Q.Find all points of discontinuity of $f$, where $f$ is defined by $f(x) = \begin{cases} 2x+3, & \text{if } x \leq 2 \\ 2x-3, & \text{if } x > 2 \end{cases}$

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.Let $f(x) = \begin{cases} 0, & x = 0 \\ 2 - x, & \text{for } 0 < x < 1 \\ 2, & \text{for } x = 1 \\ \dfrac{1}{2} - x, & \text{for } 1 < x < 2 \\ \dfrac{-3}{2}, & \text{for } x \ge 2 \end{cases}$ then which of the following is true (A) $f$ is right continuous at $x = 0$ (B) $f$ is left continuous at $x = 1$ (C) $f$ is right continuous at $x = 1$ (D) $f$ is continuous at $x = 2$

   ›Reveal solutionSolution

   Compute the left-hand limit, right-hand limit, and function value at each candidate point and compare; only at $x=2$ do all three coincide.

   Concept and Intuition

   A function is continuous at a point if the left-hand limit, right-hand limit, and the function's value there all agree. "Right continuous" only needs the right-hand limit to equal the function value; "left continuous" only needs the left-hand limit to match.

   Step-by-Step Solution

   1. At $x=0$: $f(0)=0$. Right-hand limit: $\lim_{x\to0^+}(2-x)=2$. Since $2\ne0$, $f$ is not right continuous at $0$ — (A) is false.
   2. At $x=1$: $f(1)=2$. Left-hand limit: $\lim_{x\to1^-}(2-x)=1$. Since $1\ne2$, not left continuous — (B) is false. Right-hand limit: $\lim_{x\to1^+}\left(\dfrac12-x\right)=-\dfrac12$. Since $-\dfrac12\ne2$, not right continuous either — (C) is false. …

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2. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.The function $f(x)=\begin{cases}\dfrac{2}{5-x}, & x<3\\ 5-x, & x \ge 3\end{cases}$ is (A) left discontinuous at $x=3$ (B) left continuous at $x=3$ (C) right discontinuous at $x=5$ (D) discontinuous at $x=5$

   ›Reveal solutionSolution

   Checking the one-sided limits at the junction $x=3$ shows the function jumps from $1$ to $2$ on approach from the left, i.e. it is left-discontinuous (but right-continuous) at $x=3$; there is nothing special happening at $x=5$.

   Concept and Intuition

   A piecewise function can fail to be continuous from only one side at a junction point if the junction point itself is included in one branch. Here $x=3$ is the boundary, and it belongs to the $5-x$ branch (since that branch is defined for $x\ge3$), so we must check whether the other branch's limit (from the left) agrees with the actual function value.

   Step-by-Step Solution

   1. Function value at $x=3$: since $x\ge3$ uses $f(x)=5-x$, $f(3)=5-3=2$.
   2. Left-hand limit as $x\to3^-$: uses $f(x)=\dfrac{2}{5-x}$, so $\lim_{x\to3^-}\dfrac{2}{5-x}=\dfrac{2}{5-3}=1$.
   3. Since LHL $=1\ne f(3)=2$, the function is discontinuous when approached from the left — i.e. left discontinuous at $x=3$.
   4. Right-hand limit as $x\to3^+$: still $5-x\to2=f(3)$, so it is right-continuous at $x=3$ — this rules out "left continuous" (option B, false). …

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3. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If the function $f(x) = \dfrac{\sqrt{1+x}-1}{x}$ is continuous at $x = 0$ then $f(0) =$ (A) $-\dfrac{1}{2}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{1}{2}$ (D) $-\dfrac{1}{3}$

   ›Reveal solutionSolution

   Rationalizing the numerator turns the $0/0$ form into a simple limit, which must equal $f(0)$ for continuity.

   Concept and Intuition

   For $f$ to be continuous at $x=0$, we need $f(0)=\displaystyle\lim_{x\to0}f(x)$. Since the given formula for $f(x)$ is $0/0$ at $x=0$, we must simplify it algebraically first.

