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

Correct answer: B. x+1 - [4] x+1 +1 aligned & Given, f(x)=x^2-1 , go f(x)=x- x +1 & g\ f(x)\ =x- x +1 & g(x^2-1)=x- x +1 & Let, x^2-1=t x^2=1+t x=(1+t)^ 1 / 2 & g(t)=(1+t)^ 1 / 2 -(1+t)^ 1 / 4 +1 & g(x)= 1+x - [4] 1+x +1 aligned

NDA 2023

Exam: NDA
Year: 2023
Subject: Mathematics
Chapter: Functions

NDA 2023 · Mathematics · Functions

Let f(x)=x^2-1 and gof(x)=x- x +1 Which one of the following is a...

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)?

Options

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

Correct answer

B. $\sqrt{x+1}-\sqrt[4]{x+1}+1$

Step-by-step 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}

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