Q.Find the values of $a$ and $b$ such that the function defined by $f(x) = \begin{cases} 5, & \text{if } x \leq 2 \\ ax+b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \geq 10 \end{cases}$ is a continuous function.

Showing the 12 most recent of 18 on this concept.

1. 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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2. 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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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 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If the function $f(x) = \begin{cases} \dfrac{e^{a\sqrt{x-1}} - 1}{\sqrt{x^2 - 1}}, & \text{for } x > 1 \\ \sqrt{2}, & \text{for } x = 1 \\ \log\left(\dfrac{1+bx}{1-bx}\right)^{\frac{1}{\sqrt{2}x}}, & \text{for } 0 < x < 1 \end{cases}$ is continuous at $x = 1$, then $\displaystyle\lim_{x \to a} \dfrac{x^2 - 5x + 6}{x - a} =$ (A) $b$ (B) $-b$ (C) $2b$ (D) $-2b$

   ›Reveal solutionSolution

   Continuity at $x=1$ forces $a = 2$; the required limit $\displaystyle\lim_{x\to 2}\frac{x^2-5x+6}{x-2} = -1$. The official key marks option (B) $-b$.

   Right-hand limit fixes $a$

   For $x > 1$, put $t = \sqrt{x-1} \to 0^+$, so $x = 1+t^2$ and $\sqrt{x^2-1} = t\sqrt{2+t^2}$. Using $e^{at}-1 \sim at$,

   $\lim_{x\to 1^+} \frac{e^{a\sqrt{x-1}}-1}{\sqrt{x^2-1}} = \lim_{t\to 0}\frac{at}{t\sqrt{2+t^2}} = \frac{a}{\sqrt{2}}.$

   Continuity requires this to equal $f(1)=\sqrt{2}$:

   $\frac{a}{\sqrt{2}} = \sqrt{2} \ \Rightarrow\ a = 2.$

   Left-hand limit fixes $b$

   For $0 < x < 1$, $f(x) = \dfrac{1}{\sqrt{2}\,x}\log\!\left(\dfrac{1+bx}{1-bx}\right)$. Setting its limit as $x\to 1^-$ equal to $\sqrt{2}$:

   $\frac{1}{\sqrt{2}}\log\!\left(\frac{1+b}{1-b}\right) = \sqrt{2} \ \Rightarrow\ \log\!\left(\frac{1+b}{1-b}\right) = 2 \ \Rightarrow\ b = \frac{e^2-1}{e^2+1}.$

   Evaluate the limit …

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5. 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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6. 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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7. 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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8. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If a function $f(x)=\begin{cases}\dfrac{\sqrt[3]{1+ax^2+bx^3}-\sqrt[3]{1-ax^2-bx^3}}{x^2}, & x<0\\ 5, & x=0\\ \dfrac{\tan 3x-\sin 3x}{bx^3}, & x>0\end{cases}$ is continuous at $x=0$, then the geometric mean of a and b is (A) $\dfrac{3}{2}$ (B) $\dfrac{9}{2}$ (C) $\dfrac{81}{4}$ (D) $\dfrac{9}{4}$

   ›Reveal solutionSolution

   Continuity at $x=0$ forces both one-sided limits to equal $f(0)=5$; this determines $a$ and $b$ separately, and their geometric mean is $\dfrac92$.

   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 actual value there must all agree. Here $f(0)=5$ is given, so both one-sided limits must independently equal 5. Each side involves a $\frac00$-type indeterminate form that resolves using standard small-angle/small-$u$ approximations: $\sin\phi\approx\phi$, $1-\cos\phi\approx\phi^2/2$ for the right side, and the binomial-type expansion $(1\pm u)^{1/3}\approx1\pm u/3-u^2/9$ for the left side (the leading behavior of a cube root near 1).

   Step-by-Step Solution

   Right-hand limit ($x\to0^+$): $f(x)=\dfrac{\tan3x-\sin3x}{bx^3}$.

