Q.Find whether the function is continuous or discontinuous at the indicated point: $f(x) = \begin{cases} 3x + 5, & x \ge 2 \\ x^2, & x < 2 \end{cases}$ at $x = 2$.

Showing the 12 most recent of 32 on this concept.

1. 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).
   5. At $x=5$: for all $x\ge3$ (which includes a neighbourhood of $5$), $f(x)=5-x$ is just a polynomial — continuous everywhere. So there is no discontinuity of any kind at $x=5$, ruling out options (C) and (D).

   Common Mistakes

   • Assuming the "other" piece ($2/(5-x)$) is relevant near $x=5$ — it only applies for $x<3$, nowhere near $5$.
   • Confusing "left discontinuous" with "discontinuous from the right" — here the right side actually matches the true value.

   ✓Final answer

   The correct option is (A) — left discontinuous at $x=3$.

   ANSWER: A

2. 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.
   3. At $x=2$: $f(2)=-\dfrac32$ (from the $x\ge2$ branch). Left-hand limit: $\lim_{x\to2^-}\left(\dfrac12-x\right)=\dfrac12-2=-\dfrac32$. Right-hand limit (same branch, $x\ge2$): $-\dfrac32$.
   4. All three values equal $-\dfrac32$, so $f$ is continuous at $x=2$ — (D) is true.

   Common Mistakes

   • Confusing which piece of the definition applies right at the boundary point itself (e.g. using the open-interval piece instead of the explicitly given value at $x=1$ or $x=2$).
   • Mixing up left/right limits when checking one-sided continuity.

   ✓Final answer

   The correct option is (D) — $f$ is continuous at $x = 2$.

   ANSWER: D

3. 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.
   5. For $x=1<2$, $f(x)=\dfrac{x^2-4}{\sqrt{2-x}}$ is a ratio of continuous functions with non-zero denominator ($\sqrt{2-1}=1\ne0$), so $f$ is continuous at $x=1$ — (D) is false.

   Common Mistakes

   • Assuming continuity requires checking only the given piece definitions without evaluating the actual one-sided limits.
   • Missing that $\log(x-2)\to-\infty$, mistakenly thinking it approaches $\log0$ as some finite quantity.

   ✓Final answer

   The correct option is (B) — f is left continuous at x = 2 when a = 0.

   ANSWER: B

4. 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 $f=\dfrac{t}{-t+2t^2}=\dfrac{1}{-1+2t}\to-1$.
      • Left limit ($-1$) $\ne$ right limit ($1$) $\ne$ the defined value $f(1)=1$: the two-sided limit doesn't even exist, so $f$ is discontinuous at $x=1$.
   4. (D) $f(x)=e^x+5$ is continuous everywhere (elementary function), including $x=1$.

   Common Mistakes

   • Not simplifying (A) via the Pythagorean/secant identities and instead trying to evaluate term by term.
   • Overlooking that in (C) the value $f(1)=1$ happens to match the right-hand limit, tempting one to (wrongly) call it continuous — but the two-sided limit must exist and match, which it doesn't here.

   ✓Final answer

   The correct option is (C) — $f(x)=\begin{cases}\dfrac{x-1}{|x-1|+2(x-1)^2}, & x\neq1\\ 1, & x=1\end{cases}$.

   ANSWER: C

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

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

   3. Match for continuity: Since $f$ is continuous at $0$, $f(0)$ must equal both one-sided limits:

   $f(0) = \frac12$

   Common Mistakes

   • Forgetting to rationalize/simplify the $\sqrt{x^2+x}-\sqrt{x}$ term before taking the limit, leading to an indeterminate form that seems to diverge.
   • Using only the first-order term of $\sin t \approx t$ without checking the higher-order terms vanish appropriately (they do, since we only need the leading behavior).

   ✓Final answer

   The correct option is (A) — $1/2$.

   ANSWER: A

6. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If $f(x) = \begin{cases} \dfrac{x-[x]}{x-2}, & x>2 \\ b, & x=2 \\ \dfrac{|x^2-x-2|}{a(2+x-x^2)}, & -1<x\le 2 \\ 2a-b, & x\le -1 \end{cases}$ is continuous on $\mathbb{R}$, then $\displaystyle\lim_{x\to 0}\dfrac{\sin^2 ax + x\tan bx}{x^2} =$ (A) 0 (B) 1 (C) 2 (D) 3

   ›Reveal solutionSolution

   Continuity of the piecewise function pins down $a=1,b=1$; substituting these into the limit expression and using standard small-angle limits gives $2$.

