Let $f\left( x \right)$ be a function differentiable at $x = c$. Then, $\mathop {\lim }\limits_{x \to c} f\left( x \right)$ equals

$\frac{1}{{f\left( c \right)}}$

Let $f: R \rightarrow R$ be any function. Define $g: R \rightarrow R$ by $g(x)=|f(x)|$ for $x \rightarrow R$. Then, $g$ is

differentiable if $f$ is differentiable

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JEE • MATHEMATICS • CONTINUITY

Continuity practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 114 questions

1 Easy

1 of 114

Application +4 | -1

f:R \to R

is defined by

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\cos 3x - \cos x}}{{{x^2}}},\mathrm{for}\,\,x \ne 0}\\ {\lambda ,\mathrm{for}\,\,x = 0} \end{array}} \right.

and if

f

is continuous at

x = 0

, then

\lambda

is equal to

View answer and explanation

Correct option: B

- 4

Option B (-4) is correct because the limit of the function as x approaches 0 is found using L'Hopital's rule, and for continuity, this limit must equal lambda.

2 Easy

2 of 114

Application +4 | -1

Let

f

be a continuous function satisfying

f(x) f(y)=

f(x)+f(y)+f(x y)-2

for all

x, y \in {R}

and

f(2)=5

then

\lim \limits_{{x \rightarrow 4}} f(x)

View answer and explanation

Correct option: B

Option B (17) is correct because

f(4) = 17

f(2)=5

and using the given functional equation

(f(2))^2=2 f(2)+f(4)-2

3 Easy

3 of 114

Application +4 | -1

\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}\frac{x^2+3 x-10}{x^2+2 x-15} \text {, when } x \neq-5 \\ a, \quad \text { when } x=-5\end{array}\right.

is continuous at

x=-5

, then the value of 'a' will be

View answer and explanation

Correct option: B

\frac{7}{8}

Option B (

\frac{7}{8}

) is correct because it is the limit of the function as x approaches -5, ensuring continuity at that point.

4 Easy

4 of 114

Understanding +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\sin \frac{1}{x},x \ne 0}\\ {k,x = 0} \end{array}} \right.

is continuous at

x = 0

, then the value of

k

View answer and explanation

Correct option: C

Option C (0) is correct because for the function to be continuous at x=0, f(0) must equal the limit as x approaches 0, which is 0.

5 Easy

5 of 114

Application +4 | -1

The value of

\mathrm{k}

for which the function

f(x)=\left\{\begin{array}{ll}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}, & 0 < x < \frac{\pi}{2} \\ k+\frac{2}{5} & , x=\frac{\pi}{2}\end{array}\right.

is continuous at

x=\frac{\pi}{2}

, is :

View answer and explanation

Correct option: C

\frac{3}{5}

Option C (

\frac{3}{5}

) is correct because the function is continuous at

x=\frac{\pi}{2}

\lim _{x \rightarrow \pi / 2} f(x)=f(\pi / 2)

, which leads to

k = \frac{3}{5}

6 Easy

6 of 114

Understanding +4 | -1

f(x)=\left\{\begin{array}{c}\frac{\sin x}{x}+\cos x \text {, when } x \neq 0 \\ 2, \text { when } x=0\end{array}\right.

then

View answer and explanation

Correct option: C

f(x)

is continuous at

x=0

Option C (

f(x)

is continuous at

x=0

) is correct because the limit of the function as x approaches 0 from both sides equals the function's value at x=0.

7 Easy

7 of 114

Application +4 | -1

If the function

f(x)=\left\{\begin{array}{ll}1+\sin \frac{\pi}{2} x, & \text { for }-\infty < x \leq 1 \\ a x+b & \text { for } 1 < x < 3 \\ 6 \tan \frac{\pi x}{12} & \text { for } 3 \leq x < 6\end{array}\right.

is continuous in the interval

(-\infty, 6)

, then the values of

a

and

b

are

View answer and explanation

Correct option: C

(2,0)

Option C ((2,0)) is correct because solving the equations derived from the continuity conditions at x=1 and x=3 yields a=2 and b=0.

