Q.Discuss the continuity of the function $f$ given by $f(x) = |x|$ at $x = 0$.

Showing the 12 most recent of 32 on this concept.

1. 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 $x\to0$: $\dfrac{1}{\sqrt{1}+1} = \dfrac12$.
   3. For continuity at $x=0$: $f(0) = \dfrac12$.

   Common Mistakes

   • Forgetting to rationalize and instead trying to plug $x=0$ directly (giving an indeterminate $0/0$).
   • Sign error in the conjugate multiplication.

   ✓Final answer

   The correct option is (C) — $\dfrac{1}{2}$.

   ANSWER: C

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 $f(x) = \dfrac{\cos^4 x - 1}{x^2}$, $x \ne 0$ 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

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. $e^{a}=e^{2/3}=b$.
   4. Matching exponents: $a=\dfrac23$, and correspondingly $b=e^{2/3}$.

   Common Mistakes

   • Not recognizing the $(1+u)^{1/u}\to e$ form and instead trying to directly evaluate $u^{a/u}$ (which behaves completely differently — it vanishes, losing the $a$-dependence entirely).
   • Small-angle slip: using $\tan kx\approx kx$ only for the numerator or only for the denominator.

   ✓Final answer

   The correct option is (B) — $a=\dfrac23,\ b=e^{2/3}$.

   ANSWER: B

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 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

7. 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 $x\to0$.
   5. For $f$ to be continuous at $x=0$, we must define $f(0)=\displaystyle\lim_{x\to0}f(x)=1$.

   Common Mistakes

   • Using $\tan x\sim x$ but then forgetting to also apply $\log_e(1+u)\sim u$, leading to an incorrect order-of-magnitude comparison.
   • Mixing up $\sin(x^3)$ with $(\sin x)^3$ — here it is $\sin$ evaluated at $x^3$, which is still $\sim x^3$ for small $x$.

   ✓Final answer

   The correct option is (A) — $1$.

   ANSWER: A

8. 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).
   3. So the only requirement is $a+b=1$. Testing the options: (A) $1+1=2$, (B) $-\tfrac12-\tfrac32=-2$, (C) $\tfrac32-\tfrac12=1$ ✓, (D) $\tfrac12-\tfrac32=-1$.
   4. Only (C) satisfies the condition.

   Common Mistakes

   • Mixing up which branch is active on which side of $x=\pm1$.
   • Not noticing that both conditions reduce to the same single equation, and instead searching for two independent equations.

   ✓Final answer

   The correct option is (C) — $a=\dfrac32,\ b=-\dfrac12$.

   ANSWER: C

9. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.If a function $f(x)=\begin{cases}\dfrac{\tan((\alpha+1)x)+\tan2x}{x} & \text{if } x>0\\ \beta & \text{at } x=0\\ \dfrac{\sin3x-\tan3x}{x^3} & \text{if } x<0\end{cases}$ is continuous at $x=0$ then $|\alpha|+|\beta|=$ (A) $60$ (B) $30$ (C) $15$ (D) $45$

   ›Reveal solutionSolution

   Continuity at $x=0$ forces both one-sided limits to equal $\beta=f(0)$; compute each limit via small-angle expansions and solve for $\alpha,\beta$.

   Concept and Intuition

   For a piecewise function to be continuous at a point, the left-hand limit, the right-hand limit, and the function's value there must all agree. Here both one-sided limits are $0/0$-type indeterminate forms requiring standard small-$x$ expansions of $\tan$ and $\sin$.

