Q.Show that the function $f(x) = |\sin x + \cos x|$ is continuous at $x = \pi$.

Showing the 12 most recent of 32 on this concept.

1. 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 $(\pi-x)^2 = h^2$. …

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2. 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 $\log(1+u)\approx u$). …

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3. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.If $f(x)=\begin{cases}\dfrac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}}, & x\neq -1\\[2mm] \dfrac{1}{\sqrt{\lambda\pi}}, & x=-1\end{cases}$ is right continuous at $x=-1$, then $\lambda=$ (A) $1$ (B) $\pi$ (C) $2\pi$ (D) $2$

   ›Reveal solutionSolution

   Expanding $\cos^{-1}(-1+h)$ near $h=0^+$ and simplifying the resulting $0/0$ form gives the right-hand limit $1/\sqrt{2\pi}$; matching this to $f(-1)=1/\sqrt{\lambda\pi}$ gives $\lambda=2$.

   Concept and Intuition

   Right continuity at $x=-1$ requires $\lim_{x\to-1^+}f(x)=f(-1)$. As $x\to-1^+$, both $\sqrt\pi-\sqrt{\cos^{-1}x}\to0$ and $\sqrt{x+1}\to0$, so we need a careful local expansion of $\cos^{-1}x$ near $x=-1$.

   Step-by-Step Solution

   1. Let $x=-1+h$, $h\to0^+$. Write $\cos^{-1}(-1+h)=\pi-\theta$ where $\theta\to0^+$. Then $\cos(\pi-\theta)=-\cos\theta=-1+h\Rightarrow\cos\theta=1-h$.
   2. For small $\theta$: $\cos\theta\approx1-\theta^2/2$, so $1-\theta^2/2\approx1-h\Rightarrow\theta\approx\sqrt{2h}$.
   3. So $\cos^{-1}x\approx\pi-\sqrt{2h}$, and $\sqrt{\cos^{-1}x}\approx\sqrt\pi\sqrt{1-\sqrt{2h}/\pi}\approx\sqrt\pi\left(1-\dfrac{\sqrt{2h}}{2\pi}\right)=\sqrt\pi-\dfrac{\sqrt{2h}}{2\sqrt\pi}$.
   4. So $\sqrt\pi-\sqrt{\cos^{-1}x}\approx\dfrac{\sqrt{2h}}{2\sqrt\pi}$. …

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4. 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 $e^{3/x}$: $R=\lim_{x\to0^+}\dfrac{e^{-1/x}+1}{ae^{-1/x}+b}$. …

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5. 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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6. 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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7. 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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8. 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 $\sqrt{16+\sqrt x}+4$: $\frac{\sqrt x\,(\sqrt{16+\sqrt x}+4)}{(16+\sqrt x)-16}=\frac{\sqrt x\,(\sqrt{16+\sqrt x}+4)}{\sqrt x}=\sqrt{16+\sqrt x}+4.$ …

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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 $\beta$: $\beta=-\dfrac{27}{2}$. …

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10. 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 $f=\dfrac{t}{-t+2t^2}=\dfrac{1}{-1+2t}\to-1$. …

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11. 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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12. 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 $\dfrac{\sqrt{16+t}+4}{\sqrt{16+t}+4}$: $\dfrac{t(\sqrt{16+t}+4)}{(16+t)-16}=\dfrac{t(\sqrt{16+t}+4)}{t}=\sqrt{16+t}+4$. …

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