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JEE • MATHEMATICS • CONTINUITY

Continuity practice · Mixed difficulty

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Practice format and chapter quiz

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 31-60 of 192 questions
31 Easy
31 of 192
Application +4 | -1
The value of kk for which the function f(x)={1−cos⁡4x8x2,x≠0kx=0f\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=0a = 0, is
View answer and explanation
Correct option: B
k=1k = 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.
32 Medium
32 of 192
Analyzing +4 | -1
The function
f(x)={1−cos⁡4xx2 if x<02x16+x−4 if x=0 if x>0f(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=0x=0 for
View answer and explanation
Correct option: D
no value of AA
Option D (no value of AA) 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.
33 Easy
33 of 192
Understanding +4 | -1
Let f(x)f\left( x \right) be a function differentiable at x=cx = c. Then, lim⁡x→cf(x)\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→cf(x)=f(c)\lim_{x \to c} f(x) = f(c).
34 Easy
34 of 192
Understanding +4 | -1
Let f:R→Rf: R \rightarrow R be any function. Define g:R→Rg: R \rightarrow R by g(x)=∣f(x)∣g(x)=|f(x)| for x→Rx \rightarrow R. Then, gg is
View answer and explanation
Correct option: C
continuous if ff is continuous
Option C (continuous if ff is continuous) is correct because the absolute value function is continuous, and the composition of continuous functions is continuous.
35 Easy
35 of 192
Analyzing +4 | -1
The function f(x)=1−sin⁡x+cos⁡x1+sin⁡x+cos⁡xf\left( x \right) = \frac{{1 - \sin x + \cos x}}{{1 + \sin x + \cos x}} is not defined at x=πx = \pi. The value of f(π)f\left( \pi \right), so that f(x)f\left( x \right) is continuous at x=πx = \pi, is
View answer and explanation
Correct option: C
−1- 1
Option C (−1-1) is correct because the limit of the function as xx approaches π\pi is −1-1, and for the function to be continuous, f(π)f(\pi) must equal this limit.
36 Easy
36 of 192
Analyzing +4 | -1
If f:R→Rf: R \rightarrow R is a continuous function satisfying f(0)=1f(0)=1 and f(2x)−f(x)=xf(2 x)-f(x)=x for all x∈Rx \in R and lim⁡x→∞{f(x)−f(x2n)}=P(x)\lim\limits _{x \rightarrow \infty}\left\{f(x)-f\left(\frac{x}{2^n}\right)\right\}=P(x), then P(x)P(x), is
View answer and explanation
Correct option: B
a linear polynomial in xx
Option B (a linear polynomial in xx) is correct because after simplification and taking the limit, P(x)P(x) results in a linear function of xx.
37 Easy
37 of 192
Application +4 | -1
If f(x)={ax+1,x≤π/2sin⁡x+b,x>π/2f(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=aπ2b=a \frac{\pi}{2}
Option C (b=aπ2b=a \frac{\pi}{2}) is correct because for ff to be continuous, aπ2+1=sin⁡(π2)+ba \frac{\pi}{2}+1 = \sin(\frac{\pi}{2})+b, so aπ2=ba \frac{\pi}{2} = b.
38 Medium
38 of 192
Application +4 | -1
If f(x)={cos⁡−1(cot⁡x),x<π2a(π[x]−1),x≥π2f(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 aa for ff to be continuous at x=π2x=\frac{\pi}{2} is
View answer and explanation
Correct option: A
π2(π−1)\frac{\pi}{2(\pi-1)}
Option A (π2(π−1)\frac{\pi}{2(\pi-1)}) is correct because for f to be continuous at x=π2\frac{\pi}{2}, the left and right limits must be equal, which leads to solving for a.
39 Easy
39 of 192
Application +4 | -1
The values of xx for which the tangents to the curves y=xcos⁡x,y=(sin⁡x)/xy=x \cos x, y=(\sin x) / x are parallel to the axis of xx are roots of (respectively)
View answer and explanation
Correct option: C
cot⁡x=x,tan⁡x=x\cot x=x, \tan x=x
Option C (cot⁡x=x,tan⁡x=x\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.
40 Easy
40 of 192
Understanding +4 | -1
Statement 1: A function f:R→Rf: R \rightarrow R is continuous at x0x_{0} if and only if lim⁡x→x0f(x)\lim _{x \rightarrow x_{0}} f(x) exists and lim⁡x→x0f(x)=f(x0)\lim _{x \rightarrow x_{0}} f(x)=f\left(x_{0}\right)

