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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 one
  2. B zero
  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
\(x(x+1) \geq 0...(i)\)
\(\sin ^{-1} \sqrt{x^2+x+1}\) is defined when
\(x(x+1)+1 \leq 1 \text { or } x(x+1) \leq 0...(ii)\)
From (i) and (ii),
\(x(x+1)=0 \Rightarrow x=0 \text { or }-1\)
Hence, number of solutions is 2 .