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

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(D) Given $f(x) = A \sin \left( \frac{\pi x}{2} \right) + B$. Differentiating with respect to $x$,we get $f'(x) = \frac{A \pi}{2} \cos \left( \frac{\p

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

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(D) Given $f(x) = A \sin \left( \frac{\pi x}{2} \right) + B$. Differentiating with respect to $x$,we get $f'(x) = \frac{A \pi}{2} \cos \left( \frac{\pi x}{2} \right)$. Given $f'(1/2) = \sqrt{2}$,so $\frac{A \pi}{2} \cos \left( \frac{\pi}{4} \right) = \sqrt{2}$. Since $\cos(\pi/4) = 1/\sqrt{2}$,we have $\frac{A \pi}{2 \sqrt{2}} = \sqrt{2}$,which implies $A = \frac{4}{\pi}$. Now,$\int_{0}^{1} f(x) dx = \int_{0}^{1} \left( A \sin \left( \frac{\pi x}{2} \right) + B \right) dx = \frac{2A}{\pi}$. Evaluating the integral: $\left[ -\frac{2A}{\pi} \cos \left( \frac{\pi x}{2} \right) + Bx \right]_{0}^{1} = \frac{2A}{\pi}$. Substituting the limits: $(-\frac{2A}{\pi} \cos(\pi/2) + B) - (-\frac{2A}{\pi} \cos(0) + 0) = \frac{2A}{\pi}$. Since $\cos(\pi/2) = 0$ and $\cos(0) = 1$,we get $B + \frac{2A}{\pi} = \frac{2A}{\pi}$. Therefore,$B = 0$. Thus,$A = \frac{4}{\pi}$ and $B = 0$.

If $f(x) = A \sin \left( \frac{\pi x}{2} \right) + B$,$f'(1/2) = \sqrt{2}$ and $\int_{0}^{1} f(x) dx = \frac{2A}{\pi}$,then the constants $A$ and $B$ are respectively.

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