   Concept and Intuition

   For a piecewise function to be continuous at a junction point, the one-sided limits and the defined value there must all agree. Here the junctions at $x=2$ and $x=-1$ give the equations needed to solve for the unknown constants $a,b$ before the actual limit can be evaluated.

   Step-by-Step Solution

   1. Right limit at $x=2$: for $x$ slightly $>2$, $[x]=2$, so $f(x)=\dfrac{x-2}{x-2}=1$. So $\lim_{x\to2^+}f(x)=1$.
   2. Left limit at $x=2$ (third piece): $x^2-x-2=(x-2)(x+1)$ and $2+x-x^2=-(x-2)(x+1)$. For $x$ near $2^-$: $(x-2)<0,(x+1)>0$, so $|x^2-x-2|=(2-x)(x+1)$ and $2+x-x^2=(2-x)(x+1)$ too. So the ratio simplifies to $\dfrac1a$ throughout $(-1,2)$.
   3. Continuity at $x=2$: $1=b=\dfrac1a \Rightarrow a=1,\ b=1$.
   4. Check at $x=-1$: piece 4 value $=2a-b=2-1=1$; piece 3's limit as $x\to-1^+$ is also $\dfrac1a=1$. Consistent ✓.
   5. Now compute $\displaystyle\lim_{x\to0}\dfrac{\sin^2(x)+x\tan(x)}{x^2}$ (using $a=b=1$): $=\lim_{x\to0}\left(\dfrac{\sin x}{x}\right)^2 + \lim_{x\to0}\dfrac{\tan x}{x} = 1^2+1=2$.

   Common Mistakes

   • Trying to find $a,b$ without first simplifying the absolute-value expression in the third piece.
   • Forgetting to split the limit into the two standard forms $(\sin x/x)^2$ and $\tan x/x$.

   ✓Final answer

   The correct option is (C) — 2.

   ANSWER: C

7. 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 $x$.
   5. So $\left(\frac{8}{x^2}+3\right)(-x^2)=-8-3x^2\to -8$ as $x\to0$ (the $-3x^2$ term vanishes, and the correction beyond leading order in $u$ also vanishes since it's multiplied by $x^2$ then divided again giving $O(x^2)$).
   6. So $\log L=-8\Rightarrow L=e^{-8}=k$.

   Common Mistakes

   • Forgetting the $+3$ term but assuming it changes the answer — it drops out because it is multiplied by a quantity that itself tends to zero.
   • Using $u=\frac{-2x^2}{2}=-x^2$ without checking that higher order corrections vanish in the final limit.

   ✓Final answer

   The correct option is (C) — $e^{-8}$.

   ANSWER: C

8. 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)).
   5. Since $f(0)=2\ne-2=\lim_{x\to0}f(x)$, $f$ is not continuous at $x=0$ — (C) true, (D) false.

   Common Mistakes

   • Stopping after finding the limit exists and wrongly concluding continuity, without comparing it to the given $f(0)$.
   • Sign error when factoring $\cos^4x - 1$.

   ✓Final answer

   The correct option is (C) — f is not continuous at x = 0.

   ANSWER: C

9. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If $f(x) = \begin{cases} \dfrac{(e^{ax} - 1)\log(1+x)}{\sin^2 x}, & \text{if } x > 0 \\ 2, & \text{if } x = 0 \\ \dfrac{\cos 4x - \cos bx}{\tan^2 x}, & \text{if } x < 0 \end{cases}$ is continuous at $x = 0$ then $\sqrt{b^2 - a^2} =$ (A) $4$ (B) $5$ (C) $3$ (D) $7$

   ›Reveal solutionSolution

   Continuity at $x=0$ forces both one-sided limits to equal $f(0)=2$; solving gives $a=2$, $b^2=20$, so $\sqrt{b^2-a^2}=4$.

