Continuous Function Integrals and Limits Involving Definite Integrals

To determine which of the given options are TRUE, we analyze each option step-by-step using fundamental concepts of calculus, including Rolle's Theorem, the Fundamental Theorem of Calculus, and L'Hôpital's Rule.

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Analysis of Option (A):

Consider the function $g(x)$ defined on $\left[0, \frac{\pi}{3}\right]$ by: $g(x) = \int_0^x (f(t) - 3\cos 3t) \, dt$

1. Continuity and Differentiability: Since $f(t)$ is continuous, $g(x)$ is continuous on $\left[0, \frac{\pi}{3}\right]$ and differentiable on $\left(0, \frac{\pi}{3}\right)$.
2. Boundary Values:
   • $g(0) = \int_0^0 (f(t) - 3\cos 3t) \, dt = 0$
   • $g\left(\frac{\pi}{3}\right) = \int_0^{\pi/3} f(t) \, dt - \int_0^{\pi/3} 3\cos 3t \, dt$ Using the given condition $\int_0^{\pi/3} f(t) \, dt = 0$: $g\left(\frac{\pi}{3}\right) = 0 - [\sin 3t]_0^{\pi/3} = 0 - (\sin \pi - \sin 0) = 0$

Since $g(0) = g\left(\frac{\pi}{3}\right) = 0$, by Rolle's Theorem, there exists at least one $c \in \left(0, \frac{\pi}{3}\right)$ such that $g'(c) = 0$.

By the Fundamental Theorem of Calculus: $g'(x) = f(x) - 3\cos 3x$ Thus, $g'(c) = 0 \implies f(c) - 3\cos 3c = 0$.

Therefore, Option (A) is TRUE.

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Analysis of Option (B):

Consider the function $h(x)$ defined on $\left[0, \frac{\pi}{3}\right]$ by: $h(x) = \int_0^x \left(f(t) - 3\sin 3t + \frac{6}{\pi}\right) \, dt$

1. Continuity and Differentiability: $h(x)$ is continuous on $\left[0, \frac{\pi}{3}\right]$ and differentiable on $\left(0, \frac{\pi}{3}\right)$.
2. Boundary Values:
   • $h(0) = 0$
   • $h\left(\frac{\pi}{3}\right) = \int_0^{\pi/3} f(t) \, dt - \int_0^{\pi/3} 3\sin 3t \, dt + \int_0^{\pi/3} \frac{6}{\pi} \, dt$ $h\left(\frac{\pi}{3}\right) = 0 - [-\cos 3t]_0^{\pi/3} + \frac{6}{\pi} \left(\frac{\pi}{3}\right)$ $h\left(\frac{\pi}{3}\right) = (\cos \pi - \cos 0) + 2 = (-1 - 1) + 2 = 0$

Since $h(0) = h\left(\frac{\pi}{3}\right) = 0$, by Rolle's Theorem, there exists at least one $c \in \left(0, \frac{\pi}{3}\right)$ such that $h'(c) = 0$.

By the Fundamental Theorem of Calculus: $h'(x) = f(x) - 3\sin 3x + \frac{6}{\pi}$ Thus, $h'(c) = 0 \implies f(c) - 3\sin 3c = -\frac{6}{\pi}$.

Therefore, Option (B) is TRUE.

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Analysis of Option (C):

We want to evaluate the limit: $L_1 = \lim_{x \to 0} \frac{x \int_0^x f(t) \, dt}{1 - e^{x^2}}$

Using the standard limit $\lim_{x \to 0} \frac{1 - e^{x^2}}{x^2} = -1$, we can rewrite the expression as: $L_1 = \lim_{x \to 0} \left( \frac{x^2}{1 - e^{x^2}} \cdot \frac{\int_0^x f(t) \, dt}{x} \right)$ $L_1 = -1 \cdot \lim_{x \to 0} \frac{\int_0^x f(t) \, dt}{x}$

Since $\int_0^0 f(t) \, dt = 0$, the remaining limit is an indeterminate form of type $\frac{0}{0}$. Applying L'Hôpital's Rule: $\lim_{x \to 0} \frac{\int_0^x f(t) \, dt}{x} = \lim_{x \to 0} \frac{\frac{d}{dx} \int_0^x f(t) \, dt}{\frac{d}{dx} (x)} = \lim_{x \to 0} \frac{f(x)}{1} = f(0) = 1$

Substitute this back into $L_1$: $L_1 = -1 \cdot 1 = -1$

Therefore, Option (C) is TRUE.

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Analysis of Option (D):

We want to evaluate the limit: $L_2 = \lim_{x \to 0} \frac{\sin x \int_0^x f(t) \, dt}{x^2}$

Rewrite the limit product as: $L_2 = \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \cdot \left( \lim_{x \to 0} \frac{\int_0^x f(t) \, dt}{x} \right)$

We know that:

1. $\lim_{x \to 0} \frac{\sin x}{x} = 1$
2. $\lim_{x \to 0} \frac{\int_0^x f(t) \, dt}{x} = f(0) = 1$ (by L'Hôpital's Rule)

Therefore: $L_2 = 1 \cdot 1 = 1 \neq -1$

Therefore, Option (D) is FALSE.

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Conclusion:

The correct statements are A, B, and C.