If f(x) = Asin left( frac{pi x}{2} right) + B | vedclass

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(C) Given $f(x) = A\sin \left( \frac{\pi x}{2} \right) + B$. First,we use the integral condition: $\int_0^1 \left( A\sin \left( \frac{\pi x}{2} \right

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$\frac{4}{\pi}$ and $0$

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(C) Given $f(x) = A\sin \left( \frac{\pi x}{2} \right) + B$. First,we use the integral condition: $\int_0^1 \left( A\sin \left( \frac{\pi x}{2} \right) + B \right) dx = \frac{2A}{\pi}$. Evaluating the integral: $\left[ -\frac{2A}{\pi} \cos \left( \frac{\pi x}{2} \right) + Bx \right]_0^1 = \frac{2A}{\pi}$. Substituting the limits: $\left( -\frac{2A}{\pi} \cos \left( \frac{\pi}{2} \right) + B(1) \right) - \left( -\frac{2A}{\pi} \cos(0) + B(0) \right) = \frac{2A}{\pi}$. Since $\cos(\frac{\pi}{2}) = 0$ and $\cos(0) = 1$,we get: $B - (-\frac{2A}{\pi}) = \frac{2A}{\pi} \implies B + \frac{2A}{\pi} = \frac{2A}{\pi} \implies B = 0$. Now,find $f'(x)$: $f'(x) = A \cdot \frac{\pi}{2} \cos \left( \frac{\pi x}{2} \right)$. Given $f'\left( \frac{1}{2} \right) = \sqrt{2}$,we have: $\frac{A\pi}{2} \cos \left( \frac{\pi}{4} \right) = \sqrt{2}$. Since $\cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$,we get: $\frac{A\pi}{2} \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \implies \frac{A\pi}{2} = 2 \implies A = \frac{4}{\pi}$. Thus,$A = \frac{4}{\pi}$ and $B = 0$.

If $f(x) = A\sin \left( \frac{\pi x}{2} \right) + B$,$f'\left( \frac{1}{2} \right) = \sqrt{2}$ and $\int_0^1 f(x) \, dx = \frac{2A}{\pi}$,then the constants $A$ and $B$ are respectively:

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