Question

Let $f\left(x\right)=\left\{\begin{array}{cc}\frac{1-{\sin }^{3}x}{3{\cos }^{2}x},& \mathrm{if}x<\frac{\mathrm{\pi }}{2}\\ a,& \mathrm{if}x=\frac{\mathrm{\pi }}{2}\\ \frac{b(1-\sin x)}{(\mathrm{\pi }-2\mathrm{x}{)}^2},& \mathrm{if}x>\frac{\mathrm{\pi }}{2}\end{array}.\right.$If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$, find a and b.

Answer 1

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

We have
(LHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\frac{\mathrm{\pi }}{2}-h\right)$

$$\begin{gathered} =\underset{h\to 0}{\lim }\left(\frac{1-{\sin }^3\left(\frac{\mathrm{\pi }}{2}-h\right)}{3{\cos }^2\left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{1-{\cos }^{3}h}{3{\sin }^{2}h}\right) \\ =\frac{1}{3}\underset{h\to 0}{\lim }\left(\frac{\left(1-\cos h\right)\left(1+{\cos }^{2}h+\cos h\right)}{\left(1-\cos h\right)\left(1+\cos h\right)}\right) \\ =\frac{1}{3}\underset{h\to 0}{\lim }\left(\frac{\left(1+{\cos }^{2}h+\cos h\right)}{\left(1+\cos h\right)}\right) \\ =\frac{1}{3}\left(\frac{1+1+1}{1+1}\right)=\frac{1}{2} \end{gathered}$$


(RHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\frac{\mathrm{\pi }}{2}+h\right)$
$$\begin{gathered} =\underset{h\to 0}{\lim }\left(\frac{b\left[1-\sin \left(\frac{\mathrm{\pi }}{2}+h\right)\right]}{{\left[\mathrm{\pi }-2\left({\displaystyle \frac{\mathrm{\pi }}{2}+}h\right)\right]}^2}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{b\left(1-\cos h\right)}{{\left[-2h\right]}^2}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{2b{\sin }^2{\displaystyle \frac{h}{2}}}{4{\displaystyle h^2}}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{2b{\sin }^2{\displaystyle \frac{h}{2}}}{1{\displaystyle 6}{\displaystyle \frac{{\displaystyle h^2}}{4}}}\right) \\ =\frac{b}{8}\underset{h\to 0}{\lim }{\left(\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\right)}^2 \\ =\frac{b}{8}\times 1 \\ =\frac{b}{8} \end{gathered}$$

Also, $f\left(\frac{\mathrm{\pi }}{2}\right)=a$

If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$ , then
$\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }f\left(x\right)=\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\lim }f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$

$\Rightarrow \frac{1}{2}=\frac{b}{8}=a$

$\Rightarrow a=\frac{1}{2}\mathrm{and}b=4$