If $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {3{x^2} + 12\;x - 1,\; - 1 \le x \le 2}\\ {37 - x,\;\;\;2 < x \le 3} \end{array}} \right.$, then

$f\left( x \right)$ is increasing in [-1,2]

$f\left( x \right)$ is continuous in [-1,3]

$f\left( x \right)$ is maximum at x=2

Define $f$ on $R$ into itself by $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {x\sin \frac{1}{x},{\rm{when}}\,\,\,x \ne 0}\\ {0,{\rm{when}}\,\,\,\,x = 0} \end{array}} \right.$, then

$f$ is both continuous and differentiable at 0

$f$ is differentiable but not continuous at 0

Let $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\mathop \smallint \limits_0^x \left\{ {5 + \left| {1 - t} \right|} \right\}dt,{\rm{if}}\,\,x > 2}\\ {5x + 1,{\rm{if}}\,\,x \le 2} \end{array}} \right.$

$f\left( x \right)$ is continuous at $x = 2$

$f\left( x \right)$ is everywhere differentiable

The right derivative of $f\left( x \right)at\,\,{\rm{ }}x = 2$ does not exist

At the point $x = 1$, the function $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {{x^3} - 1,1 < x < \infty }\\ {x - 1, - \infty < x \le 1} \end{array}} \right.$

Discontinuous and differentiable

Discontinuous and not differentiable

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JEE • MATHEMATICS • CONTINUITY

Continuity practice · Medium

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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 71 questions

1 Medium

1 of 71

Analyzing +4 | -1

If the function

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{{\left\{ {1 + \left| {\sin x} \right|} \right\}}^{\frac{a}{{\left| {\sin x} \right|}}}}, - \frac{\pi }{6} < x < 0}\\ {b,x = 0}\\ {{e^{\frac{{\tan 2x}}{{\tan 3x}}}},0 < x < \frac{\pi }{6}} \end{array}} \right.

Is continuous at

x = 0

View answer and explanation

Correct option: A

a = {\log _e}b,b = \frac{2}{3}

Option A (

a = \log_e b, b = \frac{2}{3}

) is correct because it satisfies the continuity condition at x=0 by equating the left-hand limit, right-hand limit, and function value.

2 Medium

2 of 71

Analyzing +4 | -1

f(x)=\left\{\begin{array}{rr}1+\cos ^{-1}\{\cot x\}, & x<\frac{\pi}{2} \\ \pi[x]+1, & x \geq \frac{\pi}{2}\end{array}\right.

Where [] denotes greatest integer and

\{ \}

denotes fractional part function, Then value of jump of discontinuity is:

View answer and explanation

Correct option: A

\frac{\pi}{2}

Option A (

\frac{\pi}{2}

) is correct because it represents the difference between the right-hand limit and the left-hand limit at the point of discontinuity.

3 Medium

3 of 71

Analyzing +4 | -1

Let

\left[ x \right]

denotes the greatest integer less than or equal to

x

and

f\left( x \right) = \left[ {{{\tan }^2}x} \right]

. Then,

View answer and explanation

Correct option: B

f\left( x \right)

is continuous at

x = 0

Option B (

f(x)

is continuous at

x = 0

) is correct because

f(x)

is a constant function in a neighborhood of

x=0

, making it continuous.

4 Medium

4 of 71

Application +4 | -1

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

and

f\left( x \right)

is continuous at

x = 0

, then the value of

k

View answer and explanation

Correct option: B

a + b

Option B (

a + b

) is correct because the limit of the given function as x approaches 0 must equal k for the function to be continuous at x=0.

5 Medium

5 of 71

Analyzing +4 | -1

If the function

f(x)=\left\{\begin{array}{ll} \frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}}, & x \neq \pi \\ k \quad, x=\pi \end{array}\right.

is continuous at

x=\pi

, then

\mathrm{k}

equals:

View answer and explanation

Correct option: D

\frac{1}{4}

Option D (1/4) is correct because the limit of the given function as x approaches pi must equal k for continuity, and this limit evaluates to 1/4.