8 Easy

8 of 114

Application +4 | -1

f(x)=\left\{\begin{array}{ll}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} & \text { for }-1 \leq x<0 \\ 2 x^2+3 x-2, & \text { for } 0 \leq x \leq 1\end{array}\right.

is continuous at

\mathrm{x}=0

, then

\mathrm{k}=

View answer and explanation

Correct option: C

-2

Option C (-2) is correct because the left-hand limit and right-hand limit must be equal for the function to be continuous at x=0.

9 Easy

9 of 114

Understanding +4 | -1

The value of

k

for which the function

f(x)= \begin{cases}\sin \frac{1}{x}, & x \neq 0 \\ k, & x=0\end{cases}

is continuous is

View answer and explanation

Correct option: D

none of these

Option D (none of these) is correct because the limit of

\sin(1/x)

x

approaches 0 does not exist, making the function discontinuous at

x=0

regardless of the value of

k

10 Easy

10 of 114

Understanding +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{1 - \cos x}}{x},x \ne 0}\\ {k,x = 0} \end{array}} \right.

is continuous at

x = 0

, then the value of

k

View answer and explanation

Correct option: A

The correct answer is A (0) because for the function to be continuous at

x=0

, the limit as

x

approaches 0 must exist and be equal to the function's value at

x=0

11 Easy

11 of 114

Application +4 | -1

Let

f(x)

and

g(x)

be continuous functions such that

f o g(x)=x

. If

g^{\prime}(b)=5

and

g(b)=a

, then

f^{\prime}(a)

is equal to

View answer and explanation

Correct option: B

\frac{1}{5}

Option B (

\frac{{1}}{{5}}

) is correct because it is found by differentiating the composite function

f(g(x)) = x

and substituting the given values.

12 Easy

12 of 114

Application +4 | -1

f(x)=\left\{\begin{array}{ll}x+a, & x \leq 0 \\ |x-4|, & x>0\end{array}\right.

and

g(x)=\left\{\begin{array}{ll}x+1 & x<0 \\ (x-4)^2+b, & x \geq 0\end{array}\right\}

are continuous on

{R}

, then (gof)

(2)+(fog)(-2)

is equal to

View answer and explanation

Correct option: D

-8

Option D (-8) is correct because it is the result of evaluating (gof)(2) + (fog)(-2) after determining the values of a and b that ensure continuity.

13 Easy

13 of 114

Application +4 | -1

Let a function

f\mathrm{ }:\mathrm{ }R \to R

be defined as

f\left( x \right) = \mathrm{ }\left\{ {\begin{array}{cccccccccccccccccccc} {\sin x\mathrm{ } - \mathrm{ }{e^x},}&{\mathrm{if}}&{x \le 0}\\ {a + \left[ { - x} \right],}&{\mathrm{if}}&{0 < x < 1}\\ {2x - b,}&{\mathrm{if}}&{x \ge 1} \end{array}} \right.

Where, [x] is the greatest integer less than or equal to x. if f is continuous on R, then (a +b) is equal to

View answer and explanation

Correct option: B

Option B (3) is correct because for the function to be continuous, the limits at the transition points must be equal, leading to a + b = 3.

14 Easy

14 of 114

Analyzing +4 | -1

Let

f:R \to R

be a function defined as

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sin (a + 1)x + \sin 2x}}{{2x}},}&{\mathrm{if}}&{x < 0}\\ {b,}&{\mathrm{if}}&{x < 0}\\ {\frac{{\sqrt {x + b{x^3}} - \sqrt x }}{{b{x^{5/2}}}},}&{\mathrm{if}}&{x > 0} \end{array}} \right.\,

f

is continuous at

x = 0

then the value of

a+b

is equal to

View answer and explanation

Correct option: D

- \frac{3}{2}

The correct answer is D (

- \frac{{3}}{{2}}

) because the continuity condition at x=0 requires the left-hand limit, right-hand limit, and function value to be equal, leading to a+b = -3/2.

