If $f(x)=x(\sqrt{x}+\sqrt{x+1})$, then

$f(x)$ is not continuous at $x=0$

$f(x)$ is continuous at $x=0$, but not differentiable at $x=0$

Let $f:\mathbb{R} \to \mathbb{R}$ be defined as $f\left( x \right) = \left[ {\begin{array}{*{20}{c}} {\left[ {{e^x}} \right],}&{x < 0} \\ {\begin{array}{*{20}{c}} {a{e^x} + \left[ {x - 1} \right],} \\ {b + \left[ {\sin \left( {\pi x} \right)} \right],} \end{array}}&{\begin{array}{*{20}{c}} {0 \leqslant x < 1} \\ {1 \leqslant x < 2} \end{array}} \\ {\left[ {{e^{ - x}}} \right] - c,}&{x \geqslant 2} \end{array}} \right.$ Where $a,b,c \in \mathbb{R}$ and $\left[ t \right]$ denotes greatest integer less than or equal to $t.$ Then, which of the following statements is true?

There exists $a,b,c \in \mathbb{R}$ such that $f$ is continuous on $\mathbb{R}.$

$f$ is discontinuous at atleast two points, for any values of $a,b\,\,{\text{and}}\,\,c$

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JEE • MATHEMATICS • CONTINUITY

Continuity practice · Mixed difficulty

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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 151-180 of 192 questions

151 Easy

151 of 192

Analyzing +4 | -1

\alpha, \beta(\alpha<\beta)

are the points of discontinuity of the function

f(f(f(x)))

, where

f(x)=\frac{1}{1-x}

, then the set of values of

a

for which the points

(\alpha, \beta)

and

\left(a, a^2\right)

lie on the same side of the line

x+2 y-3=0

, is

View answer and explanation

Correct option: A

(-3 / 2,1)

The correct answer is A ((-3 / 2,1)) because the points (α, β) and (a, a²) lie on the same side of the line x+2y-3=0, leading to the inequality (2a + 3)(a - 1) < 0.

152 Easy

152 of 192

Application +4 | -1

The value of

f(1)

so that

f(x)=\left(e^x-e\right) /(x-1)

is continuous at

x=1

View answer and explanation

Correct option: D

e

Option D (

e

) is correct because

f(x) = \frac{e(e^{x-1}-1)}{x-1}

, so

f(1) = e

153 Easy

153 of 192

Application +4 | -1

The value of

k

so that the function

f(x)=\left\{\begin{array}{cc}k(2 x-x)^2, & \text { when } x<0 \\ \cos x, & \text { when } x \geq 0\end{array}\right.

is continuous at

x=0

. Is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the left-hand limit at x=0 does not equal the right-hand limit, regardless of the value of k.

154 Easy

154 of 192

Understanding +4 | -1

f\left( x \right)

is continuous function and

g\left( x \right)

be discontinuous, then

View answer and explanation

Correct option: B

f\left( x \right) + g\left( x \right)

must be discontinuous

The correct answer is B (

f\left( x \right) + g\left( x \right)

must be discontinuous) because the sum of a continuous function and a discontinuous function is always discontinuous.

155 Hard

155 of 192

Analyzing +4 | -1

f(x)=x+\left[x+\frac{1}{3}\right]+\left[x+\frac{2}{3}\right] \forall x \in R

then the number of points of discontinuity of

f(x)

[-1,1]

, where [.] denotes greatest integer function.

View answer and explanation

Correct option: D

The correct answer is D (6) because the function f(x) has discontinuities at x = -2/3, -1/3, 0, 1/3, 2/3, and 1 within the interval [-1, 1].

156 Easy

156 of 192

Application +4 | -1

Let the function

f(x)=\left\{\begin{array}{cl}\frac{\log _e(1+5 x)-\log _e(1+\alpha x)}{x} & \text { if } x \in 0 \\ 10 & \text {; if } x=0\end{array}\right.

be continuous at

x=0

. Then

\alpha

is equal to

View answer and explanation

Correct option: D

-5

Option B (-5) is correct because for the function to be continuous at x=0, the limit as x approaches 0 must equal f(0), leading to α = -5.

