Q.If $f(x) = \begin{cases} mx + 1, & x \le \dfrac{\pi}{2} \\ \sin x + n, & x > \dfrac{\pi}{2} \end{cases}$ is continuous at $x = \dfrac{\pi}{2}$, then
(A) $m = 1,\ n = 0$
(B) $m = \dfrac{n\pi}{2} + 1$
(C) $n = \dfrac{m\pi}{2}$
(D) $m = n = \dfrac{\pi}{2}$

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If the function $f(x)$, defined below is continuous in the interval $[0, \pi]$, then ____ $f(x) = \begin{cases} x + a\sqrt{2}(\sin x), & 0 \le x < \dfrac{\pi}{4} \\ 2x(\cot x) + b, & \dfrac{\pi}{4} \le x \le \dfrac{\pi}{2} \\ a(\cos 2x) - b(\sin x), & \dfrac{\pi}{2} < x \le \pi \end{cases}$ (A) $a = \dfrac{\pi}{6}, b = \dfrac{\pi}{12}$ (B) $a = \dfrac{-\pi}{6}, b = \dfrac{\pi}{12}$ (C) $a = \dfrac{-\pi}{6}, b = \dfrac{-\pi}{12}$ (D) $a = \dfrac{\pi}{6}, b = \dfrac{-\pi}{12}$

   ›Reveal solutionSolution

   Matching the piecewise function's values at the two junction points $x=\pi/4$ and $x=\pi/2$ gives two linear equations in $a,b$, solved by $a=\pi/6,\ b=-\pi/12$.

   Concept and Intuition

   A piecewise function is continuous on an interval exactly when each piece agrees with its neighbor at every junction point. With two junctions here ($\pi/4$ and $\pi/2$), we get two equations in the two unknowns $a,b$ — a standard "match the boundary values" problem.

   Step-by-Step Solution

   1. At $x=\pi/4$: left piece $= \dfrac{\pi}{4}+a\sqrt2\sin\dfrac{\pi}{4} = \dfrac{\pi}{4}+a\sqrt2\cdot\dfrac{\sqrt2}{2}=\dfrac{\pi}{4}+a$.
   2. Middle piece at $\pi/4$: $2\left(\dfrac{\pi}{4}\right)\cot\dfrac{\pi}{4}+b = \dfrac{\pi}{2}(1)+b=\dfrac{\pi}{2}+b$.
   3. Equate: $\dfrac{\pi}{4}+a=\dfrac{\pi}{2}+b \Rightarrow a-b=\dfrac{\pi}{4}$. — (i)
   4. At $x=\pi/2$: middle piece $=2\left(\dfrac{\pi}{2}\right)\cot\dfrac{\pi}{2}+b = \pi(0)+b=b$.
   5. Right piece at $\pi/2$: $a\cos\pi-b\sin\dfrac{\pi}{2}=a(-1)-b(1)=-a-b$. …

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2. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.If $f(x) = \begin{cases} \dfrac{1 - \sin^3 x}{3\cos^2 x}, & x < \pi/2 \\ \alpha, & x = \pi/2 \\ \dfrac{\beta(1 - \sin x)}{(\pi - 2x)^2}, & x > \pi/2 \end{cases}$ is continuous at $x = \pi/2$, then $\alpha\beta =$ (A) $1$ (B) $-1$ (C) $2$ (D) $-2$

   ›Reveal solutionSolution

   Continuity at $x=\pi/2$ forces the left-hand limit, the function value $\alpha$, and the right-hand limit (in terms of $\beta$) to all coincide; algebraic factoring and a small substitution give $\alpha=1/2$, $\beta=4$.

   Concept and Intuition

   A piecewise function is continuous at a point only if the left-hand limit, the right-hand limit, and the defined value there are all equal. Here we must find both one-sided limits as $x\to\pi/2$ and set them equal to $\alpha$ (the middle value), which also pins down $\beta$.

   Step-by-Step Solution

   1. Left-hand limit ($x\to\pi/2^-$): $\displaystyle\lim_{x\to\pi/2}\frac{1-\sin^3x}{3\cos^2x}$. Factor: $1-\sin^3x=(1-\sin x)(1+\sin x+\sin^2x)$ and $\cos^2x=1-\sin^2x=(1-\sin x)(1+\sin x)$.

   $\frac{(1-\sin x)(1+\sin x+\sin^2x)}{3(1-\sin x)(1+\sin x)}=\frac{1+\sin x+\sin^2x}{3(1+\sin x)}.$

   At $x=\pi/2$, $\sin x=1$: value $=\dfrac{1+1+1}{3(2)}=\dfrac{3}{6}=\dfrac12$. So $\alpha=\dfrac12$.

