Continuity in a Domain — JEE Main Maths PYQ MCQs with Solutions
Free JEE Main Maths PYQ Continuity in a Domain 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 in a Domain · hard
Let $f : R \to R$ be a continuously differentiable function such that $f(2) = 6$ and $f'(2) = \dfrac{1}{48}$. If $\displaystyle\int_{6}^{f(x)} 4t^3\,dt = (x-2)\,g(x)$, then $\displaystyle\lim_{x\to 2} g(x)$ is equal to
A. 18 ✓ Correct
B. 24
C. 12
D. 36
Solution: $\int_6^{f(x)} 4t^3\,dt = [f(x)]^4 - 6^4 = (x-2)g(x)$. As $x\to 2$ this is a $0/0$ form, so by L'Hopital's rule, $\lim_{x\to2} g(x) = 4[f(2)]^3 f'(2) = 4(216)\left(\dfrac{1}{48}\right) = 18$.
Q2 — Continuity in a Domain · medium
If $f(x) = \begin{cases} \dfrac{\sin(p+1)x + \sin x}{x}, & x < 0 \\ q, & x = 0 \\ \dfrac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}$ is continuous at $x = 0$, then the ordered pair $(p, q)$ is equal to
A. $\left(-\dfrac{3}{2}, -\dfrac{1}{2}\right)$
B. $\left(-\dfrac{1}{2}, \dfrac{3}{2}\right)$
C. $\left(\dfrac{5}{2}, \dfrac{1}{2}\right)$
D. $\left(-\dfrac{3}{2}, \dfrac{1}{2}\right)$ ✓ Correct
Solution: Left-hand limit: $\dfrac{\sin(p+1)x}{x} + \dfrac{\sin x}{x} \to (p+1) + 1 = p+2$. Right-hand limit: $\dfrac{\sqrt{x}(\sqrt{1+x}-1)}{x^{3/2}} = \dfrac{\sqrt{1+x}-1}{x} \to \dfrac{1}{2}$. For continuity, $p+2 = q = \dfrac{1}{2}$, giving $p = -\dfrac{3}{2}$, $q = \dfrac{1}{2}$.
Q3 — Continuity in a Domain · medium
If the function $f(x) = \begin{cases} a|\pi - x| + 1, & x \le 5 \\ b|x - \pi| + 3, & x > 5 \end{cases}$ is continuous at $x = 5$, then the value of $a - b$ is
A. $\dfrac{-2}{\pi - 5}$
B. $\dfrac{2}{\pi - 5}$
C. $\dfrac{2}{\pi + 5}$
D. $\dfrac{2}{5 - \pi}$ ✓ Correct
Solution: Since $\pi < 5$, $|\pi - 5| = |5 - \pi| = 5 - \pi$. Continuity at $x=5$ requires $a(5-\pi)+1 = b(5-\pi)+3$, so $(a-b)(5-\pi) = 2$, i.e. $a - b = \dfrac{2}{5-\pi}$.
Q4 — Continuity in a Domain · medium
If $f(x) = [x] - \left[\dfrac{x}{4}\right]$, $x \in R$, where $[x]$ denotes the greatest integer function, then
A. $\lim_{x\to 4^-} f(x)$ exists but $\lim_{x\to 4^+} f(x)$ does not exist
B. $f$ is continuous at $x = 4$ ✓ Correct
C. both $\lim_{x\to 4^-} f(x)$ and $\lim_{x\to 4^+} f(x)$ exist but are not equal
D. $\lim_{x\to 4^+} f(x)$ exists but $\lim_{x\to 4^-} f(x)$ does not exist
Solution: $f(4) = 4 - 1 = 3$. As $x \to 4^-$: $[x]=3$, $[x/4]=0$, so $f\to 3$. As $x\to 4^+$: $[x]=4$, $[x/4]=1$, so $f\to 3$. Both one-sided limits equal $f(4)=3$, so $f$ is continuous at $x=4$.
Q5 — Continuity in a Domain · medium
Let $f : [-1, 3] \to R$ be defined as $f(x) = \begin{cases} |x| + [x], & -1 \le x < 1 \\ x + |x|, & 1 \le x < 2 \\ x - [x], & 2 \le x \le 3 \end{cases}$, where $[t]$ denotes the greatest integer less than or equal to $t$. Then, $f$ is discontinuous at
A. four or more points
B. only two points
C. only three points ✓ Correct
D. only one point
Solution: Checking each piece boundary and interior integer points, $f$ fails to be continuous exactly at $x = 0, 1, 2$ — three points in total.
