Q.Find the relationship between $a$ and $b$ so that the function $f$ defined by $f(x) = \begin{cases} ax+1, & \text{if } x \le 3 \\ bx+3, & \text{if } x > 3 \end{cases}$ is continuous at $x=3$.

Showing the 12 most recent of 32 on this concept.

1. 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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2. 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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3. 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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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 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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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 ✓. …

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7. 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 $\tan^2x \approx x^2$. …

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8. 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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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 $\tan^2x\sim x^2$:

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

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10. 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 $f(x)=\dfrac{(3^x-1)^2}{\sin x\log(1+x)} \to \dfrac{x^2(\ln3)^2}{x^2}=(\ln3)^2$. …

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11. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.If a real valued function $f(x)=\begin{cases}\dfrac{2x^2+(k+2)x+9}{3x^2-7x-6}, & x\neq 3\\[2mm] l, & x=3\end{cases}$ is continuous at $x=3$ and $l$ is a finite value, then $l-k=$ (A) $\frac{31}{11}$ (B) $\frac{124}{11}$ (C) $24$ (D) $32$

    ›Reveal solutionSolution

    A removable discontinuity forces the numerator to share the vanishing factor $(x-3)$; solving gives $k=-11$, $l=\dfrac{3}{11}$, so $l-k=\dfrac{124}{11}$.

    Concept and Intuition

    When a rational function's denominator vanishes at the point of interest but the function is stated to be continuous there (with a finite value), the numerator must vanish there too — otherwise the limit would blow up. This forces the numerator to contain the same factor as the denominator, which can then be cancelled.

    Step-by-Step Solution

    1. Denominator: $3x^2-7x-6$ at $x=3$ gives $27-21-6=0$. Factor: $3x^2-7x-6=(x-3)(3x+2)$.
    2. For $f$ to be continuous at $x=3$ with finite $l$, the numerator must also vanish at $x=3$: $2(3)^2+(k+2)(3)+9=18+3k+6+9=33+3k=0\Rightarrow k=-11$.
    3. With $k=-11$, numerator becomes $2x^2-9x+9$. Check $x=3$: $18-27+9=0$ ✓. Factor: $2x^2-9x+9=(x-3)(2x-3)$.
    4. So for $x\ne3$: $f(x)=\dfrac{(x-3)(2x-3)}{(x-3)(3x+2)}=\dfrac{2x-3}{3x+2}$. …

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12. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
    Q.If the function defined by $f(x) = \dfrac{\log(1+x)^{1+x}}{x^2} - \dfrac{1}{x}$, $x \neq 0$ is continuous at $x = 0$, then $6f(0) =$ ______ (A) $2$ (B) $3$ (C) $1$ (D) $6$

    ›Reveal solutionSolution

    Expand $\log(1+x)$ as a Taylor series to resolve the $0/0$-type limit and identify the continuous value $f(0)$.

    Concept and Intuition

    $f$ is defined by a formula that's indeterminate at $x=0$; continuity forces $f(0)$ to equal the limiting value as $x\to0$, which we extract via the Taylor series of $\log(1+x)$.

    Step-by-Step Solution

    1. $f(x)=\dfrac{\log\big[(1+x)^{1+x}\big]}{x^2}-\dfrac1x=\dfrac{(1+x)\log(1+x)}{x^2}-\dfrac1x$.
    2. Expand $\log(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\cdots$.
    3. $(1+x)\log(1+x)=\left(x-\dfrac{x^2}2+\dfrac{x^3}3\right)+\left(x^2-\dfrac{x^3}2\right)+O(x^4)=x+\dfrac{x^2}2-\dfrac{x^3}6+O(x^4)$.
    4. Divide by $x^2$: $\dfrac1x+\dfrac12-\dfrac{x}6+O(x^2)$.
    5. Subtract $\dfrac1x$: $f(x)=\dfrac12-\dfrac{x}6+O(x^2)\to\dfrac12$ as $x\to0$. …

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