Q191 / 270 · NDA 2023 (Phase 2) · Mathematics

Let $f(x)=x^2-1$ and $\operatorname{gof}(x)=x-\sqrt{x}+1$ Which one of the following is a possible expression for g(x)?

  • A
    $\sqrt{x+1}-\sqrt[4]{x+1}$
  • Correct answer
    B
    $\sqrt{x+1}-\sqrt[4]{x+1}+1$
  • C
    $\sqrt{x+1}+\sqrt[4]{x+1}$
  • D
    $x+1-\sqrt{x+1}+1$

Solution

$\begin{aligned} & \text { Given, } \mathrm{f}(\mathrm{x})=\mathrm{x}^2-1 \text {, go } \mathrm{f}(\mathrm{x})=\mathrm{x}-\sqrt{\mathrm{x}}+1 \\ & \Rightarrow \mathrm{~g}\{\mathrm{f}(\mathrm{x})\}=\mathrm{x}-\sqrt{\mathrm{x}}+1 \\ & \Rightarrow \mathrm{~g}\left(\mathrm{x}^2-1\right)=\mathrm{x}-\sqrt{\mathrm{x}}+1 \\ & \quad \text { Let, } \mathrm{x}^2-1=\mathrm{t} \Rightarrow \mathrm{x}^2=1+\mathrm{t} \Rightarrow \mathrm{x}=(1+\mathrm{t})^{1 / 2} \\ & \Rightarrow \mathrm{~g}(\mathrm{t})=(1+\mathrm{t})^{1 / 2}-(1+\mathrm{t})^{1 / 4}+1 \\ & \Rightarrow \mathrm{~g}(\mathrm{x})=\sqrt{1+\mathrm{x}}-\sqrt[4]{1+\mathrm{x}}+1\end{aligned}$

Hence, the correct answer is option: 'B'.

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