   Concept and Intuition

   For $f$ to be continuous at $0$, we need $\displaystyle\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)=f(0)=2$. Each one-sided piece is a $0/0$ form that resolves using the standard small-angle equivalences $e^{t}-1\sim t$, $\log(1+t)\sim t$, $\sin t\sim t$, $\tan t \sim t$, and $\cos t\approx 1-\dfrac{t^2}{2}$.

   Step-by-Step Solution

   1. Right-hand limit ($x\to0^+$): using $e^{ax}-1\sim ax$, $\log(1+x)\sim x$, $\sin^2x\sim x^2$,

   $\lim_{x\to0^+}\frac{(e^{ax}-1)\log(1+x)}{\sin^2x}=\lim_{x\to0^+}\frac{(ax)(x)}{x^2}=a.$

   Setting this equal to $f(0)=2$: $a=2$.

   2. Left-hand limit ($x\to0^-$): expand $\cos4x\approx 1-\dfrac{(4x)^2}{2}=1-8x^2$ and $\cos(bx)\approx1-\dfrac{b^2x^2}{2}$, and $\tan^2x\sim x^2$:

   $\cos4x-\cos(bx)\approx \left(1-8x^2\right)-\left(1-\frac{b^2x^2}{2}\right)=x^2\left(\frac{b^2}{2}-8\right).$

   So the limit is $\dfrac{b^2}{2}-8$. Setting this equal to $2$: $\dfrac{b^2}{2}=10\Rightarrow b^2=20$.

   3. Now $b^2-a^2 = 20-4=16$, so $\sqrt{b^2-a^2}=\sqrt{16}=4$.

   Common Mistakes

   • Mixing up which side ($x>0$ vs $x<0$) corresponds to which piece.
   • Forgetting the factor of $\tfrac12$ in the $\cos$ expansion ($1-\cos\theta\approx\theta^2/2$, not $\theta^2$).

   ✓Final answer

   The correct option is (A) — $4$.

   ANSWER: A

10. 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 $\dfrac{\sqrt{16+t}+4}{\sqrt{16+t}+4}$: $\dfrac{t(\sqrt{16+t}+4)}{(16+t)-16}=\dfrac{t(\sqrt{16+t}+4)}{t}=\sqrt{16+t}+4$.
    5. As $t\to0^+$: $\sqrt{16+0}+4=4+4=8$.
    6. Both one-sided limits equal $8$; continuity at $x=0$ requires $f(0)=a=8$.

    Common Mistakes

    • Forgetting the factor of $2$ inside $\left(\dfrac{\sin2x}{2x}\right)^2$ and getting $2$ instead of $8$ for the left limit.
    • Not substituting $t=\sqrt x$ and instead trying to rationalize directly in $x$, which is messier and error-prone.

    ✓Final answer

    The correct option is (A) — $8$.

    ANSWER: A

11. 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 $=\dfrac{x^2\ln9\ln8}{\sqrt2 x^2/8}=\dfrac{8\ln9\ln8}{\sqrt2}$.
    6. $\ln9=2\ln3,\ \ln8=3\ln2 \Rightarrow \ln9\ln8=6\ln2\ln3$, so limit $=\dfrac{48\ln2\ln3}{\sqrt2}=24\sqrt2\,\ln2\ln3$.
    7. For continuity, $k\log2\log3$ must equal this, so $k=24\sqrt2$.

    Common Mistakes

    • Missing the $72=9\times8$ factoring trick and attempting messy repeated differentiation.
    • Sign error turning $1-\cos(x/2)$ into the wrong small-angle coefficient.

    ✓Final answer

    The correct option is (D) — $24\sqrt{2}$.

    ANSWER: D

12. 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 $(\pi-x)^2 = h^2$.
    7. $k = \displaystyle\lim_{h\to 0} \frac{h^2/4}{h^2} = \frac14$.

    Common Mistakes

    • Forgetting the sign flip $\cos(\pi-h)=-\cos h$ (a common trap), which would give the wrong sign inside the square root.
    • Using a first-order (linear) approximation instead of the necessary second-order expansion of $\cos h$, which loses the $h^2$ term needed to match the denominator's order.

    ✓Final answer

    The correct option is (D) — $\dfrac{1}{4}$.

    ANSWER: D