Question

$f\left(x\right)=\frac{1-\sin x}{{(\pi -2x)}^2}.\frac{\log \sin x}{\log (1+{\pi }^2-4\pi x+4x^2)},x\ne \pi /2$ The assigned to function at $x=\pi /2$ in order that it may be continuous at $x=\pi /2$ is $-\frac{1}{m}.$ Find m

Answer 1

For continuity at $x=\pi /2$,we must have Lt= Value$f\left(x\right)=\frac{1-\sin x}{4{(\frac{\pi }{2}-x)}^2},\frac{\log \sin x}{\log \{1+4{(\frac{\pi }{2}-x)}^2\}}$ Put $\frac{\pi }{2}-x=torx=\frac{\pi }{2}-t$ As $x\to \pi /2,t\to 0$ $\therefore \stackrel{Lt}{x\to \pi /2}f\left(x\right)=\stackrel{Lt}{t\to 0}\frac{1-\cos t}{4t^2},\stackrel{Lt}{t\to 0}\frac{\log \cos t}{\log (1+4t^2)}$ $=\stackrel{Lt}{t\to 0}\frac{2{\sin }^2\frac{t}{2}}{4t^2}.\stackrel{Lt}{t\to 0}\frac{\log (1-\frac{t^2}{2!}+...)}{\log (1+4t^2)}$ $=\stackrel{Lt}{t\to 0}\frac{2}{4}.\frac{{\left(t/2\right)}^2}{t^2}.\stackrel{Lt}{t\to 0}\frac{-\frac{t^2}{2}-\frac{1}{2}{\left(\frac{t^2}{2}\right)}^2...}{4t^2-\frac{1}{2}{\left(4t^2\right)}^2}$ $=\frac{1}{8}(-\frac{1}{8})=-\frac{1}{64}$
Hence $f\left(\frac{\pi }{2}\right)=-\frac{1}{64}.$