   Step-by-Step Solution

   1. $\dfrac{\sqrt{1+x}-1}{x} = \dfrac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)} = \dfrac{(1+x)-1}{x(\sqrt{1+x}+1)} = \dfrac{x}{x(\sqrt{1+x}+1)} = \dfrac{1}{\sqrt{1+x}+1}$.
   2. As $x\to0$: $\dfrac{1}{\sqrt{1}+1} = \dfrac12$. …

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4. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.The values of $a$ and $b$ for which the function $f(x)=\begin{cases}1+|\sin x|^{a/|\sin x|}, & \dfrac{-\pi}{6}<x<0 \\ b, & x=0 \\ e^{\tan 2x/\tan 3x}, & 0<x<\dfrac{\pi}{6}\end{cases}$ is continuous at $x=0$ are (A) $a=1, b=\dfrac{2}{3}$ (B) $a=\dfrac{2}{3}, b=e^{2/3}$ (C) $a=\dfrac{2}{3}, b=\dfrac{3}{2}$ (D) $a=-1, b=e^{2/3}$

   ›Reveal solutionSolution

   Both one-sided limits must equal $b$; the right side gives $e^{2/3}$ directly, and the left side (a $(1+u)^{1/u}\to e$-type limit) matches it when $a=2/3$ — (B).

   Concept and Intuition

   Continuity at $x=0$ requires $\displaystyle\lim_{x\to0^-}f(x)=f(0)=\lim_{x\to0^+}f(x)$. The right branch is a standard $e^{\text{ratio of small angles}}$ limit, and the left branch is the classical exponential limit $(1+u)^{1/u}\to e$ as $u\to0$, raised to a power $a$.

   Step-by-Step Solution

   1. Right-hand limit: as $x\to0^+$, $\tan2x\approx2x$ and $\tan3x\approx3x$, so $\dfrac{\tan2x}{\tan3x}\to\dfrac23$. Hence $\displaystyle\lim_{x\to0^+}e^{\tan2x/\tan3x}=e^{2/3}$.
   2. Left-hand limit: let $u=|\sin x|\to0^+$ as $x\to0^-$. The left branch is $(1+u)^{a/u}=\left[(1+u)^{1/u}\right]^{a}$. Since $(1+u)^{1/u}\to e$, this tends to $e^{a}$.
   3. For continuity: left limit $=$ right limit $=f(0)=b$, i.e. $e^{a}=e^{2/3}=b$. …

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5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $f(x) = \begin{cases} \dfrac{x^2 - 4}{\sqrt{2 - x}}, & \text{for } x < 2 \\ a, & \text{for } x = 2 \\ \log(x - 2), & \text{for } x > 2 \end{cases}$ is a real valued function and 'a' is a finite real number, then (A) f is continuous at x = 2 when a = 0 (B) f is left continuous at x = 2 when a = 0 (C) f is right continuous at x = 2 when a = log 2 (D) f is not continuous at x = 1

   ›Reveal solutionSolution

   This tests one-sided limits at a piecewise junction; the left-hand limit at $x=2$ is $0$ while the right-hand limit diverges to $-\infty$, so option (B) is the only correct statement.

   Concept and Intuition

   For left continuity at $x=2$ we only need $\lim_{x\to2^-}f(x)=f(2)$; the behaviour on the other side is irrelevant. Since $\log(x-2)\to-\infty$ as $x\to2^+$, no finite value of $a$ can make $f$ right-continuous or fully continuous at $x=2$ — so those statements must be false.

   Step-by-Step Solution

   1. Left-hand limit: let $x=2-h,\ h\to0^+$. Then $x^2-4=(2-h)^2-4=h^2-4h=h(h-4)$ and $\sqrt{2-x}=\sqrt h$.
   2. So $\dfrac{x^2-4}{\sqrt{2-x}} = \dfrac{h(h-4)}{\sqrt h} = \sqrt h\,(h-4) \to 0\cdot(-4)=0$.
   3. Left continuity at $x=2$ requires $f(2)=a$ to equal this limit, i.e. $a=0$. So (B) is true.
   4. Right-hand limit: as $x\to2^+$, $x-2\to0^+$, so $\log(x-2)\to-\infty$ — unbounded, so no finite $a$ (in particular $a=\log2$) can match it. So (C) is false, and (A) (needing both sides equal) is also false. …