Statement 2: A function f:R→Rf: R \rightarrow R is discontinuous at x0x_{0} if and only if, lim⁡x→x0f(x)\lim _{x \rightarrow x_{0}} f(x) exists and lim⁡x→x0f(x)≠f(x0)\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.
41 Medium
41 of 192
Analyzing +4 | -1
Given a function f(x)f(x) continuous ∀ x∈R\forall \, x \in R such that lim⁡x→0[f(x)+log⁡(1−1ef(x))−log⁡(f(x))]=0\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)f(0) is
View answer and explanation
Correct option: A
0
Option A (0) is correct because by applying the limit properties and simplifying the expression, we find that lim⁡x→0f(x)=0\lim_{x \to 0} f(x) = 0, and since f(x)f(x) is continuous, f(0)=0f(0) = 0.
42 Easy
42 of 192
Application +4 | -1
If f:R→Rf: R \rightarrow R is defined by
f(x)={cos⁡3x−cos⁡xx2,x≠0λ,x=0f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\cos 3x - \cos x}}{{{x^2}}},}&{x \ne 0}\\ {\lambda ,}&{x = 0} \end{array}} \right. and if ff is continuous at x=0x=0, then λ\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.
43 Easy
43 of 192
Application +4 | -1
If f(x)f(x) is a continuous function satisfying f(x)f(1/x)=f(x)+f(1/x)f(x) f(1 / x)=f(x)+f(1 / x) and f(1)>0f(1)>0 then lim⁡x→1f(x)\lim \limits_{{x \rightarrow 1}} f(x) is equal to
View answer and explanation
Correct option: A
2
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).
44 Medium
44 of 192
Understanding +4 | -1
The number of points at which the function f(x)=(∣x−1∣+∣x−2∣+cos⁡x)f\left( x \right) = \left( {\left| {x - 1} \right| + \left| {x - 2} \right| + \cos x} \right) where x∈[0,4]x \in \left[ {0,4} \right] is not continuous, is
View answer and explanation
Correct option: D
0
The correct answer is D (0) because absolute value functions and cosine are continuous everywhere, and the sum of continuous functions is continuous.
45 Easy
45 of 192
Understanding +4 | -1
The function f(x)=cot⁡xf(x)=\cot x is discontinuous on the set
View answer and explanation
Correct option: A
{x=nπ,n∈Z}\{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.
46 Medium
46 of 192
Analyzing +4 | -1
Let f(x)={xPsin⁡1x,x≠00,x=0f\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)f(x) is continuous but not differentiable at x=0x = 0 if
View answer and explanation
Correct option: A
0<P⩽10 < P \leqslant 1
Option A is correct because for 0 < P <= 1, the function is continuous but not differentiable at x = 0.
47 Medium
47 of 192
Analyzing +4 | -1
If f:R→Rf:R \to R is defined by
f(x)={2sin⁡x−sin⁡2x2xcos⁡xa,if  x=0,if  x≠0,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 ff is continuous at 0 is
View answer and explanation
Correct option: D
0
Option D (0) is correct because the limit of the function as x approaches 0 is 0, which must equal f(0) for continuity.
48 Easy
48 of 192
Analyzing +4 | -1
The function f(x)=lim⁡n→∞cos⁡(2πx)−x2nsin⁡(x−1)1+x2n+1−x2nf(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 xx in
View answer and explanation
Correct option: B
R−{−1,1}\mathbb{R}-\{-1,1\}
Option B (R−{−1,1}\mathbb{R}-\{-1,1\}) is correct because the function is discontinuous at x=1x = 1 and x=−1x = -1, making it continuous for all real numbers except -1 and 1.
49 Easy
49 of 192
Application +4 | -1
If the tangent to the curve x3−y2=0x^3-y^2=0 at (m2,−m3)\left(m^2,-m^3\right) is parallel to y=−1mx−2m3y=-\frac{1}{m} x-2 m^3, then the value of m2m^2 is
View answer and explanation
Correct option: C
2/32 / 3
Option C (2/3) is correct because it is the value of m2m^2 that makes the tangent to the curve parallel to the given line.
50 Medium
50 of 192
Analyzing +4 | -1
Let f(x)={∫0x{5+∣1−t∣}dt,if  x>25x+1,if  x≤2f\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(x)f\left( x \right) is continuous but not differentiable at x=2x = 2
Option B is correct because f(x)f(x) is continuous at x=2x=2 but the left-hand derivative and right-hand derivative are not equal, making it non-differentiable.
51 Easy
51 of 192
Understanding +4 | -1
The function f(x)=(x)f(x)=(x), where (x)(x) denotes the smallest integer ≥x\geq x, is
View answer and explanation
Correct option: C
continuous on R−ZR-Z
Option C (continuous on R−ZR-Z) is correct because the function f(x)=(x)f(x)=(x) is continuous everywhere except at integer values.
52 Easy
52 of 192
Application +4 | -1
