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MHT CET · Maths · Inverse Trigonometric Functions

The number of real solutions of \(\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}\) is

  1. A zero.
  2. B one.
  3. C two.
  4. D infinite.
Verified Solution

Answer & Solution

Correct Answer

(C) two.

Step-by-step Solution

Detailed explanation

\(\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}\)
\(\tan ^{-1} \sqrt{x(x+1)}\) is defined when
\(x(x+1) \geq 0\)
\(\sin ^{-1} \sqrt{x^2+x+1}\) is defined when
\(x(x+1)+1 \leq 1\) or \(x(x+1) \leq 0\)
From (i) and (ii),
\(x(x+1)=0\) or \(x=0\) and -1 .
Hence, number of solutions is 2 .