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If the range of`y=sin^(-1)+cos^(-1)+tan^(-1) x is [k,K]`, thenA. `k=0,K=pi`B. `k=(pi)/(4),K=(3pi)/(4)`C. `k=(pi)/(2),K=pi`D. None of these |
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Answer» Correct Answer - B 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)` |
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