   Step-by-Step Solution

   1. Right-hand limit ($x\to0^+$): using $\lim_{x\to0}\tan(kx)/x=k$, $\lim_{x\to0^+}\frac{\tan((\alpha+1)x)+\tan2x}{x}=(\alpha+1)+2=\alpha+3.$ This must equal $\beta$: $\beta=\alpha+3$.
   2. Left-hand limit ($x\to0^-$): expand $\sin3x\approx3x-\dfrac{(3x)^3}{6}=3x-4.5x^3$ and $\tan3x\approx3x+\dfrac{(3x)^3}{3}=3x+9x^3$.
   3. $\sin3x-\tan3x\approx(3x-4.5x^3)-(3x+9x^3)=-13.5x^3=-\dfrac{27}{2}x^3$.
   4. So $\lim_{x\to0^-}\dfrac{\sin3x-\tan3x}{x^3}=-\dfrac{27}{2}$. This must also equal $\beta$: $\beta=-\dfrac{27}{2}$.
   5. From step 1: $\alpha=\beta-3=-\dfrac{27}{2}-3=-\dfrac{33}{2}$.
   6. $|\alpha|+|\beta|=\dfrac{33}{2}+\dfrac{27}{2}=\dfrac{60}{2}=30$.

   Common Mistakes

   • Using $\tan u\approx u+u^3/3$ but forgetting to cube the coefficient properly (using $(3x)^3=27x^3$, not $3x^3$).
   • Sign confusion between the $\sin$ and $\tan$ expansions' cubic-term coefficients ($-1/6$ vs $+1/3$).

   ✓Final answer

   The correct option is (B) — $30$.

   ANSWER: B

10. 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 $e^{3/x}$: $R=\lim_{x\to0^+}\dfrac{e^{-1/x}+1}{ae^{-1/x}+b}$.
    3. As $x\to0^+$, $-1/x\to-\infty$, so $e^{-1/x}\to0$. Hence $R=\dfrac{0+1}{0+b}=\dfrac1b$.
    4. Continuity requires $R=f(0)=a$, so $\dfrac1b=e\Rightarrow b=\dfrac1e$.
    5. $ab=e\cdot\dfrac1e=1$.

    Common Mistakes

    • Dividing by $e^{2/x}$ instead of $e^{3/x}$ — since $3/x>2/x$ for $x>0$, $e^{3/x}$ is the dominant (larger) term, and it must be the one factored out.
    • Missing that $a$ must equal BOTH the left limit and match $f(0)$, and that the right limit is a separate equation in $b$.

    ✓Final answer

    The correct option is (C) — $1$.

    ANSWER: C

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 $f(x) = \dfrac{N(x)}{D(x)} = \dfrac{-2ax}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}} \times \dfrac{\sqrt{a+x}+\sqrt{a-x}}{2x}$.
    4. Cancel $2x$: $f(x) = \dfrac{-a\left(\sqrt{a+x}+\sqrt{a-x}\right)}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}$.
    5. Take $x\to0$: numerator surd sum $\to 2\sqrt a$, denominator surd sum $\to 2\sqrt{a^2}=2a$ (taking $a>0$).
    6. $K = \lim_{x\to0} f(x) = \dfrac{-a\cdot 2\sqrt a}{2a} = -\sqrt a$.

    Common Mistakes

    • Forgetting to rationalize the denominator too, and trying L'Hôpital's rule with messy nested radicals instead.
    • Sign error when expanding $(a^2-ax+x^2)-(a^2+ax+x^2)=-2ax$.

    ✓Final answer

    The correct option is (A) — $-\sqrt{a}$.

    ANSWER: A

12. 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 $\sqrt{16+\sqrt x}+4$:

    $\frac{\sqrt x\,(\sqrt{16+\sqrt x}+4)}{(16+\sqrt x)-16}=\frac{\sqrt x\,(\sqrt{16+\sqrt x}+4)}{\sqrt x}=\sqrt{16+\sqrt x}+4.$

    3. As $x\to0^+$: $\sqrt{16+\sqrt x}+4\to\sqrt{16}+4=4+4=8$.
    4. Both one-sided limits equal $8$, so continuity at $x=0$ requires $f(0)=a=8$.

    Common Mistakes

    • Forgetting the factor of $4$ that arises from $2\sin^2(2x)/x^2 = 2\cdot(\sin2x/2x)^2\cdot4$ — easy to drop the extra $4$ from $(2x)^2$ vs $x^2$.
    • Not rationalising the surd expression and instead trying (invalid) direct substitution, which gives $0/0$.

    ✓Final answer

    The correct option is (D) — $8$.

    ANSWER: D