Question

If the functions $f,g$ and $h$ are defined from the set of real numbers $R\to R$ such that $f\left(x\right)=x^2-1$, $g\left(x\right)=\sqrt{x^2+1}$ and $h\left(x\right)=\left\{\begin{array}{ll}0,& \mathrm{if}x\le 0\\ x,& \mathrm{if}x\ge 0\end{array}\right.$ then $h\circ \left(f\circ g\right)\left(x\right)$ is defined by

• A. $1$
• B. $0$
• C. $-1$
• D. $x^2$

Answer 1

The correct option is D

$x^2$

Explanation for the correct option:

The given functions are $f\left(x\right)=x^2-1$, $g\left(x\right)=\sqrt{x^2+1}$ and $h\left(x\right)=\left\{\begin{array}{ll}0,& \mathrm{if}x<0\\ x,& \mathrm{if}x\ge 0\end{array}\right.$.

Thus, $h\circ \left(f\circ g\right)\left(x\right)=h\left[f\left\{g\left(x\right)\right\}\right]$

$$\begin{gathered} \Rightarrow h\circ \left(f\circ g\right)\left(x\right)=h\left[f\left\{\sqrt{x^2+1}\right\}\right]\mathbf{}\left[\because g\left(x\right)=\sqrt{x^2+1}\right] \\ \Rightarrow h\circ \left(f\circ g\right)\left(x\right)=h\left[{\left(\sqrt{x^2+1}\right)}^2-1\right]\left[\because f\left(x\right)=x^2-1\right] \\ \Rightarrow h\circ \left(f\circ g\right)\left(x\right)=h\left[x^2+1-1\right] \\ \Rightarrow h\circ \left(f\circ g\right)\left(x\right)=h\left(x^2\right) \end{gathered}$$

As $x^2\ge 0$, thus $h\left(x^2\right)=x^2$.

Therefore, $h\circ \left(f\circ g\right)\left(x\right)=x^2$.

Hence, option (D) is the correct answer.