$f_1(x)=2 x, f_2(x)=3 \sin x-x \cos x$ then for $x \in(0, \pi / 2)$

$f_1(x)<f_2(x)$ $f_1(x)>f_2(x)$ $f_1(|x|)<f_2(|x|)$ none of these

$f_1(x)>f_2(x)$

Let h(x) = f1(x) – f2(x) = 2x – 3 sin x + x cos x h(0) = 0 h’(x) = 2 – 2 cos x – x sin x h’’(x) = sin x – x cos x h’’’(x) = x sin x h’’’(x) > 0 ⇒ h’’(x) > 0 ⇒ h’(x) > 0 ⇒ h(x) > 0