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Mathematics Question on Inverse Trigonometric Functions

If k≤sin⁡−1x+cos⁡−1x+tan⁡−1x≤Kk \le \sin^{-1} x + \cos^{-1} x + \tan^{-1 } x \le K, then

A

k=−π,K=πk = - \pi, K = \pi

B

k=0,K=π2k = 0, K = \frac{\pi}{2}

C

k=π4,K=3π4 k = \frac{\pi}{4} , K = \frac{3 \pi}{4}

D

k=0,K=πk = 0, K = \pi

Answer

k=π4,K=3π4 k = \frac{\pi}{4} , K = \frac{3 \pi}{4}

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 Now sin⁡−1 x\sin^{-1} \, x and cos⁡−1x\cos^{-1} x are defined only if −1≤x≤1-1 \le x \le 1 So, −π4≤tan⁡−1x≤π4⇒π4≤π2+tan⁡−1x≤3π4 - \frac{\pi}{4} \le\tan^{-1} x \le\frac{\pi}{4} \Rightarrow \frac{\pi}{4} \le\frac{\pi}{2} + \tan^{-1} x \le \frac{3 \pi}{4} ∴k=π4 \therefore k = \frac{\pi}{4} and k=3π4k = \frac{3\pi}{4}