   1. $\tan3x-\sin3x=\dfrac{\sin3x}{\cos3x}-\sin3x=\sin3x\cdot\dfrac{1-\cos3x}{\cos3x}$.
   2. As $x\to0$: $\sin3x\to3x$, $1-\cos3x\to\dfrac{(3x)^2}{2}=\dfrac{9x^2}{2}$, $\cos3x\to1$.
   3. Numerator $\approx (3x)\left(\dfrac{9x^2}{2}\right)=\dfrac{27x^3}{2}$.
   4. Limit $=\dfrac{27x^3/2}{bx^3}=\dfrac{27}{2b}$.
   5. Continuity: $\dfrac{27}{2b}=5\Rightarrow b=\dfrac{27}{10}$.

   Left-hand limit ($x\to0^-$): $f(x)=\dfrac{\sqrt[3]{1+u}-\sqrt[3]{1-u}}{x^2}$ where $u=ax^2+bx^3$.

   6. $(1+u)^{1/3}\approx1+\dfrac{u}{3}-\dfrac{u^2}{9}$, $(1-u)^{1/3}\approx1-\dfrac{u}{3}-\dfrac{u^2}{9}$ (same sign on the $u^2$ term since it's even in the expansion of $(1\pm u)^{1/3}$ up to that order — more precisely the quadratic coefficient is the same for both since it depends on $u^2$, and subtracting cancels it).

   7. Difference $\approx \dfrac{2u}{3} = \dfrac{2(ax^2+bx^3)}{3}$. …

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9. 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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10. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
    Q.If a function $f(x) = \begin{cases} \dfrac{x}{|x|} - K, & x > 0 \\ \dfrac{x}{|x|} + L, & x < 0 \\ 5, & x = 0 \end{cases}$ is continuous for all real values of $x$, then $\dfrac{L + K}{L - K} =$ (A) $5$ (B) $3$ (C) $\dfrac{1}{5}$ (D) $\dfrac{1}{3}$

    ›Reveal solutionSolution

    Since $x/|x|$ is just $\pm1$, each piece is actually constant; matching both one-sided limits to $f(0)=5$ pins down $K,L$, giving $\dfrac{L+K}{L-K}=\dfrac15$.

    Concept and Intuition

    $\dfrac{x}{|x|}=1$ for $x>0$ and $=-1$ for $x<0$, so despite looking like it depends on $x$, each branch of $f$ is actually a constant function on its domain. Continuity at $x=0$ then just requires that constant to equal $f(0)=5$ from each side.

    Step-by-Step Solution

    1. For $x>0$: $f(x)=\dfrac{x}{|x|}-K = 1-K$ (constant).
    2. $\lim_{x\to0^+}f(x) = 1-K$. Continuity requires this $=f(0)=5$: $1-K=5 \Rightarrow K=-4$.
    3. For $x<0$: $f(x)=\dfrac{x}{|x|}+L=-1+L$ (constant).
    4. $\lim_{x\to0^-}f(x)=L-1$. Continuity requires $L-1=5 \Rightarrow L=6$. …

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11. 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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12. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.Let $f(x) = \begin{cases}1+\frac{2x}{a}, & 0\le x\le1\\ax, & 1<x\le2\end{cases}$. If $\lim_{x\to1}f(x)$ exists then the sum of the cubes of the possible values of a is (A) 1 (B) 5 (C) 7 (D) 9

    ›Reveal solutionSolution

    For the piecewise function's limit at the junction point to exist, the two one-sided limits (evaluated from each formula) must be equal; this gives a quadratic in $a$ whose two roots we cube and sum.

    Concept and Intuition

    At a junction point of a piecewise function, the left-hand limit uses the formula valid just below the point, and the right-hand limit uses the formula valid just above. Since both pieces are continuous polynomials/linear expressions in $x$ near $x=1$, the one-sided limits are simply the values of each formula evaluated at $x=1$. Setting them equal is exactly the condition for the overall limit to exist.

    Step-by-Step Solution

    1. As $x\to1^-$ (using $1+\dfrac{2x}{a}$ valid for $0\le x\le1$): limit $=1+\dfrac2a$.
    2. As $x\to1^+$ (using $ax$ valid for $1<x\le2$): limit $=a(1)=a$.
    3. For the limit to exist: $1+\dfrac2a=a$. …

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