Question

If the following function is continuous at $x=\frac{\pi }{2}$, then find $a$ and $b$:
$f\left(x\right)=\{\begin{array}{c}\frac{1-{\sin }^{2}x}{3{\cos }^{2}x},ifx<\frac{\pi }{2}\\ a,ifx=\frac{\pi }{2}\\ \frac{b(1-\sin x)}{(\pi -2x{)}^2},ifx>\frac{\pi }{2}\end{array}$

Answer 1

$f\left(x\right)=\{\begin{array}{c}\frac{1-{\sin }^{2}x}{3{\cos }^{2}x},ifx<\frac{\pi }{2}\\ a,ifx=\frac{\pi }{2}\\ \frac{b(1-\sin x)}{(\pi -2x{)}^2},ifx>\frac{\pi }{2}\end{array}$

Since,

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

$a={lim}_{x\to {\frac{\pi }{2}}^{-}}\frac{1-{\sin }^{2}x}{3{\cos }^{2}x}$

$a={lim}_{x\to {\frac{\pi }{2}}^{-}}\frac{{\cos }^{2}x}{3{\cos }^{2}x}$

$a={lim}_{x\to {\frac{\pi }{2}}^{-}}\frac{1}{3}$

$a=\frac{1}{3}$

Similarly,

$a={lim}_{x\to {\frac{\pi }{2}}^{+}}\frac{b(1-\sin x)}{(\pi -2x{)}^2}$

Apply L-hospital rule,

$a={lim}_{x\to {\frac{\pi }{2}}^{+}}\frac{b(0-\cos x)}{2(\pi -2x)(0-2)}$

$a={lim}_{x\to {\frac{\pi }{2}}^{+}}\frac{-b\cos x}{-4(\pi -2x)}$

$a={lim}_{x\to {\frac{\pi }{2}}^{+}}\frac{b\cos x}{4(\pi -2x)}$

Apply L-hospital rule,

$a={lim}_{x\to {\frac{\pi }{2}}^{+}}\frac{-b\sin x}{4(0-2)}$

$a={lim}_{x\to {\frac{\pi }{2}}^{+}}\frac{b\sin x}{8}$

$\frac{1}{3}=\frac{b\sin \frac{\pi }{2}}{8}$

$\frac{1}{3}=\frac{b}{8}$

$b=\frac{8}{3}$

Hence, this is the answer.