HCU MCA 2013 Paper

(b) Given that,
$$g(x) = $\underset{-10}{\overset{\mathrm{x}}{\int }}$tf'(t) dt$$x ≥ - 10.
Differentiating on both sides, we get
$$g ' (x) = xf' (x) $\times$ 1$$ (by Leibnitz rule)
⇒$$ g'(x) = xf'(x)
∴$$x ≥ - 10.
For increasing of g(x),
$$g '(x)> 0 ⇒ xf'(x) > 0
Hence, g(x) in increasing for x> 0 and decreasing for- 10 < x < 0.