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Updated 6 August 2026

NCERT Solutions

Continuity and Differentiability

CBSE · Class 12 · Mathematics

NCERT Solutions for Continuity and Differentiability — CBSE Class 12 Mathematics.

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39 Questions Solved · 1 Section

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EXERCISE 5.1

1Prove that the function

f(x) = 5x - 3

is continuous at

x = 0

, at

x = -3

and at

x = 5

.Show solution

For any real number

c

\lim_{x\to c}(5x-3)=5c-3=f(c).

f(x)=5x-3

is continuous at every real number. Hence it is continuous at

x=0

x=-3

and

x=5

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2Examine the continuity of the function

f(x) = 2x^2 - 1

x = 3

.Show solution

x=3

f(3)=2(3)^2-1=18-1=17.

Also,

\lim_{x\to 3}(2x^2-1)=2(3)^2-1=17.

Since

\lim_{x\to 3}f(x)=f(3)

, the function is continuous at

x=3

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3(a)

f(x) = x - 5

Show solution

The function is a polynomial:

f(x)=x-5.

Every polynomial function is continuous at every real number. So

f

is continuous everywhere.

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3(b)

f(x) = \frac{1}{x - 5}, x \neq 5

Show solution

The function is

f(x)=\frac{1}{x-5},\quad x\ne 5.

This is a rational function, so it is continuous wherever it is defined. Since it is not defined at

x=5

, it is continuous for all

x\ne 5

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3(c)

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

Show solution

For

x\ne -5

f(x)=\frac{x^2-25}{x+5}=\frac{(x-5)(x+5)}{x+5}=x-5.

So the function agrees with the polynomial

x-5

wherever it is defined. Hence it is continuous for all

x\ne -5

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3(d)

f(x) = |x - 5|

Show solution

We know

f(x)=|x-5|

and the modulus function is continuous everywhere. Since

x-5

is a polynomial and continuous everywhere,

|x-5|

is also continuous everywhere.

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4Prove that the function

f(x) = x^n

is continuous at

x = n

, where

n

is a positive integer.Show solution

Let

p(x)=x^n

, where

n

is a positive integer. A polynomial function is continuous at every real number. Therefore

p(x)=x^n

is continuous at

x=n

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5Is the function

f

defined byShow solution

The function is

f(x)=\begin{cases}x,& x\le 1\\5,& x>1\end{cases}

- At

x=0

, the first branch applies, so

f(x)=x

, which is continuous. Hence it is continuous at

x=0

.
- At

x=1

f(1)=1

from the first branch. The left hand limit is also

1

, but the right hand limit is

5

. Since they do not match, the function is not continuous at

x=1

.
- At

x=2

, the second branch applies and

f(x)=5

, so it is continuous at

x=2

.

So the function is continuous at

x=0

and

x=2

, but not at

x=1

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f(x) = \begin{cases} 2x + 3, & \text{if } x \leq 2 \\ 2x - 3, & \text{if } x > 2 \end{cases}

Show solution

Given

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

For

x<2

f

is linear, so continuous. For

x>2

, it is also linear, so continuous.

Check

x=2

:

- Left value:

f(2)=2(2)+3=7

- Right hand limit:

\lim_{x\to 2^+}(2x-3)=1

Since these are not equal, the function is discontinuous at

x=2

.

So the only point of discontinuity is

x=2

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f(x) = \begin{cases} |x| + 3, & \text{if } x \leq -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \geq 3 \end{cases}

Show solution

The function is

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

Each branch is continuous on its interval. So only the boundary points need checking.

At

x=-3

:
-

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

- Right hand limit from

-2x

-2(-3)=6

So it is continuous at

x=-3

.

At

x=3

:
- Left hand limit from

-2x

-2(3)=-6

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

These do not match, so it is discontinuous at

x=3

.

Hence the only point of discontinuity is

x=3

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f(x) = \begin{cases} \frac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}

Show solution

For

x>0

\frac{|x|}{x}=1

; for

x<0

\frac{|x|}{x}=-1

.

At

x=0

, the function value is defined as

0

.

The left hand limit at

0

-1

and the right hand limit at

0

1

, so the limits are not equal. Therefore the function is discontinuous at

x=0

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f(x) = \begin{cases} \frac{x}{|x|}, & \text{if } x < 0 \\ -1, & \text{if } x \geq 0 \end{cases}

Show solution

For

x<0

\frac{x}{|x|}=\frac{x}{-x}=-1.

For

x\ge 0

, the function is also

-1

.

So the function is constant equal to

-1

on its whole domain, hence continuous wherever it is defined. Since it is defined for all real numbers, it is continuous everywhere.

