Q.Find the value of $k$ so that the function $f$ is continuous at the indicated point: $f(x) = \begin{cases} \dfrac{2^{x+2} - 16}{4^x - 16}, & x \ne 2 \\ k, & x = 2 \end{cases}$ at $x = 2$.

Showing the 12 most recent of 32 on this concept.

1. 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 $x$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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 $\displaystyle\lim_{x\to0}\frac{2^x-1}{\sqrt{1+x}-1}=\lim_{x\to0}\frac{x\ln2}{x/2}=2\ln2$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $f(x) = \begin{cases} \dfrac{\sqrt{2 + \cos x} - 1}{(\pi - x)^2}, & x \neq \pi \\ k, & x = \pi \end{cases}$ is continuous at $x = \pi$, then $k =$ (A) $1$ (B) $\dfrac{1}{2}$ (C) $2$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   Continuity at $x=\pi$ requires $k$ to equal the limit of the given expression as $x\to\pi$; using a small-angle substitution, that limit is $\dfrac14$.

   Concept and Intuition

   For $f$ to be continuous at $x=\pi$, we need $k = \displaystyle\lim_{x\to\pi} \frac{\sqrt{2+\cos x}-1}{(\pi-x)^2}$. Substituting $x = \pi - h$ (so $h\to 0$ as $x\to\pi$) converts the trig limit into a small-$h$ approximation problem, where standard expansions ($\cos h \approx 1-h^2/2$, $\sqrt{1+t}\approx 1+t/2$) make the limit easy to evaluate.

   Step-by-Step Solution

   1. Let $x = \pi - h$, so as $x\to \pi$, $h\to 0$, and $\pi - x = h$.
   2. $\cos x = \cos(\pi - h) = -\cos h$.
   3. So $2+\cos x = 2 - \cos h$. Using $\cos h \approx 1 - \dfrac{h^2}{2}$ for small $h$: $2-\cos h \approx 2 - 1 + \dfrac{h^2}{2} = 1 + \dfrac{h^2}{2}$.
   4. $\sqrt{2+\cos x} \approx \sqrt{1+\dfrac{h^2}{2}} \approx 1 + \dfrac{h^2}{4}$ (using $\sqrt{1+t}\approx 1+t/2$ with $t=h^2/2$).
   5. So $\sqrt{2+\cos x} - 1 \approx \dfrac{h^2}{4}$.
   6. The denominator is $(\pi-x)^2 = h^2$. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 $=\dfrac{x^2\ln9\ln8}{\sqrt2 x^2/8}=\dfrac{8\ln9\ln8}{\sqrt2}$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If a function f defined by $f(x) = \begin{cases} \dfrac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{cases}$ is continuous at $x = 0$, then $a =$ (A) $8$ (B) $4$ (C) $3$ (D) $2$

   ›Reveal solutionSolution

   Both one-sided limits at $x=0$ evaluate to $8$ (via $1-\cos\theta=2\sin^2(\theta/2)$ on the left, and rationalizing on the right), so $a=8$.

   Concept and Intuition

   For $f$ to be continuous at $0$, the left-hand limit, right-hand limit, and $f(0)=a$ must all agree. The left piece is a classic $0/0$ trig limit solved via the half-angle identity for $1-\cos\theta$; the right piece is a classic surd limit solved by rationalizing the denominator.

   Step-by-Step Solution

   1. Left limit: $1-\cos4x=2\sin^2(2x)$, so $\dfrac{1-\cos4x}{x^2}=\dfrac{2\sin^2(2x)}{x^2}=8\left(\dfrac{\sin2x}{2x}\right)^2$.
   2. As $x\to0^-$, $\dfrac{\sin2x}{2x}\to1$, so this limit $\to8(1)^2=8$.
   3. Right limit: let $t=\sqrt x$ (so $t\to0^+$ as $x\to0^+$): $\dfrac{\sqrt x}{\sqrt{16+\sqrt x}-4}=\dfrac{t}{\sqrt{16+t}-4}$.
   4. Rationalize: multiply by $\dfrac{\sqrt{16+t}+4}{\sqrt{16+t}+4}$: $\dfrac{t(\sqrt{16+t}+4)}{(16+t)-16}=\dfrac{t(\sqrt{16+t}+4)}{t}=\sqrt{16+t}+4$. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $f(x)$ defined by $f(x) = \begin{cases} \dfrac{1 - \tan x}{4x - \pi}, & x \neq \dfrac{\pi}{4}, x \in \left[0, \dfrac{\pi}{2}\right] \\ k, & x = \dfrac{\pi}{4} \end{cases}$ is continuous in $\left[0, \dfrac{\pi}{2}\right]$, then $k =$ (A) $-1$ (B) $-\dfrac{1}{2}$ (C) $\dfrac{1}{2}$ (D) $1$

   ›Reveal solutionSolution

   Substituting $x=\pi/4+h$ and using the tangent subtraction identity, the limit of

   $\dfrac{1-\tan x}{4x-\pi}$ as $x\to\pi/4$ works out to $-\dfrac12$, which is the required

   value of $k$.

