Let $f : \mathbb{R} \to (0,1)$ be a continuous function. Then, which of the following function(s) has(have) the value zero at some point in the interval $(0, 1)$?

$f : \mathbb{R} \to (0,1) \text{ continuous}$

$0 < f(x) < 1 \ \text{for all } x$

$\text{To find: which function(s) vanish somewhere in } (0,1)$

$\text{IVT: opposite signs at the ends} \Rightarrow \text{a zero in between}$

$0 < f(x) < 1; \quad \sin t,\ \cos t > 0 \text{ on } (0, \tfrac{\pi}{2})$

$\text{(1)}\ g(x) = x^9 - f(x)$

$g(0) = -f(0) < 0, \quad g(1) = 1 - f(1) > 0 \Rightarrow (1)\ \text{true}$

$\text{(2)}\ g(x) = x - \displaystyle\int_{0}^{\frac{\pi}{2}-x} f(t)\cos t\,dt$

$g(0) = -\displaystyle\int_{0}^{\pi/2} f\cos t\,dt < 0$

$g(1) = 1 - \displaystyle\int_{0}^{\frac{\pi}{2}-1} f\cos t\,dt > 1 - \int_0^{\pi/2-1}\cos t\,dt = 1 - \cos 1 > 0$

$\text{(3)}\ g(0)=1>0,\ g'(x)=e^x - f(x)\sin x > 0 \Rightarrow g>0 \ \Rightarrow \text{false}$

$\text{(4)}\ f(x)>0 \text{ and } \int_0^{\pi/2} f\sin t\,dt > 0 \Rightarrow g(x)>0 \ \Rightarrow \text{false}$

$\boxed{\text{Answer: statements (1) and (2)}}$