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

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JEE • MATHEMATICS • CONTINUITY

Continuity practice · Mixed difficulty

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Showing 1-30 of 192 questions

1 Medium

1 of 192

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 192

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 Easy

3 of 192

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.

4 Easy

4 of 192

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

5 Easy

5 of 192

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.

6 Easy

6 of 192

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.

7 Easy

7 of 192

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}

8 Easy

8 of 192

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.

9 Medium

9 of 192

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.

10 Medium

10 of 192

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.

11 Easy

11 of 192

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.

12 Easy

12 of 192

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.

13 Hard

13 of 192

Application +4 | -1

f\left( x \right) = \frac{{{{\log }_\mathrm{e}}\left( {1 + {x^2}\tan x} \right)}}{{\sin {x^3}}},x \ne 0

, is to be continuous at

x = 0

, then

f\left( 0 \right)

must be defined as

View answer and explanation

Correct option: A

Option A (1) is correct because for the function to be continuous at x=0, f(0) must equal the limit of the function as x approaches 0, which is 1.

14 Easy

14 of 192

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

15 Medium

15 of 192

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.

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16 Easy

16 of 192

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

17 Easy

17 of 192

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.

18 Easy

18 of 192

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.

19 Medium

19 of 192

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.

20 Easy

20 of 192

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.

21 Medium

21 of 192

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.

22 Easy

22 of 192

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.

23 Medium

23 of 192

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.

24 Easy

24 of 192

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.

25 Easy

25 of 192

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.

26 Easy

26 of 192

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.

27 Medium

27 of 192

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

28 Medium

28 of 192

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.

29 Easy

29 of 192

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.

30 Medium

30 of 192

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.

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Key Takeaways

✓ This set contains 192 single-correct-answer MCQs on CONTINUITY.
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