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6. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$f(x)=\begin{cases}\dfrac{(2x^2-ax+1)-(ax^2+3bx+2)}{x+1}, & \text{if } x\neq -1\\ k, & \text{if } x=-1\end{cases}$ is a real valued function. If $a,b,k\in R$ and $f$ is continuous on $R$ then $k=$ (A) $-\dfrac{1}{3}$ (B) 6 (C) $a-2$ (D) $a-3$

   ›Reveal solutionSolution

   Continuity at the removable-type point requires the numerator to vanish there (fixing $b$) and then simplifying the quotient to evaluate $k$ as a limit. The answer is $k=a-3$.

   Concept and Intuition

   A rational expression $\dfrac{N(x)}{x+1}$ can only have a finite limit as $x\to-1$ if $N(-1)=0$ (otherwise the limit is $\pm\infty$ and no choice of $k$ can make $f$ continuous there). So the first job is to use that vanishing condition to pin down any free constant, then simplify by cancelling the common factor.

   Step-by-Step Solution

   1. Numerator: $(2x^2-ax+1)-(ax^2+3bx+2)=(2-a)x^2-(a+3b)x-1$.
   2. At $x=-1$: $(2-a)(1)+(a+3b)-1=1+3b$. For the limit (hence continuity) to exist finitely, this must be $0$: $b=-\dfrac13$.
   3. With $b=-1/3$, numerator becomes $(2-a)x^2-(a-1)x-1$.
   4. Factor out $(x+1)$: writing $(2-a)x^2-(a-1)x-1=(x+1)\left[(2-a)x-1\right]$ (verified by expansion, matching all coefficients). …

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7. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If the function $f(x)$, defined below, is continuous everywhere, then '$k$' equals ______ $f(x) = \begin{cases} \dfrac{2^x - 1}{\sqrt{1+x} - 1}, & -1 \le x < \infty, \; x \ne 0 \\ k, & x = 0 \end{cases}$ (A) $\dfrac{1}{2}\log_e 2$ (B) $\log_e 4$ (C) $\log_e 8$ (D) $\log_e 2$

   ›Reveal solutionSolution

   Continuity at $x=0$ requires $k$ to equal the limit of $f(x)$ as $x\to0$, which works out to $\ln4$ using standard small-$x$ approximations.

   Concept and Intuition

   For $f$ to be continuous at $x=0$, we need $\displaystyle\lim_{x\to0}f(x)=f(0)=k$. Both numerator and denominator vanish as $x\to0$ (this is a $0/0$ form), so we use the standard first-order approximations $2^x-1\approx x\ln2$ and $\sqrt{1+x}-1\approx x/2$ near $x=0$.

   Step-by-Step Solution

   1. Numerator: $2^x-1=e^{x\ln2}-1\approx x\ln2$ for small $x$.
   2. Denominator: $\sqrt{1+x}-1=(1+x)^{1/2}-1\approx \dfrac{x}{2}$ for small $x$ (binomial expansion).
   3. So $\displaystyle\lim_{x\to0}\frac{2^x-1}{\sqrt{1+x}-1}=\lim_{x\to0}\frac{x\ln2}{x/2}=2\ln2$. …

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8. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.$f(x) = \begin{cases} \dfrac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, & x \neq 0 \\ k\log 2 \log 3, & x = 0 \end{cases}$ Find the value of '$k$' for which the function $f$ is continuous. (A) $\sqrt{2}$ (B) 24 (C) $18\sqrt{3}$ (D) $24\sqrt{2}$

   ›Reveal solutionSolution

   Factoring the numerator as $(9^x-1)(8^x-1)$ and expanding the denominator via $1+\cos x=2\cos^2(x/2)$ gives the limit $24\sqrt2\ln2\ln3$, so $k=24\sqrt2$.

   Concept and Intuition

   Both numerator and denominator vanish as $x\to0$ — a $0/0$ form best handled by recognizing the standard small-$x$ approximations $a^x-1\approx x\log a$ and $1-\cos\theta\approx\theta^2/2$, rather than repeated L'Hôpital. Spotting that $72=9\times8$ lets the numerator factor neatly, turning a messy expression into a clean product of two standard limits.

