`f(x)={bsin^-1((x+c)/2),-1/2 lt xlt0 and 1/2 , x=0 and e^(((ax)/2)-1)/x , o lt x lt 1/2` If `f(x)` is differentiable at `x=0 and |c|<1/2`, then find thevalue of a and prove that `64b^2=(4-c^2)`.

Answer

`(a = 1)`

Explanation

Since, f(x) is differentiable at x = 0 .
`rArr` It is continuous at x = 0 .
i.e. `" "underset( x to 0) lim f(x) = underset( x to 0^(-)) lim f(x) = f(0)`
Here, ` underset( x to 0^(+)) lim f(x) = underset( h to 0) lim (e^(ah//2) -1)/h - underset ( h to 0) lim ( e^(ah//2) - 1) / (a h/2) * a/2 = a/2 `
Also, ` underset( x to 0^(-)) lim f(x) = underset( h to 0) lim b sin^(-1) ((c-h)/2) = b sin^(-1) c/2`
`:." " b sin^(-1) c/2 = a/2 = 1/2 `
`rArr" " a = 1`
Also, it is differentiable at x = 0
`R f'(0^(+)) = underset( h to 0) lim (h" "2)/h " "[:' a = 1]`
` = underset( h to 0) lim (2e^(h//2) -2-h)/(2h^(2)) = 1/8`
and `L f'(0^(-)) = underset( h to 0) lim (b sin^(-1)((c-h)/2)-1/2)/(-h) = (b//2)/sqrt(1-c^(2)/4) `
`:." " b/(sqrt(4-c^(2)) = 1/8`
`rArr" " 64b^(2) = (4-c^(2)) `
` rArr" " a=1 and 64b^(2) =(4-c^(2)) `