Q.For what value of $\lambda$ is the function defined by $f(x) = \begin{cases} \lambda(x^2-2x), & \text{if } x \le 0 \\ 4x+1, & \text{if } x > 0 \end{cases}$ continuous at $x=0$? What about continuity at $x=1$?

Showing the 12 most recent of 32 on this concept.

1. 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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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. …

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3. 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 …

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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)). …

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5. 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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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. …

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7. 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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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). …

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9. 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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10. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.If a real valued function $f(x)=\begin{cases}\dfrac{x^2+(a+3)x+(a+1)}{x+3}, & x\neq -3\\ -\dfrac{5}{2}, & x=-3\end{cases}$ is continuous at $x=-3$, then $\displaystyle\lim_{x\to a}(x^2+x+1) =$ (A) $\dfrac{7}{4}$ (B) $\dfrac{5}{2}$ (C) $\dfrac{4}{7}$ (D) $\dfrac{2}{5}$

    ›Reveal solutionSolution

    This tests continuity of a rational function with a removable-type discontinuity: matching the limit to the given value pins down the parameter $a$, then the asked limit is a trivial polynomial evaluation. Answer: $\dfrac{7}{4}$.

    Concept and Intuition

    A piecewise function is continuous at a point if the limit of the "elsewhere" formula, as $x$ approaches that point, equals the value assigned at that point. Here the formula is a rational function whose denominator vanishes at $x=-3$; for the limit to exist (and be finite, matching $-\tfrac52$), the numerator must also vanish there, so that $(x+3)$ cancels — this is the standard "$\frac00$ removable singularity" idea.

    Step-by-Step Solution

    1. Continuity at $x=-3$ requires $\displaystyle\lim_{x\to -3}\frac{x^2+(a+3)x+(a+1)}{x+3} = -\frac52$.
    2. Since the denominator $\to 0$, the numerator must vanish at $x=-3$ (else the limit is $\pm\infty$, not finite):

    $(-3)^2+(a+3)(-3)+(a+1) = 9-3a-9+a+1 = 1-2a = 0 \implies a=\frac12.$

    3. Check: with $a=\tfrac12$, numerator $=x^2+\tfrac72x+\tfrac32$. Factor out $(x+3)$: $x^2+\tfrac72x+\tfrac32 = (x+3)\left(x+\tfrac12\right)$ (verify: $(x+3)(x+\tfrac12)=x^2+\tfrac12x+3x+\tfrac32=x^2+\tfrac72x+\tfrac32$ ✓).
    4. So $\displaystyle\lim_{x\to -3}\frac{(x+3)(x+\frac12)}{x+3} = -3+\frac12 = -\frac52$, matching the given value — confirming $a=\tfrac12$. …

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11. 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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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). …

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