Q.Find whether the function is continuous or discontinuous at the indicated point: $f(x) = |x| + |x - 1|$ at $x = 1$.

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.Which one of the following function is discontinuous at x=1? (A) $f(x)=\sin^2x+\tan^2x+\cos^2x-\sec^2x$ (B) $f(x)=\dfrac{1}{1+2^{\sin x}}$ (C) $f(x)=\begin{cases}\dfrac{x-1}{|x-1|+2(x-1)^2}, & x\neq1\\ 1, & x=1\end{cases}$ (D) $f(x)=e^{x}+5$

   ›Reveal solutionSolution

   Options (A), (B), (D) simplify to functions that are continuous everywhere; option (C)'s one-sided limits at $x=1$ disagree ($+1$ from the right, $-1$ from the left), so it alone is discontinuous at $x=1$.

   Concept and Intuition

   For piecewise/rational-looking expressions, check whether the identity $\sin^2x+\cos^2x=1$ and $\sec^2x-\tan^2x=1$ collapse them to constants (continuous), whether a denominator can vanish, and — for absolute-value expressions — whether the left and right limits genuinely agree.

   Step-by-Step Solution

   1. (A) $f(x)=\sin^2x+\tan^2x+\cos^2x-\sec^2x=(\sin^2x+\cos^2x)-(\sec^2x-\tan^2x)=1-1=0$ for all $x$ where $\tan,\sec$ are defined; at $x=1$ (radian) $\cos1\ne0$, so it's fine and continuous ($\equiv0$ near $x=1$).
   2. (B) $f(x)=\dfrac{1}{1+2^{\sin x}}$: since $2^{\sin x}>0$ always, the denominator never vanishes ($\ge1+2^{-1}=1.5$); continuous everywhere, including $x=1$.
   3. (C) For $x\ne1$, let $t=x-1$: $f=\dfrac{t}{|t|+2t^2}$.
      • As $t\to0^+$ ($x\to1^+$): $|t|=t$, so $f=\dfrac{t}{t+2t^2}=\dfrac{1}{1+2t}\to1$.
      • As $t\to0^-$ ($x\to1^-$): $|t|=-t$, so $f=\dfrac{t}{-t+2t^2}=\dfrac{1}{-1+2t}\to-1$. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.Let $f(x) = \begin{cases} \dfrac{1}{|x|}, & \text{for } |x| > 1 \\ ax^2+b, & \text{for } |x| \le 1 \end{cases}$. If $\displaystyle\lim_{x\to 1} f(x)$ and $\displaystyle\lim_{x\to -1} f(x)$ exist, then the possible values for a and b are (A) $a = b = 1$ (B) $a = -\dfrac{1}{2}, b = -\dfrac{3}{2}$ (C) $a = \dfrac{3}{2}, b = -\dfrac{1}{2}$ (D) $a = \dfrac{1}{2}, b = -\dfrac{3}{2}$

   ›Reveal solutionSolution

   Both one-sided limits at $x=\pm1$ force $a+b=1$; checking the options, only $a=\tfrac32,\,b=-\tfrac12$ satisfies this.

   Concept and Intuition

   The function is piecewise, switching definition exactly at $|x|=1$. For the limit to exist at a switch-point, the two pieces must approach the same value from either side — this is the usual "match the boundary values" condition for piecewise functions.

   Step-by-Step Solution

   1. Near $x=1$: for $x$ slightly less than $1$ we're in the $ax^2+b$ branch ($|x|\le1$); for $x$ slightly more than $1$ we're in the $1/|x|$ branch.
      • Left limit: $a(1)^2+b=a+b$.
      • Right limit: $\dfrac{1}{|1|}=1$.
      • Condition: $a+b=1$.
   2. Near $x=-1$: for $x$ slightly less than $-1$ (i.e. $|x|>1$), branch is $1/|x|$; for $x$ slightly more than $-1$ (i.e. $|x|<1$), branch is $ax^2+b$.
      • Left limit: $1/|{-1}|=1$.
      • Right limit: $a(-1)^2+b=a+b$.
      • Condition: $a+b=1$ (same equation again). …

   🔒 Start your 14-day free trial to see the full solution →
3. 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. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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)). …

   🔒 Start your 14-day free trial to see the full solution →
5. 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. $e^{a}=e^{2/3}=b$. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.Let $S_n=1+3x+9x^2+27x^3+\ldots n$ terms and $-\frac{1}{3}<x<\frac{1}{3}$. If $\lim_{n\to\infty}S_n=f(x)$, then $f(x)$ is discontinuous at the point $x=$ (A) $0$ (B) $\frac{1}{3}$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   The infinite geometric series sums to $f(x)=1/(1-3x)$ on $(-1/3,1/3)$, and this function has an infinite discontinuity exactly at $x=1/3$, the edge of the interval where the series stops converging.

   Concept and Intuition

   An infinite geometric series $1+r+r^2+\dots$ converges to $\frac{1}{1-r}$ only when $|r|<1$; as $r\to1$ this sum diverges to infinity. Here $r=3x$, so the series converges for $|x|<1/3$ and the resulting closed-form function has a vertical asymptote right at the boundary $x=1/3$.