   Step-by-Step Solution

   1. Since $72=9\times8$: $72^x-9^x-8^x+1 = 9^x8^x-9^x-8^x+1=(9^x-1)(8^x-1)$.
   2. As $x\to0$: $9^x-1\sim x\ln9$, $8^x-1\sim x\ln8$, so numerator $\sim x^2\ln9\ln8$.
   3. $1+\cos x = 2\cos^2(x/2)$, so $\sqrt{1+\cos x}=\sqrt2\cos(x/2)$ (for small $x$).
   4. Denominator $=\sqrt2-\sqrt2\cos(x/2)=\sqrt2\left(1-\cos\tfrac x2\right)\sim\sqrt2\cdot\dfrac{(x/2)^2}{2}=\dfrac{\sqrt2\,x^2}{8}$.
   5. Limit $=\dfrac{x^2\ln9\ln8}{\sqrt2 x^2/8}=\dfrac{8\ln9\ln8}{\sqrt2}$. …

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9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $f(x) = \begin{cases} \left(\frac{2+3x^2}{2+5x^2}\right)^{\frac{8}{x^2}+3}, & \text{for } x \neq 0 \\ k, & \text{for } x = 0 \end{cases}$ is a continuous function at $x = 0$, then $k =$ (A) $e^{-2}$ (B) $e^{-4}$ (C) $e^{-8}$ (D) $e^{-16}$

   ›Reveal solutionSolution

   This tests the standard $1^\infty$ exponential limit technique. Taking logs and expanding to first order in $x^2$ gives $k=e^{-8}$.

   Concept and Intuition

   Expressions of the form (base $\to1$)$^{\text{power}\to\infty}$ are evaluated by writing $L=e^{\lim(\text{power})\cdot\log(\text{base})}$ and expanding $\log(1+u)\approx u$ for the small quantity $u=$ base$-1$. The constant added to the exponent (here $+3$) never survives, since it multiplies a term that itself $\to0$.

   Step-by-Step Solution

   1. For continuity at $x=0$: $k=\lim_{x\to0}\left(\frac{2+3x^2}{2+5x^2}\right)^{8/x^2+3}$.
   2. Let $L$ denote this limit. Then $\log L=\lim_{x\to0}\left(\frac{8}{x^2}+3\right)\log\left(\frac{2+3x^2}{2+5x^2}\right)$.
   3. Write the base as $1+u$ where $u=\frac{2+3x^2}{2+5x^2}-1=\frac{-2x^2}{2+5x^2}\approx -x^2$ for small $x$ (to leading order).
   4. $\log(1+u)\approx u\approx -x^2$ for small $x$. …

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10. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
    Q.If $f(x) = \dfrac{\log_e\left(1 + x^2(\tan x)\right)}{\sin x^3}$, $x \neq 0$ is to be continuous at $x = 0$, then $f(0)$ must be equal to ______ (A) $1$ (B) $0$ (C) $\dfrac{1}{2}$ (D) $-1$

    ›Reveal solutionSolution

    Using the small-angle equivalences $\log_e(1+u)\sim u$, $\tan x\sim x$, and $\sin x^3\sim x^3$ near $x=0$, the limit of $f(x)$ works out to $1$, so $f(0)=1$ is required for continuity.

    Concept and Intuition

    For a function to be continuous at a removable-looking point like $x=0$, its value there must be defined equal to the limit of the function as $x\to0$. Standard small-$x$ equivalences ($\log_e(1+u)\sim u$ for small $u$; $\tan x\sim x$; $\sin x\sim x$) let us evaluate the $\frac{0}{0}$-type limit cleanly.

