Continuity at a Point — JEE Main Maths PYQ MCQs with Solutions
Free JEE Main Maths PYQ Continuity at a Point MCQs with step-by-step solutions (9 questions). Part of Limit, Continuity and Differentiability. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Continuity at a Point · medium
If the function $f$ defined on $\left[\dfrac{\pi}{6}, \dfrac{\pi}{3}\right]$ by $f(x) = \begin{cases} \dfrac{\sqrt{2}\cos x - 1}{\cot x - 1}, & x \neq \dfrac{\pi}{4} \\ k, & x = \dfrac{\pi}{4} \end{cases}$ is continuous, then $k$ is equal to
A. $\dfrac{1}{2}$ ✓ Correct
B. $2$
C. $1$
D. $\dfrac{1}{\sqrt{2}}$
Solution: Put $x = \pi/4 + h$. Then $\sqrt{2}\cos x - 1 = \cos h - \sin h - 1 \approx -h$ and $\cot x - 1 = \dfrac{-2\sin h}{\cos h + \sin h} \approx -2h$ for small $h$. So $k = \lim_{h\to 0} \dfrac{-h}{-2h} = \dfrac{1}{2}$.
Q2 — Continuity at a Point · medium
The function $f(x) = [x]^2 - [x^2]$ (where $[x]$ is the greatest integer less than or equal to $x$), is discontinuous at
A. all integers
B. all integers except $0$ and $1$ ✓ Correct
C. all integers except $0$
D. all integers except $1$
Solution: At an integer $x = n$, $f(n) = n^2 - n^2 = 0$. As $x \to n^-$, $[x] = n-1$ and $[x^2] = n^2 - 1$ (for $n \neq 0, 1$), giving a left-hand value of $(n-1)^2 - (n^2-1) \neq 0$, so there is a jump. For $n = 0$ and $n = 1$ this jump vanishes, so $f$ is continuous there. Hence $f$ is discontinuous at all integers except $0$ and $1$.
Q3 — Continuity at a Point · easy
Let $[.]$ denote the greatest integer function and $f(x) = [\tan^2 x]$, then
A. $\lim_{x \to 0} f(x)$ does not exist
B. $f(x)$ is continuous at $x = 0$ ✓ Correct
C. $f(x)$ is not differentiable at $x = 0$
D. $f'(0) = 1$
Solution: For $x$ close to $0$ (but $x \neq 0$), $0 < \tan^2 x < 1$, so $[\tan^2 x] = 0$, and $f(0) = [\tan^2 0] = 0$. Thus $\lim_{x\to 0} f(x) = 0 = f(0)$, so $f$ is continuous at $x = 0$.
Q4 — Continuity at a Point · medium
The function $f(x) = [x]\cos\left(\dfrac{2x-1}{2}\pi\right)$, where $[.]$ denotes the greatest integer function, is discontinuous at
A. all $x$
B. all integer points
C. no $x$ ✓ Correct
D. $x$ which is not an integer
Solution: At an integer $x = n$, $\cos\left(\dfrac{2n-1}{2}\pi\right) = \cos\left(n\pi - \dfrac{\pi}{2}\right) = 0$. So even though $[x]$ jumps at integers, it is always multiplied by a cosine factor that vanishes exactly there, forcing $f(x) \to 0$ from both sides and matching $f(n) = 0$. Away from integers both factors are continuous. Hence $f$ is continuous everywhere, i.e. discontinuous at no $x$.
Q5 — Continuity at a Point · hard
If $f(x) = x\left(\sqrt{x} - \sqrt{x+1}\right)$, then
A. $f(x)$ is continuous but not differentiable at $x = 0$
B. $f(x)$ is differentiable at $x = 0$
C. $f(x)$ is not differentiable at $x = 0$ ✓ Correct
D. None of the above
Solution: Because of $\sqrt{x}$, $f$ is defined only for $x \geq 0$, so at $x = 0$ only a right-hand derivative can be examined: $f'(0^+) = \lim_{x\to 0^+} \dfrac{f(x)-f(0)}{x} = \lim_{x\to 0^+}\left(\sqrt{x}-\sqrt{x+1}\right) = -1$. Since $f$ is not even defined for $x<0$, a genuine (two-sided) derivative cannot exist at $x=0$, so $f$ is not differentiable there.
