Question:
The number of real solutions of the equation
\[
\tan^{-1}\sqrt{x(x+1)}
+
\sin^{-1}\sqrt{x^2+x+1}
=
\frac{\pi}{2}
\]
is
The number of real solutions of the equation
\[
\tan^{-1}\sqrt{x(x+1)}
+
\sin^{-1}\sqrt{x^2+x+1}
=
\frac{\pi}{2}
\]
is
Show Hint
For inverse trigonometric equations, check the domain first before solving.
Updated On: Jun 3, 2026
- $0$
- $1$
- $2$
- Infinitely many
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Use domain restrictions and the identity involving inverse trigonometric functions.
Step 2: Meaning
Since \[ \sin^{-1}\sqrt{x^2+x+1} \] is defined, we require \[ 0\le x^2+x+1\le1. \] Since \[ x^2+x+1=1+x(x+1), \] we get \[ x(x+1)\le0. \] Thus \[ -1\le x\le0. \]
Step 3: Analysis
Let \[ t=\sqrt{x(x+1)}. \] Because \[ -1\le x\le0, \] we must have \[ x(x+1)\le0. \] Hence \[ t=0. \] Therefore \[ \tan^{-1}(0)+\sin^{-1}(1)=0+\frac{\pi}{2} =\frac{\pi}{2}. \] Thus the equation is satisfied whenever \[ x(x+1)=0. \] Hence \[ x=0 \quad\text{or}\quad x=-1. \]
Step 4: Conclusion
Therefore the number of real solutions is $2$.
Final Answer: (C)
Use domain restrictions and the identity involving inverse trigonometric functions.
Step 2: Meaning
Since \[ \sin^{-1}\sqrt{x^2+x+1} \] is defined, we require \[ 0\le x^2+x+1\le1. \] Since \[ x^2+x+1=1+x(x+1), \] we get \[ x(x+1)\le0. \] Thus \[ -1\le x\le0. \]
Step 3: Analysis
Let \[ t=\sqrt{x(x+1)}. \] Because \[ -1\le x\le0, \] we must have \[ x(x+1)\le0. \] Hence \[ t=0. \] Therefore \[ \tan^{-1}(0)+\sin^{-1}(1)=0+\frac{\pi}{2} =\frac{\pi}{2}. \] Thus the equation is satisfied whenever \[ x(x+1)=0. \] Hence \[ x=0 \quad\text{or}\quad x=-1. \]
Step 4: Conclusion
Therefore the number of real solutions is $2$.
Final Answer: (C)
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