JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 2)
Let $$f, g: \mathbf{R} \rightarrow \mathbf{R}$$ be defined as :
$$f(x)=|x-1| \text { and } g(x)= \begin{cases}\mathrm{e}^x, & x \geq 0 \\ x+1, & x \leq 0 .\end{cases}$$
Then the function $$f(g(x))$$ is
neither one-one nor onto.
one-one but not onto.
both one-one and onto.
onto but not one-one.
Explanation
$$\begin{aligned} & f(x)= \begin{cases}x-1, & x \geq 1 \\ 1-x & x<0\end{cases} \\ & g(x)=\left\{\begin{array}{cc} e^x & ; \quad x \geq 0 \\ x+1 & ; \quad x \leq 0 \end{array}\right. \end{aligned}$$
_5th_April_Evening_Shift_en_2_1.png)
$$\begin{aligned} f(g(x)) & = \begin{cases}g(x)-1, & g(x) \geq 1 \\ 1-g(x) & g(x)<1\end{cases} \\ & =\left\{\begin{array}{cl} e^x-1 & x \geq 0 \\ -x & x<0 \end{array}\right. \end{aligned}$$
_5th_April_Evening_Shift_en_2_2.png)
Comments (0)
