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Question: If \(k \leq \sin ^ { - 1 } x + \cos ^ { - 1 } x + \tan ^ { - 1 } x \leq K\)then....

If k≤sin⁡−1x+cos⁡−1x+tan⁡−1x≤Kk \leq \sin ^ { - 1 } x + \cos ^ { - 1 } x + \tan ^ { - 1 } x \leq Kthen.

A

k=0,K=πk = 0 , K = \pi

B

k=0,K=π2k = 0 , K = \frac { \pi } { 2 }

C

k=π2,K=πk = \frac { \pi } { 2 } , K = \pi

D

None of these

Answer

k=0,K=πk = 0 , K = \pi

Explanation

Solution

We have

sin⁡−1x+cos⁡−1x+tan⁡−1x=π2+tan⁡−1x\sin ^ { - 1 } x + \cos ^ { - 1 } x + \tan ^ { - 1 } x = \frac { \pi } { 2 } + \tan ^ { - 1 } x

Since −π2≤tan⁡−1x≤π2⇒0≤π2+tan⁡−1x≤π\frac { - \pi } { 2 } \leq \tan ^ { - 1 } x \leq \frac { \pi } { 2 } \Rightarrow 0 \leq \frac { \pi } { 2 } + \tan ^ { - 1 } x \leq \pi

∴ K=π,k=0K = \pi , k = 0.