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f(x) = \begin{cases} x + 1, & \text{if } x \geq 1 \\ x^2 + 1, & \text{if } x < 1 \end{cases}

Show solution

The function is

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

Each branch is continuous on its interval. Check the joining point

x=1

:

- Left hand limit:

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

- Right hand limit:

\lim_{x\to 1^+}(x+1)=2

- Value:

f(1)=1+1=2

Since all three are equal, the function is continuous at

x=1

and hence everywhere.

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

Show solution

The function is

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

Each branch is continuous on its interval. Check

x=2

:

- Left value:

f(2)=2^3-3=5

- Right hand limit:

\lim_{x\to 2^+}(x^2+1)=5

They are equal, so the function is continuous at

x=2

.

Therefore the function is continuous everywhere.

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f(x) = \begin{cases} x^{10} - 1, & \text{if } x \leq 1 \\ x^2, & \text{if } x > 1 \end{cases}

Show solution

The function is

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

Each piece is continuous on its interval. Check at

x=1

:

- Left value:

1^{10}-1=0

- Right hand limit:

\lim_{x\to 1^+}x^2=1

Since these are not equal, the function is not continuous at

x=1

.

So it is continuous at all points except

x=1

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13Is the function defined byShow solution

The function is

f(x)=\begin{cases}3,& 0\le x\le 1\\4,& 1<x<3\\5,& 3\le x\le 10\end{cases}

Each piece is constant, so continuous on its interval. Check the junctions:

- At

x=1

, left value is

3

and right hand limit is

4

, so discontinuous.
- At

x=3

, left hand limit is

4

and value is

5

, so discontinuous.

Thus the points of discontinuity are

x=1

and

x=3

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f(x) = \begin{cases} 3, & \text{if } 0 \leq x \leq 1 \\ 4, & \text{if } 1 < x < 3 \\ 5, & \text{if } 3 \leq x \leq 10 \end{cases}

Show solution

The function is

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

Each part is continuous on its own interval.

- At

x=0

, left hand limit is

2(0)=0

and

f(0)=0

, so continuous.
- At

x=1

, left hand limit is

0

, but right hand limit is

4(1)=4

, so not continuous.

Hence the only point of discontinuity is

x=1

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f(x) = \begin{cases} 2x, & \text{if } x < 0 \\ 0, & \text{if } 0 \leq x \leq 1 \\ 4x, & \text{if } x > 1 \end{cases}

Show solution

The function is

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

Check the boundary points:

- At

x=-1

, left value is

-2

and right hand limit from

2x

2(-1)=-2

, so continuous.
- At

x=1

, left value is

2(1)=2

and right hand limit is

2

, so continuous.

Therefore the function is continuous everywhere; there are no points of discontinuity.

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f(x) = \begin{cases} -2, & \text{if } x \leq -1 \\ 2x, & \text{if } -1 < x \leq 1 \\ 2, & \text{if } x > 1 \end{cases}

Show solution

For continuity at

x=3

, the left and right values must be equal.

Left side at

x=3

f(3)=a(3)+1=3a+1

Right side at

x=3

3b+3

So we need

3a+1=3b+3,

which simplifies to

3a-3b=2.

Hence the required relation is

3a-3b=2

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17Find the relationship between

a

and

b

so that the function

f

defined byShow solution

For continuity at

x=0

, the left branch value at

0

must equal the right hand limit.

Left branch at

x=0

f(0)=\lambda(0^2-2\cdot 0)=0

Right hand limit at

x=0

\lim_{x\to 0^+}(4x+1)=1

Since

0\ne 1

, the function cannot be continuous at

x=0

for any value of

\lambda

.

At

x=1

, the second branch applies, and the function is continuous there because it is linear on that side and the left branch is also continuous. The question’s continuity condition at

x=0

has no solution.

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18For what value of

\lambda

is the function defined by

19Show 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

20Is the function defined by

f(x) = x^2 - \sin x + 5

continuous at

x = \pi

21(a)

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

21(b)

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

21(c)

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

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

23Find all points of discontinuity of

f

, where

24Determine if

f

defined by

25Examine the continuity of

f

, where

f

is defined by

f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases} \quad \text{at } x = \frac{\pi}{2}

f(x) = \begin{cases} kx^2, & \text{if } x \leq 2 \\ 3, & \text{if } x > 2 \end{cases} \quad \text{at } x = 2

f(x) = \begin{cases} kx + 1, & \text{if } x \leq \pi \\ \cos x, & \text{if } x > \pi \end{cases} \quad \text{at } x = \pi

f(x) = \begin{cases} kx + 1, & \text{if } x \leq 5 \\ 3x - 5, & \text{if } x > 5 \end{cases} \quad \text{at } x = 5

30Find the values of

a

and

b

such that the function defined by

31Show that the function defined by

f(x) = \cos(x^2)

is a continuous function.

32Show that the function defined by

f(x) = |\cos x|

is a continuous function.

33Examine that

\sin|x|

is a continuous function.

34Find all the points of discontinuity of

f

defined by

f(x) = |x| - |x + 1|

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