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x + a root2 sin x, 2x cot x + b, a cos 2x - b sin x continuous on [0, pi]: find a and b

The same question, asked 4 times: 10 May 2023 S1*, 10 May 2024 S1*, 12 May 2024 S1*, 22 Apr 2025 S1. Every wording below, oldest first.

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  1. Question 1

    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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  2. Question 2

    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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  3. Question 3

    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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  4. Question 4

    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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