Question

Let m and M respectively be the minimum and the maximum values of \( f(x) = {\sin ^{ - 1}}2x + \sin 2x + {\cos ^{ - 1}}2x + \cos 2x,\,x \in \left[ {0,{\pi \over 8}} \right] \). Then m + M is equal to :

• A. \( 1 + \sqrt 2 + \pi \)
• B. \( \left( {1 + \sqrt 2 } \right)\pi \)
• C. \( \pi + \sqrt 2 \)
• D. \( 1 + \pi \)

Hint:

Use \(\sin^{-1}t+\cos^{-1}t=\tfrac{\pi}{2}\) to reduce \(f\) to \(\tfrac{\pi}{2}+\sin2x+\cos2x\), then find its extreme values on the interval.

Answer 1

The Correct Option is A

Step 1: Since \(\sin^{-1}2x+\cos^{-1}2x=\tfrac{\pi}{2}\) for \(2x\in[0,\tfrac{\pi}{4}]\), \(f(x)=\tfrac{\pi}{2}+\sin2x+\cos2x=\tfrac{\pi}{2}+\sqrt2\sin\left(2x+\tfrac{\pi}{4}\right)\).
Step 2: For \(x\in[0,\tfrac{\pi}{8}]\), \(2x+\tfrac{\pi}{4}\in[\tfrac{\pi}{4},\tfrac{\pi}{2}]\), where \(\sin\) is increasing from \(\tfrac{1}{\sqrt2}\) to 1.
Step 3: Minimum \(m=\tfrac{\pi}{2}+\sqrt2\cdot\tfrac1{\sqrt2}=\tfrac{\pi}{2}+1\); maximum \(M=\tfrac{\pi}{2}+\sqrt2\).
Step 4: \(m+M=\pi+1+\sqrt2\).
Answer: (A) \(1+\sqrt2+\pi\)