JEE Main 2016 (20)

Let $h\left( x \right) =\displaystyle \int _{ a }^{ x }{ g\left( x \right) } dx=f\left( x \right) f^{ \prime }\left( x \right) \quad \Rightarrow \quad h^{ \prime }\left( x \right) =g\left( x \right)$ . Then,
$h(x) = 0 \Leftrightarrow f(x) = 0 \text{ or } f'(x) = 0$ .

In the closed interval $[a, e ]$ , due to the intermediate value theorem , $f(x)$ has at least 2 other zeros, so it has 4 zeros in $[a,e]$ .

By rolle's theorem , $f'(x)=0$ will have at least $4 -1 =3$ solutions in $(a,b)$ . Thus, $h(x)$ will have at least $4 + 3 = 7$ solutions in $[a,b ]$ .

Again by Rolle's theorem , $g(x)= 0$ will have at least $7-1=6$ solutions in $(a,b )$ .

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To show that this lower bound can be achieved, set $f(x) = -10 x ( x-2)(x-4)(x-6)$ with $0 = a < b < 2 < c < 4 < d < e = 6$ . Then, $g(x) = (f'(x))^2 + f(x) f''(x) \\= [ -40 x^3 + 360x^2 - 880x + 480]^2 + (-10 x) ( x-2)(x-4)(x-6) \times (-120x^2 + 720x - 880)$ .
We can manually verify that this has 6 zeroes in $(0,6)$ .