Question

Find the values of $p$ and $q$, for which
$f\left(x\right)=\{\begin{array}{cc}\frac{1-{\sin }^{3}x}{3{\cos }^{2}x},& ifx<\frac{\pi }{2}\\ p& ifx=\frac{\pi }{2}\\ \frac{q(1-\sin x)}{(\pi -2x{)}^2}& ifx>\frac{\pi }{2}\end{array}$

f(x) is continuous at x = $\frac{\pi }{2}$

Answer 1

Since the function needs to be continuous at $x=\frac{\pi }{2}$,

$f\left({\frac{\pi }{2}}^{-}\right)=f\left(\frac{\pi }{2}\right)=f\left({\frac{\pi }{2}}^{+}\right)$

$\Rightarrow \underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{1-{\sin }^{3}x}{3{\cos }^{2}x}=\frac{-3{\sin }^{2}x\cos x}{-6\cos x\sin x}=\frac{1}{2}$

Also, $\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{q(1-\sin x)}{(\pi -2x{)}^2}=\frac{-q\cos x}{2(\pi -2x)}=\frac{q\sin x}{-4}=\frac{-q}{4}$

Since all the limits have to be same, we get $\frac{1}{2}=p=\frac{-q}{4}$

$p=\frac{1}{2}$ and $q=-2$