   Concept and Intuition

   For continuity at $x=\pi/4$, we need $k=\lim_{x\to\pi/4}f(x)$. Since the expression is $0/0$

   at $x=\pi/4$, shifting variables via $x=\pi/4+h$ (so $h\to0$) turns $\tan x$ into

   $\tan(\pi/4+h)$, which has a clean expansion using the tangent-addition formula.

   Step-by-Step Solution

   1. Let $x=\dfrac{\pi}{4}+h$, so $h\to0$ as $x\to\pi/4$.
   2. $\tan\left(\dfrac{\pi}{4}+h\right)=\dfrac{1+\tan h}{1-\tan h}$.
   3. $1-\tan x = 1-\frac{1+\tan h}{1-\tan h}=\frac{(1-\tan h)-(1+\tan h)}{1-\tan h}=\frac{-2\tan h}{1-\tan h}$
   4. $4x-\pi = 4\left(\dfrac{\pi}{4}+h\right)-\pi = 4h$.
   5. So

   $f(x)=\frac{-2\tan h}{4h(1-\tan h)}$

   6. As $h\to0$, $\tan h\sim h$, so …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If the function f defined by $f(x) = \begin{cases} \dfrac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{cases}$ is continuous at $x = 0$, then $a =$ (A) $1$ (B) $2$ (C) $4$ (D) $8$

   ›Reveal solutionSolution

   Continuity at $x=0$ requires the left limit, the right limit, and $f(0)=a$ to all agree; both one-sided limits work out to $8$. Answer: $a=8$.

   Concept and Intuition

   For a piecewise function to be continuous at a junction point, the value defined there ($a$) must equal both the limit approaching from the left and the limit approaching from the right. Each side here needs a standard trick: the trig side uses the double-angle identity $1-\cos\theta=2\sin^2(\theta/2)$, and the surd side needs rationalisation to remove the $\sqrt{\ }-\sqrt{\ }$ indeterminate form.

   Step-by-Step Solution

   1. Left-hand limit ($x\to0^-$): $1-\cos4x=2\sin^2(2x)$, so

   $\lim_{x\to0^-}\frac{1-\cos4x}{x^2}=\lim_{x\to0^-}\frac{2\sin^2(2x)}{x^2}=2\lim_{x\to0}\left(\frac{\sin2x}{2x}\right)^2\cdot4=2\cdot1\cdot4=8.$

   2. Right-hand limit ($x\to0^+$): rationalise $\dfrac{\sqrt x}{\sqrt{16+\sqrt x}-4}$ by multiplying top and bottom by $\sqrt{16+\sqrt x}+4$: $\frac{\sqrt x\,(\sqrt{16+\sqrt x}+4)}{(16+\sqrt x)-16}=\frac{\sqrt x\,(\sqrt{16+\sqrt x}+4)}{\sqrt x}=\sqrt{16+\sqrt x}+4.$ …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If $f(x) = \dfrac{(1-\sin x)}{\log(1+\pi^2 - 4\pi x + 4x^2)}$ is continuous at $x = \pi/2$, then $f(\pi/2) =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{8}$ (C) $\dfrac{1}{16}$ (D) $\dfrac{1}{32}$

   ›Reveal solutionSolution

   Recognising $1+\pi^2-4\pi x+4x^2$ as $1+(2x-\pi)^2$ turns this into a small-angle limit; the continuity value is $1/8$.

   Concept and Intuition

   For $f$ to be continuous at $x=\pi/2$, $f(\pi/2)$ must equal $\lim_{x\to\pi/2} f(x)$. The denominator's quadratic in $x$ is a perfect "sum-of-squares" shift once you notice $\pi^2-4\pi x+4x^2=(2x-\pi)^2$, turning this into a standard small-$t$ limit using $1-\cos t\approx t^2/2$ and $\log(1+u)\approx u$.

