Let $$f : \left[-\frac{\pi}{2}, \frac{\pi}{2} \right] \rightarrow R$$ be a continuous function such that
$$f(0)= 1$$ and $$\int_{0}^{\frac{\pi}{3}} f\left(t\right)dt = 0$$
Then which of the following statements is (are) TRUE ?

Define the auxiliary function $$g(x)=f(x)-3\cos 3x,\qquad x\in\left[0,\dfrac{\pi}{3}\right]$$. Since $$f(0)=1\quad\text{and}\quad\cos 0 =1,$$ we get

$$g(0)=f(0)-3\cos 0=1-3=-2\lt 0\;.\qquad -(1)$$

Next evaluate the integral of $$g$$ on the same interval:

$$\int_{0}^{\pi/3} g(t)\,dt =\int_{0}^{\pi/3} f(t)\,dt-3\int_{0}^{\pi/3}\cos 3t\,dt.$$

The first term is given to be $$0$$ and the second integral vanishes because

$$\int_{0}^{\pi/3}\cos 3t\,dt =\left[\dfrac{\sin 3t}{3}\right]_{0}^{\pi/3} =\dfrac{\sin\pi-\sin 0}{3}=0.$$

Hence

$$\int_{0}^{\pi/3} g(t)\,dt=0.\qquad -(2)$$

If $$g(x)$$ stayed strictly negative on the whole open interval $$\left(0,\dfrac{\pi}{3}\right)$$, its integral would be negative, contradicting $$(2)$$. Therefore $$g$$ must take at least one positive value in that interval. Because $$g$$ is continuous, it passes from the negative value at $$x=0$$ to a positive value somewhere in $$\left(0,\dfrac{\pi}{3}\right)$$, and by the Intermediate Value Theorem there exists

$$c\in\left(0,\dfrac{\pi}{3}\right)\quad\text{such that}\quad g(c)=0,$$

i.e.

$$f(c)-3\cos 3c=0.$$

Thus the statement in Option A is true.

For the remaining options one cannot, in general, guarantee a solution (or the asserted limits) under the given information; hence only Option A is always correct.

Option A which is: “The equation $$f(x)-3\cos 3x=0$$ has at least one solution in $$\left(0,\dfrac{\pi}{3}\right)$$” is therefore the correct choice.