   2. Right-hand limit ($x\to\pi/2^+$): put $x=\pi/2+t$, $t\to0^+$. Then $\pi-2x=-2t$, so $(\pi-2x)^2=4t^2$.

   Also $\sin x=\sin(\pi/2+t)=\cos t\approx 1-\dfrac{t^2}{2}$, so $1-\sin x\approx \dfrac{t^2}{2}$. …

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3. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $f(x)$, defined as given below, is continuous on $R$, then the value of $a + b =$ ______

   [!FORMULA] $f(x) = \begin{cases} \sin x, & x \le 0 \\ x^2 + a, & 0 < x < 1 \\ bx + 3, & 1 \le x \le 3 \\ -3, & x > 3 \end{cases}$

   (A) $0$ (B) $2$ (C) $-2$ (D) $3$

   ›Reveal solutionSolution

   Tests matching piecewise function values at each junction point to enforce continuity, giving two equations for the two unknowns.

   Concept and Intuition

   A piecewise function is continuous at a junction point exactly when the left-hand and right-hand pieces agree in value there (since each individual piece is already continuous/smooth on its own interval). Checking each junction in turn gives one equation per unknown constant.

   Step-by-Step Solution

   1. At $x=0$: left piece ($x\le0$) gives $\sin(0)=0$. Right-approaching piece ($0<x<1$) gives $\lim_{x\to0^+}(x^2+a)=a$. Continuity requires $0=a \Rightarrow a=0$.
   2. At $x=1$: left-approaching piece ($0<x<1$) gives $\lim_{x\to1^-}(x^2+a)=1+a=1+0=1$. The piece at $x=1$ itself ($1\le x\le3$) gives $b(1)+3=b+3$. Continuity requires $1=b+3 \Rightarrow b=-2$. …

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4. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.If $f(x)$, defined below, is continuous at $x = 4$, then _______

   [!FORMULA] $f(x) = \begin{cases} \dfrac{x-4}{|x-4|} + a, & x < 4 \\ a + b, & x = 4 \\ \dfrac{x-4}{|x-4|} + b, & x > 4 \end{cases}$

   (A) $a = 0$ & $b = 0$ (B) $a = 1$ & $b = 1$ (C) $a = -1$ & $b = 1$ (D) $a = 1$ & $b = -1$

   ›Reveal solutionSolution

   This tests evaluating the sign function $\frac{x-4}{|x-4|}$ on either side of $x=4$ and matching one-sided limits to the function's value there. Answer: $a=1,\ b=-1$.

   Concept and Intuition

   The expression $\dfrac{x-4}{|x-4|}$ is $-1$ for $x<4$ and $+1$ for $x>4$ — a signum function centered at $4$. Continuity at $x=4$ needs all three (left limit, value, right limit) to agree.

   Step-by-Step Solution

   1. For $x<4$: $|x-4|=4-x$, so $\dfrac{x-4}{|x-4|}=\dfrac{x-4}{-(x-4)}=-1$. Left-hand limit $=-1+a$.
   2. At $x=4$: $f(4)=a+b$.
   3. For $x>4$: $|x-4|=x-4$, so the ratio is $+1$. Right-hand limit $=1+b$.
   4. Continuity: $-1+a = a+b = 1+b$. …

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5. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.If the function $f(x) = \begin{cases} 1+\cos x, & x \le 0 \\ a-x, & 0 < x \le 2 \\ x^2 - b^2, & x > 2 \end{cases}$ is continuous everywhere, then $a^2 + b^2 =$ (A) $4$ (B) $8$ (C) $6$ (D) $12$

   ›Reveal solutionSolution

   Match the piecewise definitions at the two junction points $x=0$ and $x=2$ to pin down $a$ and $b$. Answer: $a^2+b^2=8$.

   Concept and Intuition

   A piecewise function is continuous everywhere exactly when it is continuous at each junction between pieces — the left-hand value (from the left-side piece) must equal the right-hand value (from the right-side piece) at each breakpoint. Here there are two breakpoints, $x=0$ and $x=2$, giving one equation each for the unknowns $a$ and $b$.

   Step-by-Step Solution

   1. At $x=0$: the left piece ($x\le0$) gives $f(0^-)=1+\cos(0)=1+1=2$. The middle piece ($0<x\le2$), evaluated as $x\to0^+$, gives $f(0^+)=a-0=a$.
   2. Continuity at $0$: $a=2$.
   3. At $x=2$: the middle piece gives $f(2)=a-2=2-2=0$. The right piece ($x>2$), as $x\to2^+$, gives $f(2^+)=2^2-b^2=4-b^2$. …

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6. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.If the function $f(x)$, defined below, is continuous on the interval $[0, 8]$, then _______

   [!FORMULA] $f(x) = \begin{cases} x^2 + ax + b, & 0 \le x < 2 \\ 3x + 2, & 2 \le x \le 4 \\ 2ax + 5b, & 4 < x \le 8 \end{cases}$

   (A) $a = 3, b = -2$ (B) $a = -3, b = 2$ (C) $a = -3, b = -2$ (D) $a = 3, b = 2$

   ›Reveal solutionSolution

   This tests matching piecewise function values at the junction points to enforce continuity on a closed interval. Answer: $a=3,\ b=-2$.