   Step-by-Step Solution

   1. $S_n = 1+3x+(3x)^2+\dots+(3x)^{n-1}$, a geometric series with first term $1$ and common ratio $3x$.
   2. Sum: $S_n=\dfrac{1-(3x)^n}{1-3x}$ (for $3x\ne1$).
   3. For $-\tfrac13<x<\tfrac13$, $|3x|<1$, so $(3x)^n\to0$ as $n\to\infty$.
   4. Hence $f(x)=\lim_{n\to\infty}S_n = \dfrac{1}{1-3x}$.
   5. This function $f(x)=\frac{1}{1-3x}$ is undefined/blows up exactly where $1-3x=0$, i.e. $x=\tfrac13$ — an infinite discontinuity right at the edge of the domain of convergence. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If a real valued function $f(x) = \begin{cases} (1+\sin x)^{\operatorname{cosec}x}, & -\pi/2 < x < 0 \\ a, & x = 0 \\ \dfrac{e^{2/x}+e^{3/x}}{ae^{2/x}+be^{3/x}}, & 0 < x < \pi/2 \end{cases}$ is continuous at x = 0, then $ab =$ (A) $e$ (B) $e^2$ (C) 1 (D) $-1$

   ›Reveal solutionSolution

   The left-hand limit is the classical $1^\infty$ form giving $e$, fixing $a=e$;

   the right-hand limit needs dividing by the dominant exponential to fix $b$.

   Continuity forces both, and $ab=1$.

   Concept and Intuition

   A piecewise function is continuous at a point only if the left-hand limit,

   right-hand limit, and the function's value there all agree. Here the left

   piece is a $1^\infty$ indeterminate form (standard trick: exponentiate and use

   $\log(1+u)\sim u$), and the right piece is a ratio of two exponentials growing at

   different rates as $x\to0^+$ (since $1/x\to+\infty$), so the faster-growing

   exponential $e^{3/x}$ dominates and everything else becomes negligible after

   dividing through by it.

   Step-by-Step Solution

   1. Left limit: $L=\lim_{x\to0^-}(1+\sin x)^{\operatorname{cosec}x}$. Take logs: $\log L=\lim \operatorname{cosec}x\cdot\log(1+\sin x)=\lim\dfrac{\sin x}{\sin x}\left(1+O(\sin x)\right)\to1$. So $L=e$. Continuity requires $f(0)=a=L=e$.
   2. Right limit: $R=\lim_{x\to0^+}\dfrac{e^{2/x}+e^{3/x}}{ae^{2/x}+be^{3/x}}$. Divide numerator and denominator by $e^{3/x}$: $R=\lim_{x\to0^+}\dfrac{e^{-1/x}+1}{ae^{-1/x}+b}$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 $x\to0$: $\dfrac{1}{\sqrt{1}+1} = \dfrac12$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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). …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.If $f(x)=\begin{cases}\dfrac{\sqrt{a^2-ax+x^2}-\sqrt{x^2+ax+a^2}}{\sqrt{a+x}-\sqrt{a-x}}, & x\ne0\\K, & x=0\end{cases}$ is continuous at $x=0$, then $K=$ (A) $-\sqrt{a}$ (B) $\sqrt{a}$ (C) $-1$ (D) $a+\sqrt{a}$

    ›Reveal solutionSolution

    This is a $0/0$ form at $x=0$; rationalizing both the numerator and denominator turns it into a clean limit that evaluates to $-\sqrt a$.

    Concept and Intuition

    Whenever both numerator and denominator vanish at the point of interest, multiplying each by its conjugate surd converts the difference-of-square-roots into a simple polynomial difference, which then cancels the common factor causing the indeterminacy.

    Step-by-Step Solution

    1. Let $N(x)=\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}$. Multiply and divide by the conjugate:

    $N(x) = \frac{(a^2-ax+x^2)-(a^2+ax+x^2)}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}} = \frac{-2ax}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}$

    2. Let $D(x)=\sqrt{a+x}-\sqrt{a-x}$. Similarly,

    $D(x) = \frac{(a+x)-(a-x)}{\sqrt{a+x}+\sqrt{a-x}} = \frac{2x}{\sqrt{a+x}+\sqrt{a-x}}$

    3. So $f(x) = \dfrac{N(x)}{D(x)} = \dfrac{-2ax}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}} \times \dfrac{\sqrt{a+x}+\sqrt{a-x}}{2x}$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.If a function $f(x)$ defined on $[a, b]$ is discontinuous at $x = \alpha \in (a, b)$, then (A) $\lim\limits_{x \to \alpha^-} f(x) = \lim\limits_{x \to \alpha^+} f(x) = f(\alpha)$ (B) $\lim\limits_{x \to \alpha} f(x) \neq f(\alpha)$ (C) $\lim\limits_{x \to a^-} f(x) = f(a)$ (D) $\lim\limits_{x \to b^+} f(x) = f(b)$

    ›Reveal solutionSolution

    Discontinuity at an interior point is, by definition, the failure of "limit equals function value" there.

    Concept and Intuition

    A function is continuous at $\alpha$ iff $\lim_{x\to\alpha}f(x)=f(\alpha)$ (which subsumes both one-sided limits agreeing with each other and with $f(\alpha)$). Discontinuity is simply the negation of this statement.

    Step-by-Step Solution

    1. Continuity at $\alpha\in(a,b)$ means $\lim_{x\to\alpha^-}f(x)=\lim_{x\to\alpha^+}f(x)=f(\alpha)$ — this is option (A), which must be false since $f$ is given to be discontinuous at $\alpha$.
    2. The negation of continuity is exactly: the limit at $\alpha$ (however it behaves) is not equal to $f(\alpha)$ — option (B). …

    🔒 Start your 14-day free trial to see the full solution →