Ans: (b) $-1$
-
$\begin{array}{cl}
& f\left(\dfrac{\pi}{2}+h\right)-f\left(\dfrac{\pi}{2}\right) \\
= & -\left(\dfrac{\pi}{2}+h\right)\sin \left(\dfrac{\pi}{2}+h\right)+\dfrac{\pi}{2}\sin \dfrac{\pi}{2} \\
= & -\left(\dfrac{\pi}{2}+h\right)\left(\sin\dfrac{\pi}{2}\cos h + \sin h \cos \dfrac{\pi}{2}\right)+\dfrac{\pi}{2} \\
= & -\left(\dfrac{\pi}{2}+h\right)\cos h +\dfrac{\pi}{2} \\
= & \dfrac{\pi}{2} -\dfrac{\pi}{2}\cos h -h\cos h\\
= & \dfrac{\pi}{2}(1-\cos h)-h\cos h \\
= & \dfrac{\pi}{2} \times 2\sin^2 \left(\dfrac{h}{2}\right) -h\cos h \\
= & \pi\sin^2\left(\dfrac{h}{2}\right)-h\cos h
\end{array}$ -
$\begin{array}{cl}
& f’\left(\dfrac{\pi}{2}\right) \\
= & \dlim_{h \to 0} \dfrac{1}{h} \left[ f\left(\dfrac{\pi}{2}+h\right)-f\left(\dfrac{\pi}{2}\right)\right] \\
= & \dlim_{h \to 0} \dfrac{1}{h} \left[ \pi\sin^2\left(\dfrac{h}{2}\right)-h\cos h\right] \\
= & \dlim_{h \to 0} \left\{\dfrac{1}{h} \left[ \pi\sin^2\left(\dfrac{h}{2}\right)\right]-\cos h\right\} \\
= & \dlim_{h \to 0} \left[\dfrac{\pi\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\times \dfrac{\sin\left(\frac{h}{2}\right)}{2} -\cos h\right] \\
= & \pi \times 1 \times \dfrac{0}{2}-\cos 0 \\
= & -1
\end{array}$
