JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 10)
Let $$f(x)=\left\{\begin{array}{ccc}-\mathrm{a} & \text { if } & -\mathrm{a} \leq x \leq 0 \\ x+\mathrm{a} & \text { if } & 0< x \leq \mathrm{a}\end{array}\right.$$ where $$\mathrm{a}> 0$$ and $$\mathrm{g}(x)=(f(|x|)-|f(x)|) / 2$$. Then the function $$g:[-a, a] \rightarrow[-a, a]$$ is
neither one-one nor onto.
both one-one and onto.
one-one.
onto
Explanation
$$\begin{aligned} & f(x)=\left\{\begin{array}{l} -a \quad \text { if }-a \leq x \leq 0 \\ x+a \quad \text { if } 0< x \leq a \end{array}\right. \\ & f(|x|)=\left\{\begin{array}{cc} -a & -a \leq|x| \leq 0 \\ |x|+a & \text { if } 0 < |x| \leq a \end{array}\right. \end{aligned}$$
$$|x|<0$$ is not possible, so
$$\begin{aligned} & f(|x|)= \begin{cases}x+a & -a \leq x \leq 0 \\ -x+a & 0 < x \leq a\end{cases} \\ & |f(x)|= \begin{cases}a & -a \leq x \leq 0 \\ x+a & 0 < x \leq a\end{cases} \\ & h(x)= \begin{cases}-\frac{x}{2} & -a \leq x \leq 0 \\ 0 & 0 < x \leq a\end{cases} \end{aligned}$$
_8th_April_Evening_Shift_en_10_1.png)
Neither one-one nor onto.
Comments (0)