15 Easy

15 of 114

Analyzing +4 | -1

Assertion(A): The function

f(x)=\cos \left[\frac{\pi}{x}\right] \cdot \cos \left(\frac{\pi}{2}(2 x-1)\right),[

.] denotes GIF is continuous in

\mathbb{Z}-\{0\}

.

Reason (R) : For

n \in \mathbb{Z}, \cos \frac{(2 n-1) \pi}{2}=0

Which of the following is true?

View answer and explanation

Correct option: A

Both

A, R

are true and

R

is the correct reason of

A

Option A is correct because the function is continuous due to the cosine term becoming zero at integer values, and the reason correctly explains why that cosine term is zero.

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16 Easy

16 of 114

Analyzing +4 | -1

If the function

f(x)=\left\{\begin{array}{cc} \frac{\log _e\left(1-x+x^2\right)+\log _e\left(1+x+x^2\right)}{\sec x-\cos x}, & x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)-\{0\} \\ k & x=0 \end{array}\right.

is continuous at

x=0

, then

k

is equal to

View answer and explanation

Correct option: A

1

Option A (1) is correct because by applying L'Hopital's rule and logarithmic properties, the limit of the function as x approaches 0 is found to be 1, which equals k for continuity.

17 Easy

17 of 114

Application +4 | -1

Let

f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right)

is continuous in

\left[0, \frac{\pi}{2}\right)

then

f\left(\frac{\pi}{4}\right)

, is

View answer and explanation

Correct option: C

-\frac{1}{2}

Option C (-

\frac{1}{2}

) is correct because it is the limit of the given function as x approaches

\frac{\pi}{4}

, ensuring continuity.

18 Easy

18 of 114

Analyzing +4 | -1

The value of

f\left( 0 \right)

, so that the function

f\left( x \right) = \frac{{{{\left( {27 - 2x} \right)}^{1/3}} - 3}}{{9 - 3{{\left( {243 + 5x} \right)}^{1/5}}}}\left( {x \ne 0} \right)

is continuous is given by

View answer and explanation

Correct option: C

Option C (2) is correct because it is the limit of the function as x approaches 0, calculated using L'Hopital's rule to ensure continuity.

19 Easy

19 of 114

Application +4 | -1

The value of

k

for which the function

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{1 - \cos 4x}}{{8{x^2}}},x \ne 0}\\ {kx = 0} \end{array}} \right.

is continuous at

a = 0

, is

View answer and explanation

Correct option: B

k = 1

Option B (k = 1) is correct because the limit of the function as x approaches 0 must equal k for the function to be continuous.

20 Easy

20 of 114

Understanding +4 | -1

Let

f\left( x \right)

be a function differentiable at

x = c

. Then,

\lim\limits_{x \to c} f\left( x \right)

equals

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because differentiability at a point implies continuity at that point, so

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

21 Easy

21 of 114

Understanding +4 | -1

Let

f: R \rightarrow R

be any function. Define

g: R \rightarrow R

g(x)=|f(x)|

for

x \rightarrow R

. Then,

g

View answer and explanation

Correct option: C

continuous if

f

is continuous

Option C (continuous if

f

is continuous) is correct because the absolute value function is continuous, and the composition of continuous functions is continuous.

22 Easy

22 of 114

Analyzing +4 | -1

The function

f\left( x \right) = \frac{{1 - \sin x + \cos x}}{{1 + \sin x + \cos x}}

is not defined at

x = \pi

. The value of

f\left( \pi \right)

, so that

f\left( x \right)

is continuous at

x = \pi

, is

View answer and explanation

Correct option: C

- 1

Option C (

-1

) is correct because the limit of the function as

x

approaches

\pi

-1

, and for the function to be continuous,

f(\pi)

must equal this limit.

