Existence of Zeroes for Continuous Functions over Interval

To determine which of the given functions must attain the value zero at some point in the open interval $(0, 1)$, we can apply the Intermediate Value Theorem (IVT). The theorem states that if a function $g(x)$ is continuous on a closed interval $[a, b]$ and $g(a)$ and $g(b)$ have opposite signs, then there exists at least one $c \in (a, b)$ such that $g(c) = 0$.

Given that $f: \mathbb{R} \rightarrow (0, 1)$ is continuous, we have $0 < f(x) < 1$ for all $x \in \mathbb{R}$.

---

Option (A): $g(x) = x^9 - f(x)$

1. Continuity: Since $x^9$ and $f(x)$ are continuous functions, $g(x) = x^9 - f(x)$ is continuous on the interval $[0, 1]$.

2. Evaluation at endpoints:

   • At $x = 0$: $g(0) = 0^9 - f(0) = -f(0)$ Since $0 < f(0) < 1$, we have $g(0) < 0$.
   • At $x = 1$: $g(1) = 1^9 - f(1) = 1 - f(1)$ Since $f(1) < 1$, we have $1 - f(1) > 0$, so $g(1) > 0$.

3. Conclusion: Since $g(x)$ is continuous on $[0, 1]$ with $g(0) < 0$ and $g(1) > 0$, by the Intermediate Value Theorem, there exists at least one $c \in (0, 1)$ such that $g(c) = 0$.

Thus, Option (A) is correct.

---

Option (B): $g(x) = x - \int_0^{\frac{\pi}{2}-x} f(t) \cos t \, dt$

1. Continuity: The definite integral function is continuous with respect to its upper limit, making $g(x)$ continuous on $[0, 1]$.

2. Evaluation at endpoints:

   • At $x = 0$: $g(0) = 0 - \int_0^{\frac{\pi}{2}} f(t) \cos t \, dt = -\int_0^{\frac{\pi}{2}} f(t) \cos t \, dt$ For $t \in \left(0, \frac{\pi}{2}\right)$, both $f(t) > 0$ and $\cos t > 0$. Therefore, the integral is strictly positive: $\int_0^{\frac{\pi}{2}} f(t) \cos t \, dt > 0 \implies g(0) < 0$
   • At $x = 1$: $g(1) = 1 - \int_0^{\frac{\pi}{2}-1} f(t) \cos t \, dt$ For $t \in \left[0, \frac{\pi}{2}-1\right]$, we know $f(t) < 1$ and $\cos t \le 1$, which gives $f(t) \cos t < 1$. Thus: $\int_0^{\frac{\pi}{2}-1} f(t) \cos t \, dt < \int_0^{\frac{\pi}{2}-1} 1 \, dt = \frac{\pi}{2} - 1$ Substituting this back into $g(1)$: $g(1) > 1 - \left(\frac{\pi}{2} - 1\right) = 2 - \frac{\pi}{2}$ Since $\pi \approx 3.14159$, we have $\frac{\pi}{2} \approx 1.57$, so $2 - \frac{\pi}{2} > 0$. Hence, $g(1) > 0$.

3. Conclusion: Since $g(x)$ is continuous on $[0, 1]$ with $g(0) < 0$ and $g(1) > 0$, by the Intermediate Value Theorem, there exists at least one $c \in (0, 1)$ such that $g(c) = 0$.

Thus, Option (B) is correct.

---

Option (C): $g(x) = e^x - \int_0^x f(t) \sin t \, dt$

1. For $x \in (0, 1)$, since $f(t) < 1$ and $\sin t < 1$ for $t \in (0, 1)$: $\int_0^x f(t) \sin t \, dt < \int_0^x 1 \, dt = x$
2. Since $e^x > 1 + x > x$ for all $x > 0$: $g(x) = e^x - \int_0^x f(t) \sin t \, dt > x - x = 0$ Therefore, $g(x) > 0$ for all $x \in (0, 1)$, so $g(x)$ can never attain zero in $(0, 1)$.

Thus, Option (C) is incorrect.

---

Option (D): $g(x) = f(x) + \int_0^{\frac{\pi}{2}} f(t) \sin t \, dt$

1. Since $f(x) > 0$ for all $x \in \mathbb{R}$, the term $f(x)$ is strictly positive.
2. For $t \in \left(0, \frac{\pi}{2}\right)$, both $f(t) > 0$ and $\sin t > 0$, so $\int_0^{\frac{\pi}{2}} f(t) \sin t \, dt > 0$.
3. Sum of two strictly positive quantities is strictly positive: $g(x) > 0 \quad \text{for all } x \in (0, 1)$ Therefore, $g(x)$ can never attain zero in $(0, 1)$.

Thus, Option (D) is incorrect.

---

Final Answer:

The correct options are A and B.