Let ff be a continuous function satisfying f(x+y)=f(x+y)= f(x)+f(y)f(x)+f(y) for all x,y∈Rx, y \in \mathbf{R} and f(1)=5f(1)=5 then lim⁡x→4\lim \limits_{{x \rightarrow 4}} f(x)f(x) is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because f(4)=20f(4)=20 as f(x+y)=f(x)+f(y)f(x+y) = f(x) + f(y) and f(1)=5f(1)=5.
53 Easy
53 of 192
Analyzing +4 | -1
If f(x)={∣x−a∣x−a, when x≠a1, when x=af(x)=\left\{\begin{array}{c}\frac{|x-a|}{x-a}, \text { when } x \neq a \\ 1, \text { when } x=a\end{array}\right., then
View answer and explanation
Correct option: B
f(x)f(x) is discontinuous at x=ax=a
Option B is correct because the left-hand limit and right-hand limit at x=a are not equal, indicating a discontinuity.
54 Medium
54 of 192
Application +4 | -1
If the function
f(x)={x2−2x+asin⁡x when x≠0b when x=0f(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=0x=0, then
View answer and explanation
Correct option: A
a=0,b=−2a=0, b=-2
Option A (a=0,b=−2a=0, b=-2) is correct because for the function to be continuous at x=0x=0, the limit as xx approaches 0 must exist and be equal to f(0)f(0), which leads to these values.
55 Medium
55 of 192
Analyzing +4 | -1
Let f=(−π4,  π4)→Rf = \left( { - \frac{\pi }{4},\,\,\frac{\pi }{4}} \right) \to R be defined as
f(x)={(1+∣sin⁡x∣)3a∣sin⁡x∣,−π4<x<0b,x=0ecot⁡4xcot⁡2x,0<x<π4f(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.
If ff is continuous at x=0x = 0, then the value of 6 a+b26\,a + {b^2} is equal to
View answer and explanation
Correct option: C
1+e1 + e
Option C (1+e1 + e) is correct because it is the result of evaluating the limits and applying the continuity condition at x=0.
56 Easy
56 of 192
Analyzing +4 | -1
If f(x)={((a−n)nx−tan⁡x)sin⁡nxx2 at x≠0,0 at x=0f(x)=\left\{\begin{array}{cl}\frac{((a-n) n x-\tan x) \sin n x}{x^2} & \text { at } x \neq 0, \\ 0 & \text { at } x=0\end{array}\right. where nn is a non-zero real number, and ff is continuous at x=0x=0, then a is equal to
View answer and explanation
Correct option: D
n+1nn+\frac{1}{n}
Option D (n+1nn+\frac{1}{n}) is correct because for f(x)f(x) to be continuous at x=0x=0, the limit of f(x)f(x) as xx approaches 0 must equal f(0)f(0), which leads to a=n+1na = n + \frac{1}{n}.
57 Easy
57 of 192
Analyzing +4 | -1
If for p≠q≠0p \neq q \neq 0, the function f(x)=p(729+x)−37729+qx−93f(x)=\frac{\sqrt[7]{p(729+x)-3}}{\sqrt[3]{729+q x-9}} is continuous at x=0x=0, then
View answer and explanation
Correct option: B
63qf(0)−p2=063 q f(0)-p^2=0
Option B (63qf(0)−p2=063 q f(0)-p^2=0) is correct because by applying L'Hopital's rule and the continuity condition, the relationship between p, q, and f(0) is correctly derived.
58 Easy
58 of 192
Application +4 | -1
If f(x)={mx+1,x≤π2sin⁡x+n,x>π2f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {mx + 1,x \le \frac{\pi }{2}}\\ {\sin x + n,x > \frac{\pi }{2}} \end{array}} \right. is continuous at x=π2x = \frac{\pi }{2}, then
View answer and explanation
Correct option: C
n=mπ2n = \frac{{m\mathrm{ }\pi }}{2}
Option C (n=mπ2n = \frac{{m\mathrm{ }\pi }}{2}) is correct because it ensures the continuity of the function at x=π2x = \frac{\pi}{2} by making the left-hand limit equal to the right-hand limit.
59 Easy
59 of 192
Analyzing +4 | -1
Let f:[1,3]→Rf:[1,3] \rightarrow R be a function satisfying x[x]≤f(x)≤6−x\frac{x}{[x]} \leq f(x) \leq \sqrt{6-x}, for all x≠2x \neq 2 and f(2)=1f(2)=1, where RR is the set of all real numbers and [x][x] denotes the largest integer less than or equal to xx.

Statement 1: lim⁡x→2−f(x)\lim \limits_{x \rightarrow 2^{-}} f(x) exists.
Statement 2: ff is continuous at x=2x=2.
View answer and explanation
Correct option: D
Statement 1 is true, Statement 2 is false.
Option D (Statement 1 is true, Statement 2 is false) is correct because the left-hand limit exists at x=2, but the function is not continuous at x=2 since the right-hand limit does not exist.
60 Easy
60 of 192
Analyzing +4 | -1
If f(x)={xe1/x+1, when x≠00, when x=0f(x)=\left\{\begin{array}{cc}\frac{x}{e^{1 / x}+1}, & \text { when } x \neq 0 \\ 0 & \text {, when } x=0\end{array}\right., then
View answer and explanation
Correct option: C
f(x)f(x) is continuous at x=0x=0
Option B (f(x) is discontinuous at x=ax=a) is correct because the left-hand limit and right-hand limit are not equal at x=ax=a.
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