Question

Find the values of $a$ and $b$ so that the function, $f\left(x\right)=\{\begin{array}{c}\frac{1-{\sin }^{2}x}{3{\cos }^{2}x},x<\pi /2\\ a,x=\pi /2\\ \frac{b(1-\sin x)}{{(\pi -2x)}^2},x>\pi /2\end{array}$ is continuous.

Answer 1

For f(x) to be continuous $\underset{x\to {\pi /2}^{-}}{lim}f\left(x\right)=f\left(\pi /2\right)=\underset{x\to {\pi /2}^{+}}{lim}f\left(x\right)$

$\underset{x\to {\pi /2}^{-}}{lim}f\left(x\right)=\underset{x\to {\pi /2}^{-}}{lim}\frac{1-{\sin }^{2}x}{3{\cos }^{2}x}$

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

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

$\underset{x\to {\pi /2}^{+}}{lim}f\left(x\right)=\underset{x\to {\pi /2}^{+}}{lim}\frac{b(1-\sin x)}{(\pi -2x{)}^2}$

Multiply and divide by $(1+\sin x)$, then put $(1-\sin x)(1+\sin x)=1-{\sin }^{2}x={\cos }^{2}x$

$\Rightarrow \underset{x\to {\pi /2}^{+}}{lim}\frac{b\left({\cos }^{2}x\right)}{(\pi -2x{)}^2(1+\sin x)}$

$=\underset{x\to {\pi /2}^{+}}{lim}\frac{b\left({\sin }^2\right(\pi /2-x\left)\right)}{(\pi -2x{)}^2(1+\sin x)}$

$=\underset{x\to {\pi /2}^{+}}{lim}\frac{b\left({\sin }^2\right(\pi /2-x\left)\right)}{4(\pi /2-x{)}^2(1+\sin x)}$

Putting limit for $(1+\sin x)$ because it's value is not affecting the limit,

$=\underset{x\to {\pi /2}^{+}}{lim}\frac{b\left({\sin }^2\right(\pi /2-x\left)\right)}{4(\pi /2-x{)}^2(2)}$

$=\frac{1}{8}\underset{x\to {\pi /2}^{+}}{lim}\frac{b\left({\sin }^2\right(\pi /2-x\left)\right)}{(\pi /2-x{)}^2}$

Put $h=\pi /2-x$, we get

$=\frac{1}{8}\underset{h\to 0^{+}}{lim}\frac{b\left({\sin }^2\right(h\left)\right)}{(h{)}^2}$

$=\frac{b}{8}$

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

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

Hence, $a=\frac{1}{3},b=\frac{8}{3}$