GUJCET 2025 Math Paper

Concept:Use domain restrictions of inverse trigonometric functions and solve the resulting algebraic condition.Explanation:The given equation is $\tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^2+x+1}=\frac{\pi}{2}$.For $\sqrt{x(x+1)}$ to be real, we need $x(x+1)\ge 0$, so $x^2+x\ge 0$.For $\sin^{-1}\sqrt{x^2+x+1}$, the argument must lie in $[0,1]$.Since $\sqrt{x^2+x+1}\ge 0$, we only need $x^2+x+1\le 1$.This gives $x^2+x\le 0$.Combining $x^2+x\ge 0$ and $x^2+x\le 0$, we get $x^2+x=0$.Then $\sqrt{x(x+1)}=0$ and $\sqrt{x^2+x+1}=1$.Substitute these values: $\tan^{-1}0+\sin^{-1}1=0+\frac{\pi}{2}=\frac{\pi}{2}$.So the equation is satisfied exactly when $x^2+x=0$.Factorising: $x(x+1)=0$, giving $x=0$ or $x=-1$.Both values are valid real solutions.Thus, the number of real solutions is $2$.Answer:The equation has $2$ real solutions, so the correct option is C.