NIMCET 2022 — Mathematics PYQ

Solution of the Equation 

The given equation is ${\tan }^{-1}\sqrt{x^2+x}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$. 

1. The identity ${\sin }^{-1}y+{\cos }^{-1}y=\frac{\pi }{2}$ is used. 

2. The equation is rewritten as ${\tan }^{-1}\sqrt{x^2+x}=\frac{\pi }{2}-{\sin }^{-1}\sqrt{x^2+x+1}$. 

3. This simplifies to ${\tan }^{-1}\sqrt{x^2+x}={\cos }^{-1}\sqrt{x^2+x+1}$. 

4. Let $\theta ={\tan }^{-1}\sqrt{x^2+x}$. Then $\tan \theta =\sqrt{x^2+x}$. 

5. A right-angled triangle is considered where the opposite side is $\sqrt{x^2+x}$ and the adjacent side is $1$. [3, 4]  

6. The hypotenuse is calculated using the Pythagorean theorem: $H=\sqrt{(\sqrt{x^2+x}{)}^2+1^2}=\sqrt{x^2+x+1}$. 

7. From this triangle, $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{1}{\sqrt{x^2+x+1}}$. 

8. Therefore, $\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{x^2+x+1}}\right)$. 

9. Equating the two expressions for $\theta$, it is obtained that ${\cos }^{-1}\left(\frac{1}{\sqrt{x^2+x+1}}\right)={\cos }^{-1}\sqrt{x^2+x+1}$. [5]  

10. This implies $\frac{1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}$. [1, 2, 5]  

11. Cross-multiplication yields $1=(\sqrt{x^2+x+1}{)}^2$. 

12. This simplifies to $1=x^2+x+1$. 

13. Rearranging the terms gives $x^2+x=0$. 

14. Factoring out $x$ results in $x(x+1)=0$. 

15. The solutions for $x$ are $x=0$ or $x=-1$. 

16. These values are checked for validity in the original equation's domain. For ${\tan }^{-1}\sqrt{x^2+x}$, $x^2+x\ge 0$, which is satisfied by $x=0$ and $x=-1$. For ${\sin }^{-1}\sqrt{x^2+x+1}$, $0\le x^2+x+1\le 1$. 

• If $x=0$, $0\le 0^2+0+1\le 1⟹0\le 1\le 1$, which is true. 

• If $x=-1$, $0\le (-1{)}^2+(-1)+1\le 1⟹0\le 1-1+1\le 1⟹0\le 1\le 1$, which is true. 

17. Both $x=0$ and $x=-1$ are valid solutions. 

Final Answer 

The solutions of the equation are $0$ and $-1$. Therefore, option (c) is the correct answer.