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Find a and b making a function with three pieces (two joining points) continuous

This type of question, asked 10 times: May 2022 S1, 4 May 2023 S1, 10 May 2023 S1, 15 May 2023 S2, 10 May 2024 S1, 11 May 2024 S1, 12 May 2024 S1, 13 May 2024 S2, 22 Apr 2025 S1, 12 May 2026 S1. All 10 questions below, oldest first.

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  1. Question 1

    May 2022 S1

    122
    If f(x)=ax2+bx+1f(x) = ax^2 + bx + 1 if ∣2x−3∣≥2|2x-3| \geq 2, f(x)=3x+2f(x) = 3x + 2 if 12<x<52\frac{1}{2} < x < \frac{5}{2}, is continuous on its domain, then a+ba+b has the value

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  2. Question 2

    4 May 2023 S1

    148
    Let a,b∈R (a≠0)a,b\in\mathbb{R}\,(a\neq0). If the function ff is defined as f(x)={2x2a,0≤x<1a,1≤x<22b2−4bx,2≤x<∞f(x)=\begin{cases}\frac{2x^2}{a},&0\leq x<1\\a,&1\leq x<\sqrt{2}\\\frac{2b^2-4b}{x},&\sqrt{2}\leq x<\infty\end{cases} is continuous in [0,∞)[0,\infty), then an ordered pair (a,b)(a,b) is

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  3. Question 3

    10 May 2023 S1

    120
    The values of a and b, so that the function f(x)={x+a2sin⁡x,0≤x≤π42xcot⁡x+b,π4≤x≤π2acos⁡2x−bsin⁡x,π2<x≤πf(x)=\begin{cases}x+a\sqrt{2}\sin x,&0\leq x\leq\frac{\pi}{4}\\2x\cot x+b,&\frac{\pi}{4}\leq x\leq\frac{\pi}{2}\\a\cos2x-b\sin x,&\frac{\pi}{2}<x\leq\pi\end{cases} is continuous for 0≤x≤π0\leq x\leq\pi, are respectively given by

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  4. Question 4

    15 May 2023 S2

    120
    If the function f(x)={−2sin⁡xx≤−π/2Asin⁡x+B−π/2<x<π/2cos⁡xx≥π/2f(x) = \begin{cases} -2\sin x & x \leq -\pi/2 \\ A\sin x+B & -\pi/2 < x < \pi/2 \\ \cos x & x \geq \pi/2 \end{cases} is continuous everywhere, then the values of AA and BB are respectively

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  5. Question 5

    10 May 2024 S1

    139
    f(x)f(x) continuous on [0,π][0,\pi] (piecewise: x+a2sin⁡xx+a\sqrt{2}\sin x on [0,π/4)[0,\pi/4), 2xcot⁡x+b2x\cot x+b on [π/4,π/2][\pi/4,\pi/2], acos⁡2x−bsin⁡xa\cos2x-b\sin x on (π/2,π](\pi/2,\pi]). Value of 2a+3b2a+3b is

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  6. Question 6

    11 May 2024 S1

    128
    If f(x)={1+mx−1−mxx,−1≤x<02x+1x−2,0≤x≤1f(x)=\begin{cases}\dfrac{\sqrt{1+mx}-\sqrt{1-mx}}{x}, & -1\leq x<0\\ \dfrac{2x+1}{x-2}, & 0\leq x\leq1\end{cases} is continuous in the interval [−1,1][-1,1], then m is equal to

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  7. Question 7

    12 May 2024 S1

    104
    The values of aa and bb, so that the function f(x)={x+a2sin⁡x,0≤x≤π42xcot⁡x+b,π4≤x≤π2acos⁡2x−bsin⁡x,π2<x≤πf(x)=\begin{cases}x+a\sqrt{2}\sin x, & 0\le x\le\frac{\pi}{4}\\ 2x\cot x+b, & \frac{\pi}{4}\le x\le\frac{\pi}{2}\\ a\cos2x-b\sin x, & \frac{\pi}{2}<x\le\pi\end{cases} is continuous, are respectively given by

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  8. Question 8

    13 May 2024 S2

    121
    If f(x)f(x) is continuous on its domain [−2,2][-2,2], where f(x)={sin⁡axx+3,−2≤x<02x+7,0≤x≤1x2+8−b,1<x≤2f(x)=\begin{cases}\frac{\sin ax}{x}+3, & -2\leq x<0\\ 2x+7, & 0\leq x\leq1\\ \sqrt{x^2+8}-b, & 1<x\leq2\end{cases} then the value of 2a+3b2a+3b is:

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  9. Question 9

    22 Apr 2025 S1

    120
    If the function f(x)={x+a2sin⁡x if 0≤x≤π42xcotx+b if π4<x≤π2acos2x−bsinx if π2<x≤πf(x) =\left\{ \begin{matrix} x+a\sqrt{2}\sin x & \text{~if~}0 \leq x\leq\frac{\pi}{4} \\ 2xcotx+b & \text{~if~}\frac{\pi}{4}<x\leq\frac{\pi}{2} \\ acos2x-bsinx & \text{~if~}\frac{\pi}{2}<x\leq\pi \end{matrix} \right. is continuous in [0,π]\lbrack 0,\pi\rbrack then a−b=a - b =

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  10. Question 10

    12 May 2026 S1

    70
    If the function defined by f(x)={ax2+bx+1,∣2x−3∣≥23x+2,12<x<52f(x) = \begin{cases} ax^2 + bx + 1, & |2x - 3| \ge 2 \\ 3x + 2, & \dfrac{1}{2} < x < \dfrac{5}{2} \end{cases} is continuous in its domain, then the values of aa and bb respectively are

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