`Let f(x)=(sin^(-1)(1-{x})xxcos^(-1)(1-{x}))/(sqrt(2{x})xx(1-{x}))`, where `{x}` denotes the fractional part of x.
`R=lim_(xto0+) f(x)` is equal to

Options

A. `np(1-n)`
B. `-np(1+n)`
C. `n^(2)p`
D. `np(1+n)`

Answer

A

Explanation

We have `f(x)=(sin^(-1)(1-{x})cos^(-1)(1-{x}))/(sqrt(2{x})(1-{x}))`
`:.underset(xto0^(+))limf(x)=underset(hto0)limf(0+h)`
`=underset(hto0)lim(sin^(-1)(1-{0+h})cos^(-1)(1-{0+h}))/(sqrt(2{0+h})(1-{0+h}))`
`=underset(hto0)lim(sin^(-1)(1-h)cos^(-1)(1-h))/(sqrt(2h)(1-h))`
`=underset(hto0)lim(sin^(-1)(1-h))/((1-h))underset(hto0)lim(cos^(-1)(1-h))/(sqrt(2)h)`
In second limit, put `1-h=costheta.` Then
`underset(xto0^(+))limf(x)=underset(hto0)lim(sin^(-1)(1-h))/((1-h))underset(hto0)lim(cos^(-1)(costheta))/(sqrt(2(1-costheta)))`
`=underset
(hto0)lim(sin^(-1)(1-h))/((1-h))underset(thetato0)lim(theta)/(2sin(theta//2))(becausethetagt0)`
`=sin^(-1)1xx1=pi//2`
and `underset(xto0^(-))limf(x)=underset(hto0)limf(0-h)`
`=underset(hto0)lim(sin^(-1)(1-{0-h})cos^(-1)(1-{0-h}))/(sqrt(2{0-h}")")(1-{0-h}))`
`=underset(hto0)lim(sin^(-1)(1+h-1)cos^(-1)(1+h-1))/(sqrt(2(-h+1))(1+h-1))`
`=underset(hto0)lim(sin^(-1)h)/(h)underset(hto0)lim(cos^(-1)h)/(sqrt(2(1-h)))`
`=1(pi//2)/(sqrt(2))=(pi)/(2sqrt(2))`