Q42 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics

Define the functions $f, g$ and $h$ from R to R such that $f(x)=x^2-1, g(x)=\sqrt{x^2+1}$ and $h(x)=\left\{\begin{array}{l}0, \text { if } x \leq 0 \\ x, \text { if } x \geq 0\end{array}\right.$ consider the following statements

  • A
    fog is invertible
  • B
    $h$ is an identify function
  • Correct answer
    C
    fog is not invertible
  • D
    $(h \circ f \circ g) x=x^2$

Solution

$f \circ g(x)=f\left(\sqrt{x^2+1}\right)=\left(x^2+1\right)-1=x^2$ $\because$ Codomain of $\operatorname{fog}(x)=\mathrm{R}$ and Range of fog $(x)=[0, \infty)$ $\therefore$ fog is not onto function $\quad \Rightarrow$ fog is not invertible. Now $(h \circ f \circ g) x=h \circ[f \circ g(x)]=h\left(x^2\right)$ $\because x^2 \geq 0 \forall x \in \mathrm{R}$. Hence, $(h \circ$ fog $)(x)=x^2$ also given, $h(x)=\left\{\begin{array}{ll}0, & \text { if } x \leq 0 \\ x, & \text { if } x \geq 0\end{array}\right.$ is not identity function because for $x \lt 0, h(x) \neq x$

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

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