Let $f:R \rightarrow(0,1)$ be a continuous function. Then which of the following functions must have the value zero at some point in the interval $(0,1)$ ?
(A) $x^{7}-f(x)$
(B) $x^{2}-\int_{0}^{x} f(t) d t$
(C) $\int_{0}^{\frac{\pi}{2} - x} f(t) \sin 2 t d t-x$
(D) $\sqrt{x}-\int_{0}^{x^2} \sqrt{f(t)} d t$.
What i considered was for A) lets say we have a maximum value of the function be "c" and minimum be "d" , as we know both lie in between 0 and 1, $x^7$ will have that value c and d by intermediate value theorem , but i was not able to prove that it would be at same x . ( Though it gave me a clue to try using MVT( mean value theorem) but not working). For others i think of same strategy as above but wasnot able to give a proof , though i find out a counterexamole in case of B that is f(x) = x .
In your explanation, if by "the function" you mean $x^7 - f(x)$, then I would disagree with your claim that the minimum lies between $0$ and $1$. In fact, evaluating the function at $x= 0$ and $x=1$ should give you enough to make the conclusion in the case of (A). As an additional hint, I think you should try using the Intermediate value theorem, which states that if a continuous function $g : [a,b] \to \mathbb R$ has minimum $c$ and maximum $d,$ then it takes on every value $y \in [c,d]$ on $[a,b]$.
I shall provide some hints for the rest:
What values can it take at $x= 0$ and at $x= 1$? Is it true that for all continuous functions $f : \mathbb R \longrightarrow (0,1)$, you can guarantee that the function is negative at $0$ and positive at $1$ (or vice versa)?
As you have already noticed, it is enough to find a single continuous function for which the statement does not hold to be able to say the answer is no.