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
- A zero.
- B one.
- C two.
- D infinite.
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 .
\(\tan ^{-1} \sqrt{x(x+1)}\) 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 .
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