Let us test each option using only the given fact:

• $f:\mathbb R\to (0,1)$ is continuous.
• So for every real $x$, we have $0<f(x)<1.$

We must determine which given functions must attain the value $0$ at some point in $(0,1)$.

1. Option A

The function is $A(x)=e^x-\int_0^x f(t)\sin t\,dt.$

For $x\in(0,1)$, since $0<f(t)<1$ and $\sin t>0$ on $(0,1)$, $0< f(t)\sin t< \sin t.$ Therefore, $0<\int_0^x f(t)\sin t\,dt<\int_0^x \sin t\,dt=1-\cos x.$ Hence, $A(x)>e^x-(1-\cos x)=e^x-1+\cos x.$ Now for $x\in(0,1)$,

• $e^x>1$,
• $\cos x>0$. So $e^x-1+\cos x>0.$ Thus $A(x)>0$ for all $x\in(0,1)$.

So A does not have a zero in $(0,1)$.

2. Option B

The function is $B(x)=f(x)+\int_0^{\pi/2} f(t)\sin t\,dt.$

Here,

• $f(x)>0$,
• $f(t)\sin t\ge 0$, and in fact positive on a set of positive length, so $\int_0^{\pi/2} f(t)\sin t\,dt>0.$ Therefore, $B(x)>0 \quad \text{for all } x.$

So B does not have a zero in $(0,1)$.

3. Option C

The function is $C(x)=f(x)-\int_0^{\pi/2-x} f(t)\cos t\,dt.$

We want to show this has a zero for some $x\in(0,1)$.

Since $f$ is continuous, $C$ is continuous on $(0,1)$ (indeed on all relevant points). So it is enough to find two points in $(0,1)$ where $C$ takes opposite signs.

Step 1: Evaluate near $x=0$

At $x=0$ (for sign reference), $C(0)=f(0)-\int_0^{\pi/2} f(t)\cos t\,dt.$ Since $0<f(t)<1$, $\int_0^{\pi/2} f(t)\cos t\,dt>0.$ But this alone does not force sign. So let us instead use endpoint comparison cleverly.

Step 2: Evaluate at $x=\pi/2$

At $x=\pi/2$, $C\left(\frac\pi2\right)=f\left(\frac\pi2\right)>0.$

Step 3: Evaluate at $x=0$ more carefully

Observe $\int_0^{\pi/2} f(t)\cos t\,dt$ is a weighted average-type quantity. Since $0<f(t)<1$ and $\int_0^{\pi/2}\cos t\,dt=1,$ we have $0<\int_0^{\pi/2} f(t)\cos t\,dt<1.$ Also $0<f(0)<1$. This still does not guarantee $C(0)$ negative.

So let us instead define $G(x)=\int_0^{\pi/2-x} f(t)\cos t\,dt.$ Then $C(x)=f(x)-G(x).$ Now note:

• $G(\pi/2)=0$, hence $C(\pi/2)=f(\pi/2)>0$.
• $G(0)=\int_0^{\pi/2}f(t)\cos t\,dt$.

To prove existence of a zero in $(0,1)$ for all such $f$, consider the special point $x=1$ (since $1<\pi/2$): $C(1)=f(1)-\int_0^{\pi/2-1} f(t)\cos t\,dt.$ This again is not automatically of fixed sign.

So we need a stronger structural argument.

Let us inspect whether the statement can be proved by choosing a continuous auxiliary function and using IVT. Define $H(x)=\int_0^{\pi/2-x} f(t)\cos t\,dt.$ Then $H$ is continuous, decreasing, with $H(0)=\int_0^{\pi/2} f(t)\cos t\,dt,\qquad H\left(\frac\pi2\right)=0.$ Also $f(x)$ is continuous and takes values in $(0,1)$. Now compare at endpoints: $f(0)\in(0,1),\qquad H(0)\in(0,1).$ No definite sign at $x=0$.

However, since the stored answer includes C, let us test with a simple example to verify universality. Take constant function $f(t)=c$ with $0<c<1$. Then $C(x)=c-c\int_0^{\pi/2-x}\cos t\,dt=c-c\sin\left(\frac\pi2-x\right)=c-c\cos x=c(1-\cos x).$ This is $0$ only at $x=0$, not in $(0,1)$. So for this valid continuous $f$, there is no zero in $(0,1)$.

Hence C is not necessarily zero at some point in $(0,1)$.

So C is not correct.

4. Option D

The printed option appears as x9 - f(x), which is evidently a formatting issue. The intended expression is almost surely $D(x)=x-f(x).$ (Otherwise the question makes no mathematical sense in the context.)

Now define $D(x)=x-f(x).$ This is continuous on $[0,1]$.

Check endpoint values:

• At $x=0$, $D(0)=0-f(0)<0$ because $f(0)>0$.
• At $x=1$, $D(1)=1-f(1)>0$ because $f(1)<1$.

Since $D$ is continuous on $[0,1]$, by the Intermediate Value Theorem there exists some $c\in(0,1)$ such that $D(c)=0.$ That is, $c-f(c)=0.$

So D does have a zero in $(0,1)$.

Final conclusion

The functions which must have value zero at some point in $(0,1)$: $\boxed{D}$

Option C fails, for example when $f(x)=c\in(0,1)$ is constant.

Comparison with stored answer

Stored correct answer: $C, D$

My derived answer: $D$ only.

So I do not agree with the stored answer. The stored answer likely treats option C incorrectly; the counterexample $f(x)=c$ with $0<c<1$ shows that C need not vanish in $(0,1)$, since then $C(x)=c(1-\cos x)>0\quad \text{for }x\in(0,1).$