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 one
- B zero
- 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
\(x(x+1) \geq 0...(i)\)
\(\sin ^{-1} \sqrt{x^2+x+1}\) 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 .
\(\tan ^{-1} \sqrt{x(x+1)}\) 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 .
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