157 Easy

157 of 192

Analyzing +4 | -1

Let

f:R \to R

be defined as

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\lambda \left| {{x^2} - 5x + 6} \right|}}{{\mu (5x - {x^2} - 6)}}}&,&{x < 2}\\ {{e^{\frac{{\tan (x - 2)}}{{x - \left[ x \right]}}\,}}}&,&{x > 2}\\ \mu &,&{x = 2}\\ \,&\,&\, \end{array}} \right.

Where,

\left[ x \right]

is the greatest integer less than or equal to

x

. If

f

is continuous at

x = 2,

then

\lambda + \mu

is equal to

View answer and explanation

Correct option: A

e\,( - e + 1)

Option A (

e(-e+1)

) is correct because, for continuity at x=2, the left-hand limit, right-hand limit, and the function's value at x=2 must be equal, leading to

\lambda = -e^2

and

\mu = e

158 Medium

158 of 192

Analyzing +4 | -1

f(x)=x(\sqrt{x}+\sqrt{x+1})

, then

View answer and explanation

Correct option: C

f(x)

is differentiable at

x=0

Option C,

f(x)

is differentiable at

x=0

, is correct because the limit of the difference quotient exists at x=0, implying differentiability.

159 Medium

159 of 192

Understanding +4 | -1

The function

f\left( x \right) = \frac{{2{x^2} + 7}}{{{x^3} + 3{x^2} - x - 3}}

is discontinuous for

View answer and explanation

Correct option: C

x = 1,x = - 1,x = - 3

only

Option C (

x = 1,x = -1,x = -3

only) is correct because these are the values that make the denominator zero, leading to discontinuity.

160 Easy

160 of 192

Analyzing +4 | -1

Let

f:R \to R

be defined as

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{{x^3}}}{{{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {\frac{{1 + 2x{e^{ - 2x}}}}{{{{(1 - x{e^{ - x}})}^2}}}} \right)}&,&{x \ne 0}\\ \alpha &,&{x = 0} \end{array}} \right.

f

is continuous at

x = 0

, then

\alpha

is equal to

View answer and explanation

Correct option: A

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

161 Easy

161 of 192

Understanding +4 | -1

x = \frac{3}{2}

the function

f\left( x \right) = \frac{{\left| {2x - 3} \right|}}{{2x - 3}}

View answer and explanation

Correct option: B

Discontinuous

The correct answer is B (Discontinuous) because the left-hand limit and right-hand limit at

x = \frac{3}{2}

are not equal.

162 Easy

162 of 192

Analyzing +4 | -1

The value of

f\left( 0 \right)

so that the function

f\left( x \right) = \frac{{2 - {{\left( {256 - 7x} \right)}^{1/8}}}}{{{{\left( {5x + 32} \right)}^{1/5}} - 2}}\left( {x \ne 0} \right)

is continuous everywhere, is given by

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the limit as x approaches 0 of the given function is 7/64, which is not any of the other options.

163 Medium

163 of 192

Analyzing +4 | -1

f(x)=\lim \limits_{{n \rightarrow \infty}} \cos ^{2 n}\left(\pi x^2\right)+[x](

where, [.]) denote the greatest integer function and

n \in N

View answer and explanation

Correct option: C

Discontinuous at

{x}=1

and

{x}=0

Option C (Discontinuous at

{x}=1

and

{x}=0

) is correct because the limit of the function as x approaches 0 and 1 from the left and right are not equal to the function's value at those points.

164 Medium

164 of 192

Analyzing +4 | -1

Let

f\left( x \right)

be a function satisfying

f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)

and

f\left( x \right) = x\mathrm{ }g\left( x \right)

for all

x,y \in R

, where

g\left( x \right)

is continuous. Then,

View answer and explanation

Correct option: C

f'\left( x \right) = g\left( 0 \right)

Option C (

f'(x) = g(0)

) is correct because the derivative of f(x) is equal to g(0), derived from the given functional equation and the continuity of g(x).

165 Easy

165 of 192

Application +4 | -1

The value of

f

x = 0

so that function

f\left( x \right) = \frac{{{2^x} - {2^{ - x}}}}{x},x \ne 0

is continuous at

x = 0

, is

View answer and explanation

Correct option: D

\log 4

Option D (

\log 4

) is correct because the limit of the function as x approaches 0, calculated using L'Hopital's rule, is equal to

\log 4

, which is the value of f(0) for continuity.

