Question

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

Answer 1

Solving the given function for continuity,

Now solving left hand limit: $f\left(x\right)=\frac{1-{sin}^{2}x}{3{cos}^{2}x}$

At $x=\frac{\Pi }{2},f\left(x\right)=\frac{0}{0}$ form, so using L-hospital rule, we get

$f\left(x\right)=\frac{-2sinxcosx}{-6cosxsinx}=\frac{1}{3}$

Now solving right hand limit:\ $f\left(x\right)=\frac{b(1-sinx)}{{(\pi -2x)}^2}$

This is again of the form $f\left(x\right)=\frac{0}{0}$ form, so using L-hospital rule, we get

$f\left(x\right)=\frac{b(-cosx)}{2(\Pi -2x)(-2)}=\frac{bcosx}{4(\Pi -2x)}$

This is again of the form $f\left(x\right)=\frac{0}{0}$ form, so using L-hospital rule, we get

$f\left(x\right)=\frac{-bsinx}{4(-2)}=\frac{bsinx}{8}$

At $x=\frac{\Pi }{2},f\left(x\right)=\frac{b}{8}$

$L.H.L=a=R.H.L$

$⟹\frac{1}{3}=a=\frac{b}{8}$

We get, $a=\frac{1}{3}$ and $b=\frac{8}{3}$