If $f(x)= (x-a)(x-b)$ for then the minimum number of roots of equation $\pi(f'(x))^2 \cos(\pi(f(x))) + \sin(\pi(f(x)))f''(x) =0$

Question

If $f(x)= (x-a)(x-b)$ for $a,b$ $\in \mathbb{R}$ then the minimum number of roots of equation $$\pi(f'(x))^2 \cos(\pi(f(x))) + \sin(\pi(f(x)))f''(x) =0$$

in $(\alpha,\beta)$ where $f(\alpha) =+3 = f(\beta)$ and $\alpha <a<b<\beta$ will be:

Answer 1

Hint You can rewrite that equation as $$\frac{d}{dx}\left(\sin(\pi f(x))f'(x)\right)|_{x=x^*} = 0$$

Consider the function $g(x) = \sin (2\pi x -\pi(a+b))(2x - (a+b))$

Between any two roots of this function, there will be at least one point at which derivative vanishes (Rolle's theorem). Find the number of roots in $(\alpha, \beta)$ and it would be one less than that

Answer 2

As a hint: $$(uv')=u'v'+u.v''$$ so $$y=(\sin(\pi f(x))\times f'(x))'=\\\pi f'(x)\cos(\pi f(x))\times f'(x)+f''(x)\times\sin(\pi f(x))$$ and $f'=2x-a-b$ so you need to find to roots of $$(\sin(\pi (x-a)(x-b))\times (2x-a-b))'=0$$