6 Medium

6 of 71

Analyzing +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {3{x^2} + 12\;x - 1,\; - 1 \le x \le 2}\\ {37 - x,\;\;\;2 < x \le 3} \end{array}} \right.

, then

View answer and explanation

Correct option: D

All the above

Option D (All the above) is correct because the function is continuous, increasing in the specified interval, and has a maximum at x=2.

7 Medium

7 of 71

Understanding +4 | -1

If a function

f

is differentiable from left and also from right at

a

, then

f

View answer and explanation

Correct option: A

Continuous at

a

Option A (Continuous at

a

) is correct because if a function is differentiable from both left and right at a point, it must be continuous at that point.

8 Medium

8 of 71

Analyzing +4 | -1

Let

f\left( x \right) = \left[ {{x^3} - x} \right]

, where

\left[ x \right]

the greatest integer function is. Then the number of points in the interval

\left( {1,2} \right)

, where function is discontinuous is

View answer and explanation

Correct option: C

Option C (6) is correct because the greatest integer function will have discontinuities at integer values of x³-x. In the interval (1,2), x³-x takes on integer values 6 times.

9 Medium

9 of 71

Analyzing +4 | -1

Define

f

R

into itself by

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

, then

View answer and explanation

Correct option: A

f

is continuous at 0 but not differentiable at 0

Option A (

f

is continuous at 0 but not differentiable at 0) is correct because the limit of

f(x)

x

approaches 0 exists and equals

f(0)

, but the limit of the difference quotient does not exist at

x = 0

10 Medium

10 of 71

Analyzing +4 | -1

Let

f(x)= \begin{cases}x\left(1+\frac{1}{3} \sin \left(\log x^2\right)\right) & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases}

Then

View answer and explanation

Correct option: D

f

is continuous at

x=0

, but not differentiable at

x=0

Option D (

f

is continuous at

x=0

, but not differentiable at

x=0

) is correct because the limit of the function as x approaches 0 is 0, making it continuous, but the derivative does not exist at x=0.

11 Medium

11 of 71

Application +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{{2^x} - 1}}{{\sqrt {1 + x} - 1}}, - 1 \le x < \infty ,x \ne 0}\\ {k,x = 0} \end{array}} \right.

is continuous everywhere, then

k

is equal to

View answer and explanation

Correct option: B

{\log _e}4

Option B (

{\log _e}4

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

12 Medium

12 of 71

Analyzing +4 | -1

The function

f(x)= \begin{cases}\frac{1-\cos 4 x}{x^2} & \text { if } x < 0 \\ \frac{2 \sqrt{x}}{\sqrt{16+\sqrt{x}}-4} & \text { if } x=0 \\ \text { if } x > 0\end{cases}

is continuous at

x=0

for

View answer and explanation

Correct option: D

no value of

A

Option D (no value of

A

) is correct because the left-hand limit and right-hand limit at x=0 are not equal, so the function cannot be continuous at x=0 for any value of A.

13 Medium

13 of 71

Application +4 | -1

f(x)= \begin{cases}\cos ^{-1}(\cot x), & x < \frac{\pi}{2} \\ a(\pi[x]-1), & x \geq \frac{\pi}{2}\end{cases}

The value of

a

for

f

to be continuous at

x=\frac{\pi}{2}

View answer and explanation

Correct option: A

\frac{\pi}{2(\pi-1)}

Option A (

\frac{\pi}{2(\pi-1)}

) is correct because for f to be continuous at x=

\frac{\pi}{2}

, the left and right limits must be equal, which leads to solving for a.

14 Medium

14 of 71

Analyzing +4 | -1

Given a function

f(x)

continuous

\forall \, x \in R

such that

\lim \limits_{{x \rightarrow 0}}\left[f(x)+\log \left(1-\frac{1}{e^{f(x)}}\right)-\log (f(x))\right]=0

, then

f(0)

View answer and explanation

Correct option: A

Option A (0) is correct because by applying the limit properties and simplifying the expression, we find that

\lim_{x \to 0} f(x) = 0

, and since

f(x)

is continuous,

f(0) = 0

15 Medium

15 of 71

Understanding +4 | -1

The number of points at which the function

f\left( x \right) = \left( {\left| {x - 1} \right| + \left| {x - 2} \right| + \cos x} \right)

where

x \in \left[ {0,4} \right]

is not continuous, is

View answer and explanation

Correct option: D

The correct answer is D (0) because absolute value functions and cosine are continuous everywhere, and the sum of continuous functions is continuous.