Q6 — Continuity in a Domain · hard
Let $f : R \to R$ be a function defined as $f(x) = \begin{cases} 5, & x \le 1 \\ a + bx, & 1 < x < 3 \\ b + 5x, & 3 \le x < 5 \\ 30, & x \ge 5 \end{cases}$. Then, $f$ is
A. continuous if $a = -5$ and $b = 10$
B. continuous if $a = -5$ and $b = 5$
C. continuous if $a = 0$ and $b = 5$
D. not continuous for any values of $a$ and $b$ ✓ Correct
Solution: At $x=1$: $a+b=5$. At $x=3$: $a+3b = b+15 \Rightarrow a+2b=15$. At $x=5$: $b+25=30 \Rightarrow b=5$, so $a=15-10=5$; but then $a+b=10 \ne 5$, contradicting the $x=1$ condition. No values of $a,b$ satisfy all three junctions simultaneously, so $f$ cannot be made continuous.
Q7 — Continuity in a Domain · hard
If $f(x) = \dfrac{1}{2}|x| - 1$, then on the interval $[0, \pi]$
A. $\tan[f(x)]$ and $1/f(x)$ are both continuous
B. $\tan[f(x)]$ and $1/f(x)$ are both discontinuous ✓ Correct
C. $\tan[f(x)]$ and $f^{-1}(x)$ are both continuous
D. $\tan[f(x)]$ is continuous but $1/f(x)$ is not continuous
Solution: On $[0,\pi]$, $f(x) = \frac{x}{2}-1$ is zero at $x=2$, making $1/f(x)$ undefined (discontinuous) there. Also $[f(x)]$ jumps from $-1$ to $0$ as $x$ crosses $2$, so $\tan[f(x)]$ also jumps — both are discontinuous.
Q8 — Continuity in a Domain · hard
The following functions are continuous on $(0, \pi)$:
(a) $\tan x$
(b) $\displaystyle\int_0^x t \sin\dfrac{1}{t}\,dt$
(c) $f(x) = \begin{cases} 1, & 0 \le x \le \frac{3\pi}{4} \\ 2\sin\frac{2x}{9}, & \frac{3\pi}{4} < x < \pi \end{cases}$
(d) $f(x) = \begin{cases} x\sin x, & 0 \le x \le \frac{\pi}{2} \\ \frac{\pi}{2}\sin(\pi - x), & \frac{\pi}{2} < x < \pi \end{cases}$
A. $\tan x$
B. $\int_0^x t\sin\frac{1}{t}\,dt$ ✓ Correct
C. piecewise function in (c) ✓ Correct
D. piecewise function in (d)
Solution: $\tan x$ has a discontinuity at $x=\pi/2 \in (0,\pi)$. Since $t\sin(1/t) \to 0$ as $t\to 0^+$, its integral is continuous on $(0,\pi)$. In (c), both pieces equal $1$ at $x=3\pi/4$ (as $2\sin(\pi/6)=1$), so it is continuous. In (d), the two pieces disagree at the junction $x=\pi/2$, so it is discontinuous there.
Q9 — Continuity in a Domain · hard
Let $[x]$ denote the greatest integer less than or equal to $x$. If $f(x) = [x\sin(\pi x)]$, then $f(x)$ is
A. continuous at $x = 0$ ✓ Correct
B. continuous in $(-1, 0)$ ✓ Correct
C. differentiable at $x = 1$
D. differentiable in $(-1, 1)$ ✓ Correct
Solution: Near $x=0$, $x\sin(\pi x) \to 0$ without reaching $\pm1$, so $f\equiv 0$ near $0$, giving continuity there. On $(-1,0)$, $x\sin(\pi x)$ stays strictly within $(0,1)$, so $[x\sin \pi x]=0$ throughout, giving continuity (hence differentiability) there and on all of $(-1,1)$. At $x=1$, $x\sin(\pi x)=0$ exactly, which is not an interior point of the domain analysed, so differentiability there is not guaranteed.