UPSC NDA II 2023 Mathematics Solved Paper

Concept:Given $f(x)=x^2-1$ and $(g \circ f)(x)=x-\sqrt{x}+1$, test each candidate $g(x)$ by substituting $f(x)$ to see if it yields the same expression.Explanation:Check each option by computing $g(f(x))$ and comparing to $x-\sqrt{x}+1$.For option 1: $g(x)=\sqrt{x+1}-\sqrt[4]{x+1}$.$g(f(x)) = \sqrt{(x^2-1)+1} - \sqrt[4]{(x^2-1)+1} = \sqrt{x^2} - \sqrt[4]{x^2} = x - \sqrt{x}$.This is $x-\sqrt{x}$, missing the $+1$. So not correct.For option 2: $g(x)=\sqrt{x+1}-\sqrt[4]{x+1}+1$.$g(f(x)) = \sqrt{(x^2-1)+1} - \sqrt[4]{(x^2-1)+1} + 1 = \sqrt{x^2} - \sqrt[4]{x^2} + 1 = x - \sqrt{x} + 1$.This matches exactly. So option 2 is correct.For option 3: $g(x)=\sqrt{x+1}+\sqrt[4]{x+1}$.$g(f(x)) = \sqrt{(x^2-1)+1} + \sqrt[4]{(x^2-1)+1} = \sqrt{x^2} + \sqrt[4]{x^2} = x + \sqrt{x}$.This is $x+\sqrt{x}$, not matching. So not correct.For option 4: $g(x)=x+1-\sqrt{x+1}+1 = x - \sqrt{x+1} + 2$.$g(f(x)) = (x^2-1) - \sqrt{(x^2-1)+1} + 2 = x^2 - 1 - \sqrt{x^2} + 2 = x^2 - x + 1$.This is $x^2 - x + 1$, not $x - \sqrt{x} + 1$. So not correct.Answer:Option B: $\sqrt{x+1}-\sqrt[4]{x+1}+1$