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16 Medium

16 of 71

Analyzing +4 | -1

Let

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

then

f(x)

is continuous but not differentiable at

x = 0

View answer and explanation

Correct option: A

0 < P \leqslant 1

Option A is correct because for 0 < P <= 1, the function is continuous but not differentiable at x = 0.

17 Medium

17 of 71

Analyzing +4 | -1

f:R \to R

is defined by

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{2\sin x - \sin 2x}}{{2x\cos x}}}\\ {a,\mathrm{if}\,\,x = 0} \end{array}} \right.,\mathrm{if}\,\,x \ne 0,

Then the value of a so that

f

is continuous at 0 is

View answer and explanation

Correct option: D

Option D (0) is correct because the limit of the function as x approaches 0 is 0, which must equal f(0) for continuity.

18 Medium

18 of 71

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\int \limits_0^x \left\{ {5 + \left| {1 - t} \right|} \right\}dt,\mathrm{if}\,\,x > 2}\\ {5x + 1,\mathrm{if}\,\,x \le 2} \end{array}} \right.

View answer and explanation

Correct option: B

f\left( x \right)

is continuous but not differentiable at

x = 2

Option B is correct because

f(x)

is continuous at

x=2

but the left-hand derivative and right-hand derivative are not equal, making it non-differentiable.

19 Medium

19 of 71

Application +4 | -1

If the function

f(x)=\left\{\begin{array}{cc} \frac{x^2-2 x+a}{\sin x} & \text { when } x \neq 0 \\ b & \text { when } x=0 \end{array}\right.

is continuous at

x=0

, then

View answer and explanation

Correct option: A

a=0, b=-2

Option A (

a=0, b=-2

) is correct because for the function to be continuous at

x=0

, the limit as

x

approaches 0 must exist and be equal to

f(0)

, which leads to these values.

20 Medium

20 of 71

Analyzing +4 | -1

Let

f = \left( { - \frac{\pi }{4},\,\,\frac{\pi }{4}} \right) \to R

be defined as

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {{{(1 + \left| {\sin \mathrm{ }x} \right|)}^{\frac{{3a}}{{\left| {\sin \mathrm{ }x} \right|}}}},}&{ - \frac{\pi }{4} < x < 0}\\ {b,}&{x = 0}\\ {{e^{\frac{{\cot \mathrm{ }4x}}{{\cot \mathrm{ }2x}}}},}&{0 < x < \frac{\pi }{4}} \end{array}} \right.

f

is continuous at

x = 0

, then the value of

6\,a + {b^2}

is equal to

View answer and explanation

Correct option: C

1 + e

Option C (

1 + e

) is correct because it is the result of evaluating the limits and applying the continuity condition at x=0.

21 Medium

21 of 71

Analyzing +4 | -1

f(x)

is a differentiable function such that

f: R \rightarrow r

and

f\left(\frac{1}{n}\right)=0 \forall n \geq 1, n \in l

then

View answer and explanation

Correct option: B

f(x)=0 \forall x \in(0,1)

Option B is correct because if f(1/n) = 0 for all n >= 1, then f(x) = 0 for all x in (0, 1) due to the differentiability of f(x).

22 Medium

22 of 71

Application +4 | -1

f\left( x \right) = {\left| x \right|^3}

, then

f'\left( 0 \right)

equals

View answer and explanation

Correct option: A

Option A (0) is correct because the derivative of |x|^3 at x=0, calculated using the limit definition, is 0.

