If `f(x)=(1-sinx)/((pi-2x)^2),w h e nx!=pi/2a n df(pi/2)=lambda,` the `f(x)` will be continuous function at `x=pi/2,w h e r elambda=` `1/8` (b) `1/4` (c) `1/2` (d) none of these

Explanation

`LHL=lim_(h→0)​f((π/2​)−h)`

Given,

`f(x)=(1-sinx)/((pi-2x)^2)`


`=lim_(h→0)​(1-sin((π/2​)−h))/((pi-2((π/2​)−h))^2)`

`=lim_(h→0)1−cosh​/((π−π+2h)^2)×(1+cosh​)/(1+cosh)`


`=lim_(h→0)(sin^2h)/(​(4h^2(1+cosh)​)`

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