Q6 — Continuity at a Point · easy
The function $f(x) = \dfrac{\log(1+ax) - \log(1-bx)}{x}$ is not defined at $x = 0$. The value which should be assigned to $f$ at $x = 0$, so that it is continuous at $x = 0$, is
A. $a - b$
B. $a + b$ ✓ Correct
C. $\log a - \log b$
D. None of these
Solution: Using $\lim_{t\to 0} \dfrac{\log(1+t)}{t} = 1$: $\lim_{x\to 0} \dfrac{\log(1+ax)}{x} = a$ and $\lim_{x\to 0} \dfrac{-\log(1-bx)}{x} = b$. So the limit of $f(x)$ as $x\to 0$ is $a+b$, which must be assigned as $f(0)$.
Q7 — Continuity at a Point · hard
Let $[x]$ be the greatest integer less than or equal to $x$. Then, at which of the following point(s) is the function $f(x) = x\cos(\pi(x + [x]))$ discontinuous?
A. $x = -1$ ✓ Correct
B. $x = 1$ ✓ Correct
C. $x = 0$
D. $x = 2$ ✓ Correct
Solution: Writing $x = n + \{x\}$ with $n=[x]$, the argument becomes $\pi(2n+\{x\})$, so $\cos(\pi(x+[x])) = \cos(\pi\{x\})$ using the value of $[x]$ just below/above an integer. Checking one-sided limits of $x\cos(\pi(x+[x]))$ at $x=-1,0,1,2$ shows the jump in $[x]$ flips the sign of the cosine term at $x=-1,1,2$ (giving unequal left/right limits there) but not at $x=0$ where the factor $x$ itself kills the jump. So $f$ is discontinuous at $x=-1,1,2$.
Q8 — Continuity at a Point · hard
For every pair of continuous functions $f, g : [0,1] \to \mathbb{R}$ such that $\max\{f(x) : x \in [0,1]\} = \max\{g(x) : x \in [0,1]\}$, the correct statement(s) is (are)
A. $[f(c)]^2 + 3f(c) = [g(c)]^2 + 3g(c)$ for some $c \in [0,1]$ ✓ Correct
B. $[f(c)]^2 + f(c) = [g(c)]^2 + 3g(c)$ for some $c \in [0,1]$
C. $[f(c)]^2 + 3f(c) = [g(c)]^2 + g(c)$ for some $c \in [0,1]$
D. $[f(c)]^2 = [g(c)]^2$ for some $c \in [0,1]$ ✓ Correct
Solution: Let $M$ be the common maximum, attained by $f$ at $c_1$ and by $g$ at $c_2$, so $f(c_1) \geq g(c_1)$ and $g(c_2) \geq f(c_2)$. For $h(x) = [f(x)]^2 + 3f(x) - [g(x)]^2 - 3g(x)$ (increasing in $f$ and decreasing in $g$), $h(c_1) \geq 0$ and $h(c_2) \leq 0$; by IVT some $c$ gives $h(c) = 0$, proving (a). The same IVT argument applied to $[f(x)]^2 - [g(x)]^2$ proves (d).
Q9 — Continuity at a Point · hard
For every integer $n$, let $a_n$ and $b_n$ be real numbers. Let function $f : \mathbb{R} \to \mathbb{R}$ be given by $f(x) = \begin{cases} a_n + \sin \pi x, & x \in [2n, 2n+1] \\ b_n + \cos \pi x, & x \in (2n-1, 2n) \end{cases}$ for all integers $n$. If $f$ is continuous, then which of the following hold(s) for all $n$?
A. $a_{n-1} - b_{n-1} = 0$
B. $a_n - b_n = 1$ ✓ Correct
C. $a_n - b_{n+1} = 1$
D. $a_{n+1} - b_n = -1$ ✓ Correct
Solution: At $x = 2n$: from the right, $f = a_n + \sin(2n\pi) = a_n$; from the left, $f = b_n + \cos(2n\pi) = b_n + 1$. Continuity gives $a_n - b_n = 1$. At $x = 2n+1$: from the left, $f = a_n + \sin((2n+1)\pi) = a_n$; from the right, $f = b_{n+1} + \cos((2n+1)\pi) = b_{n+1} - 1$. Continuity gives $a_n - b_{n+1} = -1$, i.e. (shifting index) $a_{n+1} - b_n = -1$.