JEE MAIN - Mathematics (2024 - 4th April Morning Shift - No. 1)
$$\text { Let } f(x)=\left\{\begin{array}{lr}
-2, & -2 \leq x \leq 0 \\
x-2, & 0< x \leq 2
\end{array} \text { and } \mathrm{h}(x)=f(|x|)+|f(x)| \text {. Then } \int_\limits{-2}^2 \mathrm{~h}(x) \mathrm{d} x\right. \text { is equal to: }$$
2
6
4
1
Explanation
$$f(x)=\left\{\begin{array}{cc} -2 & -2 \leq x \leq 0 \\ x-2 & 0< x \leq 2 \end{array} h(x)=f|x|+|f(x)|\right.$$
_4th_April_Morning_Shift_en_1_1.png)
$$\begin{aligned} & h(x)=\left\{\begin{array}{cc} -x-2+2=-x & -2 \leq x \leq 0 \\ 0 & 0< x \leq 2 \end{array}\right. \\ & \therefore \int_\limits{-2}^2 h(x) d x=\int_\limits{-2}^0-x d x+\int_\limits0^2 0 d x \\ & \left.\frac{x^2}{2}\right|_{-2} ^0=\frac{4}{2}=2 \end{aligned}$$
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