    Step-by-Step Solution

    1. As $x\to0$, let $u=x^2\tan x$. Since $\tan x\sim x$ for small $x$, $u\sim x^2\cdot x=x^3\to0$.
    2. Using $\log_e(1+u)\sim u$ for small $u$: $\log_e(1+x^2\tan x)\sim x^2\tan x\sim x^3$.
    3. For the denominator, $\sin(x^3)\sim x^3$ as $x\to0$ (since $\sin\theta\sim\theta$ for small $\theta$, here $\theta=x^3\to0$).
    4. So $f(x)=\dfrac{\log_e(1+x^2\tan x)}{\sin(x^3)}\to\dfrac{x^3}{x^3}=1$ as $x\to0$. …

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11. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.If $f(x) = \dfrac{\cos^4 x - 1}{x^2}$, $x \ne 0$ and $f(0) = 2$ is a real valued function, then (A) $\displaystyle\lim_{x \to 0} f(x)$ does not exist (B) $\displaystyle\lim_{x \to 0} f(x) = 1$ (C) f is not continuous at x = 0 (D) f is continuous at x = 0

    ›Reveal solutionSolution

    This tests factoring $\cos^4x-1$ and using the standard limit $\frac{\sin x}{x}\to1$; the limit exists and equals $-2$, but $f(0)=2$, so $f$ fails continuity at $0$.

    Concept and Intuition

    $\cos^4x-1$ factors as a difference of squares twice: $\cos^4x-1=(\cos^2x-1)(\cos^2x+1)=-\sin^2x(\cos^2x+1)$. This lets us isolate the familiar $\left(\frac{\sin x}{x}\right)^2\to1$ building block.

    Step-by-Step Solution

    1. $\cos^4x - 1 = (\cos^2x-1)(\cos^2x+1) = -\sin^2x\,(\cos^2x+1)$.
    2. $f(x) = \dfrac{-\sin^2x(\cos^2x+1)}{x^2} = -\left(\dfrac{\sin x}{x}\right)^2(\cos^2x+1)$.
    3. As $x\to0$: $\left(\dfrac{\sin x}{x}\right)^2\to1$ and $\cos^2x+1\to2$.
    4. So $\displaystyle\lim_{x\to0}f(x) = -1\times2 = -2$ — this limit exists (rules out (A)) and is not $1$ (rules out (B)). …

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12. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.If $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=\begin{cases} \dfrac{\sin x - \sin\dfrac{x}{2}}{x}, & x<0 \\[2mm] \dfrac{\sqrt{x^2+x}-\sqrt{x}}{x^{3/2}}, & x>0 \end{cases}$ is continuous on $\mathbb{R}$, then $f(0)=$ (A) $1/2$ (B) $3/2$ (C) $1$ (D) $-1$

    ›Reveal solutionSolution

    This tests continuity of a piecewise function at the junction point using standard small-$x$ approximations; both one-sided limits equal $1/2$.

    Concept and Intuition

    A piecewise function is continuous at a boundary point only if the left-hand limit, right-hand limit, and the defined value at that point all agree. Here $f(0)$ isn't given directly by either branch (both blow up as $0/0$), so we must compute the limits from each side and set $f(0)$ equal to their common value.

    Step-by-Step Solution

    1. Left-hand limit ($x\to 0^-$):

    $\lim_{x\to0^-}\frac{\sin x-\sin(x/2)}{x}$

    Use the small-angle expansion $\sin t = t - \dfrac{t^3}{6}+\cdots$:

    $\sin x - \sin\frac{x}{2} = \left(x-\frac{x^3}{6}\right)-\left(\frac{x}{2}-\frac{x^3}{48}\right)+\cdots = \frac{x}{2} + O(x^3)$

    Dividing by $x$: the limit is $\dfrac12$.

    2. Right-hand limit ($x\to 0^+$):

    $\lim_{x\to0^+}\frac{\sqrt{x^2+x}-\sqrt{x}}{x^{3/2}} = \lim_{x\to0^+}\frac{\sqrt{x}\left(\sqrt{x+1}-1\right)}{x^{3/2}} = \lim_{x\to0^+}\frac{\sqrt{x+1}-1}{x}$

    Using $\sqrt{1+x}\approx 1+\dfrac{x}{2}-\dfrac{x^2}{8}+\cdots$:

    $\frac{\sqrt{x+1}-1}{x}\approx \frac{x/2}{x} = \frac12$ …

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