$f:[-2,2] \rightarrow[-2,2]$ and $g:[-2,2] \rightarrow[0,4]$ are two functions defined as $f(x)=\left\{\begin{array}{l}-2,-2 \leq x \leq 0 \\ x^2-2,0 \leq x \leq 2\end{array}\right.$ and $g(x)=|f(x)|+f(|x|)$, then

• A

  f and g are injective mappings
• Correct answer
  B

  f and g are surjective mappings
• C

  f is bijective mapping and g is injective mapping
• D

  f is not bijective mapping and g is surjective mapping

Solution

Given, $f(x)=\left\{\begin{array}{cc}-2, & -2 \leq x \leq 0 \\ x^2-2, & 0 \leq x \leq 2\end{array}\right.$ $\begin{gathered}\therefore \quad|f(x)|=\left\{\begin{array}{cc}2, & -2 \leq x \leq 0 \\ 2-x^2, & 0 \leq x \leq \sqrt{2} \\ x^2-2, & \sqrt{2} < x \leq 2\end{array}\right. \\ \therefore g(x)=|f(x)|+f(|x|)=\left\{\begin{array}{cc}x^2, & -2 \leq x \leq 0 \\ 0, & 0 \leq x \leq \sqrt{2} \\ 2\left(x^2-2\right), & \sqrt{2} < x \leq 2\end{array}\right.\end{gathered}$
Here, $g(x)$ is not one-one but onto because range $=[0,4]=$ codomain .

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