23 Medium

23 of 71

Application +4 | -1

f:R \to R

given by

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {2\cos x,\mathrm{if}\,\,x \le - \frac{\pi }{2}}\\ {a + \sin x + b,\mathrm{if}\,\, - \frac{\pi }{2} < x < \frac{\pi }{2}}\\ {1 + {{\cos }^2}x,\mathrm{if}\,\,x \ge \frac{\pi }{2}} \end{array}} \right.

is a continuous Function on

R

, then

\left( {a,b} \right)

is equal to

View answer and explanation

Correct option: A

\left( {\frac{1}{2},\frac{1}{2}} \right)

Option A (

\left( {\frac{1}{2},\frac{1}{2}} \right)

) is correct because for the function to be continuous, the left-hand limit and right-hand limit at

x = -\frac{\pi}{2}

must be equal, and only this option satisfies the condition a = b.

24 Medium

24 of 71

Analyzing +4 | -1

At the point

x = 1

, the function

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^3} - 1,1 < x < \infty }\\ {x - 1, - \infty < x \le 1} \end{array}} \right.

View answer and explanation

Correct option: B

Continuous and not differentiable

Option B (Continuous and not differentiable) is correct because the function is continuous at

x=1

but the left and right derivatives are not equal.

25 Medium

25 of 71

Analyzing +4 | -1

Let

f(x)=\lim\limits _{n \rightarrow \infty} \frac{(2 \sin x)^{2 n}}{3^n-(2 \cos x)^{2 n}}, n \in

Z. Then,

View answer and explanation

Correct option: D

all the above

Option D (all the above) is correct because the function is discontinuous at

x=n \pi \pm \frac{\pi}{6}

f(\frac{\pi}{3})=1

, and

f(0)=0

, all of which can be verified by evaluating the limit.

26 Medium

26 of 71

Analyzing +4 | -1

Let

f(x)=\left\{\begin{array}{cc}x^{2} \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{array}\right.

Then at

x=0

View answer and explanation

Correct option: D

f

is continuous but

f^{\prime}

is not continuous

Option D is correct because the function f(x) is continuous at x=0, but its derivative f'(x) is not continuous at x=0 due to the oscillatory behavior of the cosine term.

27 Medium

27 of 71

Analyzing +4 | -1

f\left( x \right)

defined by

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

then

f\left( x \right)

is continuous for all

View answer and explanation

Correct option: D

x

except at

x = 0

and

x = 1

Option D (

x

except at

x = 0

and

x = 1

) is correct because the function is discontinuous at these two points.

28 Medium

28 of 71

Analyzing +4 | -1

If the function

f(x)=\left\{\begin{array}{cc} (1+|\cos x|) \frac{\lambda}{|\cos x|} , & 0 < x < \frac{\pi}{2} \\ \mu, & x=\frac{\pi}{2} \\ \frac{\cot 6 x}{e^{\cot 4 x}}, & \frac{\pi}{2} < x < \pi \end{array}\right.

Is continuous at

x=\frac{\pi}{2}

, then

9 \lambda+6 \log _{e} \mu+\mu^{6}-e^{6 \lambda}

is equal to

View answer and explanation

Correct option: C

Option C (10) is correct because by evaluating the limits and using the continuity condition, we find λ and μ, and then substitute these values into the given expression.

29 Medium

29 of 71

Analyzing +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1\,\,\mathrm{for}{\mkern 1mu} {\mkern 1mu} x < 0}\\ {1 + \sin x{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathrm{for}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0 \le x \le \pi /2} \end{array}} \right.

, then at

x = 0

, the derivative

f'\left( x \right)

View answer and explanation

Correct option: D

Does not exist

Option D (Does not exist) is correct because the left and right derivatives are not equal at

x=0

30 Medium

30 of 71

Analyzing +4 | -1

Let

f(x)=\left[x^2-x\right]+|-x+[x]|

, where

x \in \mathbb{R}

and

[t]

denotes the greatest integer less than or equal to

t

. Then,

f

View answer and explanation

Correct option: B

continuous at

x=1

, but not continuous at

x=0

Option B is correct. The function is continuous at x=1 because the left-hand limit, right-hand limit, and the function value are equal at x=1, but it is discontinuous at x=0.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 71 single-correct-answer MCQs on CONTINUITY.
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