=1$$

Also,$$f(-1)=1$$

Therefore, $f$ is continuous at $-1$.

At $x=0$

$$\lim_{x\to0^-}f(x)=-2(0)-1=-1$$

and$$\lim_{x\to0^+}f(x)=-1$$

Also,$$f(0)=-1$$

Therefore, $f$ is continuous at $0$.

Final Answer

There are no points of discontinuity. f(x)=∣x∣−∣x+1∣ is continuous for every real x.​

Quick Answer Key

Q.	Answer
1	Continuous at $0,-3,5$
2	Continuous at $x=3$
3(a)	Continuous on $\mathbb R$
3(b)	Continuous on $\mathbb R-\{5\}$
3(c)	Continuous on $\mathbb R-\{-5\}$
3(d)	Continuous on $\mathbb R$
4	Continuous at $x=n$
5	Discontinuous at $x=1$
6	Discontinuous at $x=2$
7	Discontinuous at $x=3$
8	Discontinuous at $x=0$
9	Continuous everywhere
10	Continuous everywhere
11	Continuous everywhere
12	Discontinuous at $x=1$
13	Discontinuous at $x=1$
14	Discontinuous at $x=1,3$
15	Discontinuous at $x=1$
16	Continuous everywhere
17	$a-b=\frac23$
18	$\lambda=0$ for continuity at $0$; continuous at $1$ for every $\lambda$
19	Discontinuous at every integer
20	Continuous at $x=\pi$
21	All three continuous on $\mathbb R$
22	Continuous on their respective domains
23	No point of discontinuity
24	Continuous everywhere
25	Continuous everywhere
26	$k=6$
27	$k=\frac34$
28	$k=-\frac2\pi$
29	$k=\frac95$
30	$a=2,\ b=1$
31	Continuous everywhere
32	Continuous everywhere
33	Continuous everywhere
34	No point of discontinuity

Common Mistakes

❌ Forgetting to check whether the function is defined.

❌ Assuming every rational function is continuous everywhere.

❌ Not comparing LHL and RHL in piecewise functions.

❌ Ignoring denominator equal to zero.

❌ Missing modulus function properties.

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Exam Tips

✅ Always write:

Function Value

↓

LHL

↓

RHL

↓

Conclusion

✅ Remember:

Polynomial → Continuous Everywhere

Rational → Continuous except denominator = 0

Modulus → Continuous Everywhere

Greatest Integer → Discontinuous at Integers

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Practice MCQs

1.

Which function is continuous everywhere?

A. 1/x

B. |x|

C. 1/(x−2)

D. [x]

Answer: B

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2.

The function 1/(x−3) is discontinuous at

A. 1

B. 2

C. 3

D. 4

Answer: C

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3.

Every polynomial function is

A. Discontinuous

B. Continuous

C. Undefined

D. None

Answer: B

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4.

The greatest integer function is discontinuous at

A. Rational numbers

B. Irrational numbers

C. Integers

D. Positive numbers

Answer: C

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FAQ Section

Q1. What is continuity?

A function is continuous if

LHL = RHL = Function Value.

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Q2. Are polynomial functions always continuous?

Yes. Every polynomial function is continuous for all real numbers.

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Q3. Where are rational functions discontinuous?

At points where the denominator becomes zero.

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Q4. Is modulus function continuous?

Yes, |x| is continuous everywhere.

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Q5. Which questions are most important for CBSE Boards?

Piecewise functions, rational functions and greatest integer functions.

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