23 Easy

23 of 114

Analyzing +4 | -1

f: R \rightarrow R

is a continuous function satisfying

f(0)=1

and

f(2 x)-f(x)=x

for all

x \in R

and

\lim\limits _{x \rightarrow \infty}\left\{f(x)-f\left(\frac{x}{2^n}\right)\right\}=P(x)

, then

P(x)

, is

View answer and explanation

Correct option: B

a linear polynomial in

x

Option B (a linear polynomial in

x

) is correct because after simplification and taking the limit,

P(x)

results in a linear function of

x

24 Easy

24 of 114

Application +4 | -1

f(x)=\left\{\begin{array}{cc}a x+1, & x \leq \pi / 2 \\ \sin x+b, & x>\pi / 2\end{array}\right.

is continuous then

View answer and explanation

Correct option: C

b=a \frac{\pi}{2}

Option C (

b=a \frac{\pi}{2}

) is correct because for

f

to be continuous,

a \frac{\pi}{2}+1 = \sin(\frac{\pi}{2})+b

, so

a \frac{\pi}{2} = b

25 Easy

25 of 114

Application +4 | -1

The values of

x

for which the tangents to the curves

y=x \cos x, y=(\sin x) / x

are parallel to the axis of

x

are roots of (respectively)

View answer and explanation

Correct option: C

\cot x=x, \tan x=x

Option C (

\cot x=x, \tan x=x

) is correct because these equations represent the conditions where the derivatives of the given functions are equal to zero, indicating a tangent parallel to the x-axis.

26 Easy

26 of 114

Understanding +4 | -1

Statement 1: A function

f: R \rightarrow R

is continuous at

x_{0}

if and only if

\lim _{x \rightarrow x_{0}} f(x)

exists and

\lim _{x \rightarrow x_{0}} f(x)=f\left(x_{0}\right)

Statement 2: A function

f: R \rightarrow R

is discontinuous at

x_{0}

if and only if,

\lim _{x \rightarrow x_{0}} f(x)

exists and

\lim _{x \rightarrow x_{0}} f(x) \neq f\left(x_{0}\right)

View answer and explanation

Correct option: D

Statement 1 is true, Statement 2 is false.

Option D is correct. Statement 1 is the definition of continuity, while Statement 2 is incorrect because discontinuity can also occur when the limit does not exist.

27 Easy

27 of 114

Application +4 | -1

f: R \rightarrow R

is defined by

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\cos 3x - \cos x}}{{{x^2}}},}&{x \ne 0}\\ {\lambda ,}&{x = 0} \end{array}} \right.

and if

f

is continuous at

x=0

, then

\lambda

is equal to

View answer and explanation

Correct option: B

-4

Option B (-4) is correct because it is the limit of the given function as x approaches 0, ensuring continuity.

28 Easy

28 of 114

Application +4 | -1

f(x)

is a continuous function satisfying

f(x) f(1 / x)=f(x)+f(1 / x)

and

f(1)>0

then

\lim \limits_{{x \rightarrow 1}} f(x)

is equal to

View answer and explanation

Correct option: A

Option A (2) is correct because the limit can be found by using the continuity of f(x) and solving the equation f(1)f(1/1) = f(1) + f(1/1).

29 Easy

29 of 114

Understanding +4 | -1

The function

f(x)=\cot x

is discontinuous on the set

View answer and explanation

Correct option: A

\{x=n \pi, n \in Z\}

Option A ({x=nπ, n ∈ Z}) is correct because cot(x) = cos(x)/sin(x), and sin(x) = 0 when x = nπ, where n is an integer.

30 Easy

30 of 114

Analyzing +4 | -1

The function

f(x) = \lim\limits_{n \to \infty } \frac{{\cos (2\pi x) - {x^{2n}}\sin (x - 1)}}{{1 + {x^{2n + 1}} - {x^{2n}}}}

is continuous for all

x

View answer and explanation

Correct option: B

\mathbb{R}-\{-1,1\}

Option B (

\mathbb{R}-\{-1,1\}

) is correct because the function is discontinuous at

x = 1

and

x = -1

, making it continuous for all real numbers except -1 and 1.

Revision notes and key takeaways

Review CONTINUITY in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 114 single-correct-answer MCQs on CONTINUITY.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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