QJEEAdv19Mab36a: JEE Advanced 2019 Mathematics - Class 12 JEEAdv19Mab36a Solution: Let f : R R be given by . Define , x > 0 Then which of the…

Given, f : R $\to$ and

f(x) = (x $-$ 1)(x $-$ 2)(x $-$ 5)

Since, $F(x) = \int_0^x {f(t)dt}$ , x > 0

So, $F'(x) = f(x) = (x - 1)(x - 2)(x - 5)$

According to wavy curve method



F'(x) changes, it's sign from negative to positive at x = 1 and 5, so F(x) has minima at x = 1 and 5 and as F'(x) changes, it's sign from positive to negative at x = 2, so F(x) has maxima at x = 2.

$\because$ $F'(2) = \int\limits_0^2 {f(t)dt = \int\limits_0^2 {({t^3} - 8{t^2}} } + 17t - 10)dt$

$= \left[ {{{{t^4}} \over 4} - 8{{{t^3}} \over 3} + 17{{{t^2}} \over 2} - 10t} \right]_0^2$

$= 4 - {{64} \over 3} + 34 - 20 = 38 - {{124} \over 3} = - {{10} \over 3}$

$\because$ AT the point of maxima x = 2, the functional value F(2), = $- {{10} \over 3}$ , is negative for the interval, x $\in$ (0, 5), so F(x) $\ne$ 0 for any value of x $\in$ (0, 5),

Hence, options (a), (b) and (d) are correct.