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166 Hard

166 of 192

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{1}{{\left| x \right|}}\mathrm{for}\,\,\left| x \right| \ge 1}\\ {a{x^2} + b\,\,\mathrm{for}\,\left| x \right| < 1} \end{array}} \right.

f\left( x \right)

is continuous and differentiable at any point, then

View answer and explanation

Correct option: B

a = - \frac{1}{2},b = \frac{3}{2}

Option B is correct because the conditions for continuity and differentiability at x = 1 lead to the equations a + b = 1 and 2a = -1, which are satisfied by a = -1/2 and b = 3/2.

167 Medium

167 of 192

Analyzing +4 | -1

The values of

a

and

b

so that the function

f(x)= \begin{cases}x+a \sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\ 2 x \cot x+b, & \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi\end{cases}

is continuous for

0 \leq x \leq \pi

are, respectively,

View answer and explanation

Correct option: A

\frac{\pi}{3}, \frac{\pi}{12}

Option A (

\frac{\pi}{3}, \frac{\pi}{12}

) is correct because these values of a and b make the given piecewise function continuous at x =

\frac{\pi}{4}

and x =

\frac{\pi}{2}

168 Easy

168 of 192

Application +4 | -1

The function

f(x)

defined by

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\left\{ {{x^2} + {e^{\frac{1}{{2 - x}}}}} \right\}}&{x \ne 2}\\ {k,}&{x = 2} \end{array}} \right.

is continuous from right at

x = 2,

then

k

is equal to

View answer and explanation

Correct option: B

\frac{1}{4}

Option B (

\frac{1}{4}

) is correct because the right-hand limit of the function as x approaches 2 is equal to

\frac{1}{4}

, and for the function to be continuous from the right, f(2) must equal this limit.

169 Easy

169 of 192

Application +4 | -1

f(x)=\left\{\begin{array}{cc}\frac{1-\sqrt{2} \sin x}{\pi-4 x}, & x \neq \frac{\pi}{4} \\ a, & x=\frac{\pi}{4}\end{array}\right.

is continuous at

x=\frac{\pi}{4}

, then

a=

View answer and explanation

Correct option: D

1 / 4

Option D (

\frac{1}{4}

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

\frac{\pi}{4}

, ensuring continuity.

170 Easy

170 of 192

Analyzing +4 | -1

Let

f:\mathbb{R} \to \mathbb{R}

be defined as

f\left( x \right) = \left[ {\begin{array}{cccccccccccccccccccc} {\left[ {{e^x}} \right],}&{x < 0} \\ {\begin{array}{cccccccccccccccccccc} {a{e^x} + \left[ {x - 1} \right],} \\ {b + \left[ {\sin \left( {\pi x} \right)} \right],} \end{array}}&{\begin{array}{cccccccccccccccccccc} {0 \leqslant x < 1} \\ {1 \leqslant x < 2} \end{array}} \\ {\left[ {{e^{ - x}}} \right] - c,}&{x \geqslant 2} \end{array}} \right.

Where

a,b,c \in \mathbb{R}

and

\left[ t \right]

denotes greatest integer less than or equal to

t.

Then, which of the following statements is true?

View answer and explanation

Correct option: C

f

is discontinuous at exactly one point, then

a + b + c \ne 1

The correct answer is 'If

f

is discontinuous at exactly one point, then

a + b + c \ne 1

' because continuity conditions lead to constraints on a, b, and c, and the function cannot be continuous at all points.

171 Easy

171 of 192

Application +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sin \pi x}}{{5x}},x \ne 0}\\ {k,x = 0} \end{array}} \right.

