If the range of`y=sin^(-1)+cos^(-1)+tan^(-1) x is [k,K]`, then

Options

A. `k=0,K=pi`
B. `k=(pi)/(4),K=(3pi)/(4)`
C. `k=(pi)/(2),K=pi`
D. None of these

Answer

B

Explanation

We have,
`sin^(-1)x+cos^(-1)x+tan^(-1)x=(pi)/(2)+tan^(-1)x`
Domain of above function is `[-1,1]`
Since, `-1le x le 1 implies -(pi)/(4) le tan^(-1) x le (pi)/(4)`
` implies (pi)/(2)-(pi)/(4) le (pi)/(2) + tan^(-1) x le (pi)/(2)+(pi)/(4)`
So, `k=(pi)/(4), K=(3pi)/(4)`