IIT-JEE 1995 ScreeningQuestion 1
If $f(x)=A\sin\left(\dfrac{\pi x}{2}\right)+B$, $f'\left(\dfrac12\right)=\sqrt2$ and $\displaystyle\int_0^1f(x)\,dx=\dfrac{2A}{\pi}$, then constants $A$ and $B$ are
Definite Integration
Solution
Concept. The derivative condition fixes $A$; the integral condition fixes $B$. Formulas used. (1) $f'(x)=\frac{A\pi}2\cos\frac{\pi x}2$(2) $\int_0^1\sin\frac{\pi x}2\,dx=\Big[-\frac2\pi\cos\frac{\pi x}2\Big]_0^1=\frac2\pi$Step 1 - $A$. By (1), $f'\big(\frac12\big)=\frac{A\pi}2\cdot\frac{\sqrt2}2=\sqrt2$, so $A=\frac4\pi$. Step 2 - $B$. By (2), $\int_0^1f=\frac{2A}\pi+B$. This equals $\frac{2A}\pi$, so $B=0$. Answer: (D)