Let f: R → R be given by f(x) = (x - 1)(x - 2)(x - 5) . Define F(x)...

$F\left(x\right) = \mathop{∫}_{0}^{x}f\left(t\right) . dt$
${F}^{'}\left(x\right) = f$\left(x\right) = \left(x - 1\right)\left(x - 2\right)\left(x - 5\right)$<br/><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/7ee4711f-f024-4fd5-9bd0-8574b04c363c-image2.png"/><br/>$⇒ F\left(x\right)$has maxima at$x = 2$<br/>and minima at$x = 1$and$x = 5$<br/>now$F\left(2\right) = \mathop{∫}_{0}^{2}\left({x}^{3} - 8{x}^{2} + 17x - 10\right) . dx$<br/>$= {\left[\dfrac{{x}^{4}}{4} - \dfrac{8{x}^{3}}{3} + \dfrac{17{x}^{2}}{2} - 10x\right]}_{0}^{2}$<br/>$F\left(2\right) = \dfrac{- 10}{3}$which is a maxima in$x \in \left(0, 5\right)$<br/>$F\left(0\right) = 0$<br/><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/98bf389d-7b99-411d-bc48-e31281a3378b-image3.png"/><br/>Hence$F\left(x\right) < 0 \forall x \in \left(0, 5\right)