Let f ( x ) = { - 2 , - 2 ? x ? 0 x - 2 , 0 < x ? 2 and h ( x ) = f ( | x | ) + | f ( x ) | . Then ? - 2 2 h ( x ) ? d x is equal to [2024]

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Solution

Q.
Let $f (x)$ $= {\begin{matrix} - 2, & - 2 \leq x \leq 0 \\ x - 2, & 0 < x \leq 2 \end{matrix}$ and $h (x) = f (| x |) + | f (x) | .$ Then $\int_{- 2}^{2} h (x) d x$ is equal to [2024]

1 6

2 2

3 4

4 1

Ans.
(2)

$f (x)$ $= {\begin{matrix} - 2, & - 2 \leq x \leq 0 \\ x - 2, & 0 < x \leq 2 \end{matrix}$

$h (x) = f (| x |) + | f (x) |$

$f (| x |)$ $= {\begin{matrix} - x - 2, & - 2 \leq x \leq 0 \\ x - 2, & 0 < x \leq 2 \end{matrix}$

$| f (x) |$ $= {\begin{matrix} 2, & - 2 \leq x \leq 0 \\ - (x - 2), & 0 < x \leq 2 \end{matrix}$

So, $h (x)$ $= {\begin{matrix} - x - 2 + 2 = - x, & - 2 \leq x \leq 0 \\ x - 2 - (x - 2) = 0, & 0 < x \leq 2 \end{matrix}$

$∴$ $\int_{- 2}^{2} h (x) d x = \int_{- 2}^{0} - x d x + \int_{0}^{2} 0 d x$

$= - {[\frac{x^{2}}{2}]}_{- 2}^{0} = - [0 - \frac{4}{2}] = 2$

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