If `f(x)={(sin(cosx)-cosx)/((pi-2x)^2),x!=pi/2; k, x=pi/2` is continuous at `x=pi/2,` then `k` is equal to
(a)0 (b) `1/2` (c) 1 (d) `-1`

Explanation

Given:

`f(x)=(sin(cos x)-cosh)/((pi-2x)^(2))
x!=pi//2`

`k ,x=pi//2`

`l t_(x->((pi)/(2))^(-))f(x)=lim_(x->((pi)/(2))^(+))f(x)=f(pi//2)=k`

let `x->(pi)/(2)-h `,
` x->(pi)/(2)implies h ->0`

`Lt_(h->0)(sincos((pi)/(2)-h)-cos((pi)/(2)-h))/[(pi-(2)((pi)/(2)-h)]^(2))`

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