   Step-by-Step Solution

   1. Rewrite the denominator: $1+\pi^2-4\pi x+4x^2 = 1+(2x-\pi)^2$.
   2. Let $t=x-\pi/2$, so $x\to\pi/2 \iff t\to 0$, and $2x-\pi=2t$.
   3. Numerator: $1-\sin x = 1-\sin(\pi/2+t) = 1-\cos t$. For small $t$, $1-\cos t \approx \dfrac{t^2}{2}$.
   4. Denominator: $\log(1+(2t)^2) = \log(1+4t^2) \approx 4t^2$ for small $t$ (since $\log(1+u)\approx u$). …

   🔒 Start your 14-day free trial to see the full solution →
9. 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 →
10. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
    Q.If a function $f(x) = \begin{cases} \dfrac{2^{\sin^2 x} - \cos bx}{\tan^2 x}, & \text{for } -\dfrac{\pi}{2} < x < 0 \\ 2, & \text{for } x = 0 \\ \dfrac{\sin^2 ax - \sin^2 bx}{1 - \cos x}, & \text{for } 0 < x < \dfrac{\pi}{2} \end{cases}$ is continuous at $x = 0$, then $a^2 + b^2 =$ (A) $9$ (B) $9 - \log 16$ (C) $9 - \log 8$ (D) $1$

    ›Reveal solutionSolution

    Continuity at $x=0$ forces both one-sided limits to equal $f(0)=2$; expanding each piece to second order in $x$ gives two equations in $a^2,b^2$ whose sum is $9-\log16$.

    Concept and Intuition

    A piecewise function is continuous at a boundary point exactly when the left-hand limit, right-hand limit, and the defined value all coincide. Here each branch is a $\frac00$-type expression as $x\to0$, so we expand numerator and denominator to matching (second) order in $x$ and read off the limiting constant.

    Step-by-Step Solution

    1. Right piece, $0<x<\pi/2$: $f(x)=\dfrac{\sin^2(ax)-\sin^2(bx)}{1-\cos x}$. Using $\sin u \approx u$ for small $u$: $\sin^2(ax)-\sin^2(bx)\approx a^2x^2-b^2x^2=(a^2-b^2)x^2$. Using $1-\cos x \approx \dfrac{x^2}{2}$: limit $= \dfrac{(a^2-b^2)x^2}{x^2/2}=2(a^2-b^2)$. Continuity requires this $=f(0)=2$, so $a^2-b^2=1$. — (i)
    2. Left piece, $-\pi/2<x<0$: $f(x)=\dfrac{2^{\sin^2x}-\cos(bx)}{\tan^2x}$. $2^{\sin^2x}=e^{\sin^2x\ln2}\approx 1+x^2\ln2$ (using $\sin^2x\approx x^2$). $\cos(bx)\approx 1-\dfrac{b^2x^2}{2}$. Numerator $\approx x^2\ln2 + \dfrac{b^2x^2}{2} = x^2\left(\ln2+\dfrac{b^2}{2}\right)$. Denominator $\tan^2x \approx x^2$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.If a function defined by $f(x) = \dfrac{(3^x - 1)^2}{\sin x \log(1+x)}$, $x \ne 0$, is continuous at $x = 0$, then $f(0) =$ (A) $2\log 3$ (B) $\log 3^2$ (C) $2 + \log 3$ (D) $(\log 3)^2$

    ›Reveal solutionSolution

    Standard small-$x$ equivalents give $f(x)\to(\ln3)^2$ as $x\to0$, so $f(0)=(\log3)^2$.

    Concept and Intuition

    For continuity at $x=0$, $f(0)$ must equal $\lim_{x\to0}f(x)$. Use the standard limits $\lim_{x\to0}\dfrac{a^x-1}{x}=\ln a$, $\lim_{x\to0}\dfrac{\sin x}{x}=1$, $\lim_{x\to0}\dfrac{\log(1+x)}{x}=1$.

    Step-by-Step Solution

    1. $(3^x-1)^2 = x^2\left(\dfrac{3^x-1}{x}\right)^2 \to x^2(\ln3)^2$ as $x\to0$.
    2. $\sin x \log(1+x) = x\cdot\dfrac{\sin x}{x}\cdot x\cdot\dfrac{\log(1+x)}{x} = x^2\cdot\dfrac{\sin x}{x}\cdot\dfrac{\log(1+x)}{x} \to x^2$ as $x\to0$.
    3. So $f(x)=\dfrac{(3^x-1)^2}{\sin x\log(1+x)} \to \dfrac{x^2(\ln3)^2}{x^2}=(\ln3)^2$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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). …

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