   Concept and Intuition

   A piecewise function is continuous at a junction point exactly when the values from each adjoining piece agree there (the left-hand and right-hand pieces must meet without a jump).

   Step-by-Step Solution

   1. At $x=2$: from the first piece (as $x\to 2^-$), $f\to 2^2+2a+b = 4+2a+b$. From the middle piece, $f(2)=3(2)+2=8$.
   2. Continuity at $x=2$: $4+2a+b=8 \Rightarrow 2a+b=4$. — (i)
   3. At $x=4$: the middle piece gives $f(4)=3(4)+2=14$. From the third piece (as $x\to4^+$), $f\to 2a(4)+5b=8a+5b$.
   4. Continuity at $x=4$: $8a+5b=14$. — (ii) …

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7. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If $f(x)=\begin{cases}3ax-2b, & x>1\\ ax+b+1, & x<1\end{cases}$ and $\lim\limits_{x \to 1} f(x)$ exists, then the relation between $a$ and $b$ is (A) $3a-2b=1$ (B) $2a-3b=1$ (C) $2a+3b=1$ (D) $2a+3b=-1$

   ›Reveal solutionSolution

   Existence of the limit at a piecewise junction forces the two one-sided limits to match, giving $2a-3b=1$.

   Concept and Intuition

   For a piecewise function, $\lim_{x\to c}f(x)$ exists only when the value approached from the left equals the value approached from the right — the two pieces must "meet" at that point (the function value at $x=1$ itself doesn't matter here since neither branch is defined at $x=1$, only $\lim$).

   Step-by-Step Solution

   1. Left-hand limit as $x\to1^-$: uses the branch $ax+b+1$ (valid for $x<1$), giving $a(1)+b+1=a+b+1$. …

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8. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $\displaystyle\lim_{x \to a^+} f(x) = p$, $\displaystyle\lim_{x \to a^-} f(x) = m$ and $f(a) = k$, then which one of the following is true? (A) When $p - k \neq 0$ and $m - k \neq 0$, then $f(x)$ is continuous at $x = a$ (B) When $p - k = 0$ and $m - k \neq 0$, then $f(x)$ is left continuous at $x = a$ (C) When $p - k \neq 0$ and $m - k = 0$, then $f(x)$ is right continuous at $x = a$ (D) When $p - m = 0$ and $p - k = 0$, then $f(x)$ is right continuous at $x = a$

   ›Reveal solutionSolution

   Only option (D) correctly matches its stated hypothesis to its continuity conclusion; (B) and (C) swap "left" and "right", and (A) is simply false.

   Concept and Intuition

   Right continuity at $a$ is precisely $\lim_{x\to a^+}f(x)=f(a)$ (i.e. $p=k$); left continuity is $\lim_{x\to a^-}f(x)=f(a)$ (i.e. $m=k$). Full continuity needs both plus $p=m$. Careful bookkeeping of which equality corresponds to which side is the whole content of this question.

   Step-by-Step Solution

   1. (A): $p-k\ne0$ means the right-hand limit differs from $f(a)$, so $f$ is NOT right continuous — hence certainly not continuous. (A) is false.
   2. (B): $p-k=0\Rightarrow p=k$, which is exactly the definition of right continuity, not left. Since (B) claims "left continuous", it is false (regardless of $m$).
   3. (C): $m-k=0\Rightarrow m=k$, which is exactly left continuity, not right. (C) claims "right continuous", so it is false. …

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9. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If $f(x)=\begin{cases} 1+6x-3x^2, & x\le 1 \\ x+\log_2(b^2+7), & x>1 \end{cases}$ is continuous at all real x, then b = (A) $\pm 1$ (B) $0$ (C) $\pm 5$ (D) $\pm 2$

   ›Reveal solutionSolution

   Both pieces are continuous on their own; matching them at the junction $x=1$ gives $\log_2(b^2+7)=3$, so $b=\pm1$.

   Concept and Intuition

   A piecewise function built from continuous pieces (a polynomial and a $\log$-plus-linear expression) is automatically continuous everywhere except possibly at the boundary point where the definition switches — here $x=1$. So the entire "continuous for all real $x$" condition reduces to one equation: the value approaching from the left must equal the value approaching from the right (and both must equal $f(1)$, which is given by the $x\le1$ branch).