, if

f\left( x \right)

is continuous at

x = 0

, then

k

is equal to

View answer and explanation

Correct option: A

\frac{\pi }{5}

Option A (

\frac{\pi}{5}

) is correct because for f(x) to be continuous at x=0, the limit of f(x) as x approaches 0 must equal f(0), which leads to k =

\frac{\pi}{5}

172 Medium

172 of 192

Application +4 | -1

Let

f(x)

be continuous function such that

f(0)=1

and

f(x)-f\left(\frac{x}{7}\right)=\frac{x}{7} \, \forall \, x \in R

, then

f(42)=

View answer and explanation

Correct option: A

Option A (8) is correct because the limit definition of the derivative leads to f(x) = x/6 + 1, and therefore f(42) = 8.

173 Easy

173 of 192

Analyzing +4 | -1

f(x)=\left\{\begin{array}{r}\sin ^{-1}|x|, \text { when } x \neq 0 \\ 0, \text { when } x=0\end{array}\right.

then

View answer and explanation

Correct option: C

\mathrm{f}(\mathrm{x})

is continuous at

\mathrm{x}=0

Option C is correct because the limit of the function as x approaches 0 is equal to the function's value at x=0, satisfying the condition for continuity.

174 Medium

174 of 192

Application +4 | -1

Let

f(x)=\left\{\begin{array}{cc}\frac{\tan x-\cot x}{x-\frac{\pi}{4},} & x \neq \frac{\pi}{4} \\ a, & x=\frac{\pi}{4}\end{array}\right.

The value of

a

so that

f(x)

is continuous at

x=\pi / 4

, is

View answer and explanation

Correct option: B

Option B (4) is correct because it is the limit of the given function as x approaches

\frac{\pi}{4}

, ensuring continuity.

175 Easy

175 of 192

Analyzing +4 | -1

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\left( {\left[ {\frac{1}{x}} \right] + \left[ {\frac{2}{x}} \right] + \ldots + \left[ {\frac{n}{x}} \right]} \right),}&{x \ne 0}\\ {k,}&{x = 0} \end{array}} \right.

and

n \in N

. Then the value of

k

for which

f(x)

is continuous at

x=0

, is

View answer and explanation

Correct option: D

\frac{n(n+1)}{2}

Option D (

\frac{n(n+1)}{2}

) is correct because by using the squeeze theorem and properties of the greatest integer function, the limit of

f(x)

x

approaches 0 is found to be

\frac{n(n+1)}{2}

, which must equal

k

for continuity.

176 Easy

176 of 192

Application +4 | -1

For the function

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{{x^3} - {a^3}}}{{x - a}},x \ne a}\\ {b,x = a} \end{array}} \right.

, if

f\left( x \right)

is continuous at

x = a

, then

b

is equal to

View answer and explanation

Correct option: C

3{a^2}

Option C (

3a^2

) is correct because for the function to be continuous at x=a, the limit of the function as x approaches a must equal b.

177 Easy

177 of 192

Application +4 | -1

For the parabola

y^2=16 x

, the ratio of the length of the subtangent to the abscissa is

View answer and explanation

Correct option: A

2: 1

Option A (2:1) is correct because the length of the subtangent is twice the abscissa for the given parabola.

178 Easy

178 of 192

Application +4 | -1

f(x)=\left\{\begin{array}{cc}\frac{x^5-32}{x-2}, & \text { where } x \neq 2 \\ x, & \text { where } x=2\end{array}\right.

is continuous at

x=2

, then the value of

k

View answer and explanation

Correct option: D

Option D (80) is correct because the limit of the function as x approaches 2 is 80, and for continuity, f(2) must equal this limit.

179 Medium

179 of 192

Analyzing +4 | -1

If the function

f

defined on

\left( {\frac{\pi }{6},\,\frac{\pi }{3}} \right)

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sqrt 2 \cos x - 1}}{{\cot x - 1}},}&{x \ne \frac{\pi }{4}}\\ {k,}&{x = \frac{\pi }{4}} \end{array}} \right.

is continuous, then

k

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

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

\frac{\pi}{4}

must equal k for the function to be continuous, and the limit is calculated to be

\frac{1}{2}

180 Medium

180 of 192

Analyzing +4 | -1

Let

f(x)=-1+|x-2|

, and

g(x)=1-|x|

; then the set of all points where

f o g

is discontinuous is :

View answer and explanation

Correct option: D

an empty set

Option D (an empty set) is correct because the composite function fog simplifies to |x|, which is continuous everywhere.

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