Question

If \(\cos^{-1}p + \cos^{-1}q + \cos^{-1}r = 3\pi\), then \(p^2 + q^2 + r^2 + 2pqr\) is equal to

• A. 3
• B. 1
• C. 2
• D. -1

Hint:

The principal value range of inverse cosine determines possible sums.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The range of \(\cos^{-1}x\) is \([0, \pi]\). For the sum to be \(3\pi\), each must be \(\pi\).

Step 2: Detailed Explanation:
Since \(0 \le \cos^{-1}x \le \pi\), the maximum sum of three such terms is \(3\pi\), achieved only when each term equals \(\pi\). Thus \(\cos^{-1}p = \pi \Rightarrow p = \cos\pi = -1\). Similarly \(q = -1\), \(r = -1\). Then \(p^2 + q^2 + r^2 + 2pqr = (-1)^2 + (-1)^2 + (-1)^2 + 2(-1)(-1)(-1) = 1+1+1 -2 = 1\).

Step 3: Final Answer:
1, which corresponds to option (B).