   Step-by-Step Solution

   1. For $x\le1$: $f(x)=1+6x-3x^2$, continuous everywhere (polynomial). $f(1)=1+6-3=4$.
   2. For $x>1$: $f(x)=x+\log_2(b^2+7)$, continuous on its domain (as long as $b^2+7>0$, always true). …

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10. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.If a function f(x) defined by $f(x)=\begin{cases} ax^2+bx+c, & x\le-1\\ 2x^2+4x+1, & -1<x<1\\ cx^2+bx+a, & x\ge1\end{cases}$ is continuous on $\mathbb{R}$, and $\lim\limits_{x\to\frac{3}{2}} f(x)=14$, then $\lim\limits_{x\to-2} f(x)=$ (A) 6 (B) $-8$ (C) 5 (D) 1

    ›Reveal solutionSolution

    Using continuity at the two junction points plus the given limit at $x=3/2$ pins down all three constants $a,b,c$; the requested limit then evaluates to $-8$.

    Concept and Intuition

    A piecewise function is continuous on $\mathbb R$ exactly when its pieces agree at every junction point. Each junction gives one linear equation in the unknown coefficients. Combined with the extra numerical condition given ($\lim_{x\to3/2}f(x)=14$), we get enough equations to solve for $a,b,c$ uniquely.

    Step-by-Step Solution

    1. Continuity at $x=-1$: $a(-1)^2+b(-1)+c=2(-1)^2+4(-1)+1\Rightarrow a-b+c=-1$.
    2. Continuity at $x=1$: $2(1)^2+4(1)+1=c(1)^2+b(1)+a\Rightarrow a+b+c=7$.
    3. Subtracting: $2b=8\Rightarrow b=4$; adding relations gives $a+c=3$.
    4. Since $x=3/2\ge1$, $f(3/2)=c(3/2)^2+b(3/2)+a=\tfrac94c+6+a=14\Rightarrow \tfrac94c+a=8$. …

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11. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.The value of '$k$' ($k>0$), for which the function $f(x)=\dfrac{(e^{x}-1)^4}{\sin\left(\dfrac{x^2}{k^2}\right)\log\left(1+\dfrac{x^2}{2}\right)}$, where $x\neq 0$ and $f(0)=8$ is continuous, is ______. (A) $1$ (B) $4$ (C) $2$ (D) $3$

    ›Reveal solutionSolution

    Replace each factor by its leading small-$x$ equivalent and match the resulting constant to $f(0)$. Answer: $k=2$.

    Concept and Intuition

    Near $x=0$, standard small-angle/small-argument equivalents apply: $e^x-1\sim x$, $\sin\theta\sim\theta$, $\log(1+\theta)\sim\theta$. Substituting these turns the limit into a simple ratio of leading powers of $x$, all of which cancel, leaving a constant in $k$.

    Step-by-Step Solution

    1. $(e^x-1)^4\sim x^4$ as $x\to0$.
    2. $\sin\left(\dfrac{x^2}{k^2}\right)\sim\dfrac{x^2}{k^2}$.
    3. $\log\left(1+\dfrac{x^2}{2}\right)\sim\dfrac{x^2}{2}$.
    4. So $f(x)\to\dfrac{x^4}{\left(\frac{x^2}{k^2}\right)\left(\frac{x^2}{2}\right)}=\dfrac{x^4}{\frac{x^4}{2k^2}}=2k^2$. …

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12. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.$f(x)=\begin{cases}\dfrac{e^{\alpha x}-e^{x}-x}{x^2}, & x\neq 0\\[2mm] \dfrac{3}{2}, & x=0\end{cases}$ Find the value of '$\alpha$' for which the function $f$ is continuous. (A) $1$ (B) $0$ (C) $4$ (D) $2$

    ›Reveal solutionSolution

    Continuity forces the numerator's Taylor expansion to have zero linear term (fixing $\alpha$) and the resulting quadratic term to equal $f(0)$. Answer: $\alpha=2$.

    Concept and Intuition

    For $f$ to be continuous at $x=0$, $\displaystyle\lim_{x\to0}\frac{e^{\alpha x}-e^x-x}{x^2}$ must equal $f(0)=\frac32$. Since the denominator is $x^2\to0$, the numerator must vanish to second order — its constant and linear Taylor terms must be zero, and the surviving quadratic term must match $\frac32$.

    Step-by-Step Solution

    1. Expand: $e^{\alpha x}=1+\alpha x+\frac{\alpha^2}{2}x^2+O(x^3)$, $e^{x}=1+x+\frac12x^2+O(x^3)$.
    2. Numerator $=e^{\alpha x}-e^x-x=(\alpha-2)x+\frac{\alpha^2-1}{2}x^2+O(x^3)$.
    3. For the limit of (numerator)/$x^2$ to exist and be finite, the $x^1$ term must vanish: $\alpha-2=0\Rightarrow\alpha=2$. …

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