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(i)

(ii)
$0 \le x \le 2$

(iii)
$\int_{-1}^1 f(|x|) dx = \int_1^a |f(x)| dx$

$2 \int_0^1 2-x dx = \int_1^2 2-x dx + \int_2^a -(2-x) dx$

$2[2x - \frac{x^2}{2}]\bigl|_0^1 = [2x - \frac{x^2}{2}]\bigl|_1^2 + [\frac{x^2}{2} - 2x]\bigl|_2^a$

$3 = \frac{1}{2} + \frac{a^2}{2} - 2a + 2$

$a = 2 \pm \sqrt{5}$

$\therefore a = 2 + \sqrt{5}$ since $a > 2$

KS Comments:

There are other alternative methods to solving the integration like using the area of trapezoid.

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