Question

If \(a\) and \(b\) are respectively the minimum and maximum values of the expression \(f(x)=(\sin ^{-1}x+2\cos ^{-1}x)\), then \(a+b\) is equal to...

• A. \(\frac{\pi }{2}\)
• B. \(\pi\)
• C. \(\frac{3\pi }{2}\)
• D. \(2\pi\)

Hint:

Use the identity connecting sin inverse and cos inverse to write the expression in terms of cos inverse alone, then use its range.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Use \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) to reduce the expression to a single inverse function.

Step 2: Simplify
\(f(x)=\sin^{-1}x+2\cos^{-1}x=\left(\sin^{-1}x+\cos^{-1}x\right)+\cos^{-1}x=\frac{\pi}{2}+\cos^{-1}x\).

Step 3: Find the extreme values
For \(x\in[-1,1]\), \(\cos^{-1}x\) varies from 0 to \(\pi\).
So the minimum is \(a=\frac{\pi}{2}\) and the maximum is \(b=\frac{3\pi}{2}\).

Step 4: Add
\(a+b=\frac{\pi}{2}+\frac{3\pi}{2}=2\pi\).

Final Answer:
The sum of the minimum and maximum values is 2π. \[ \boxed{\text{(D) }2\pi} \]