Let f: left[-frac{pi}{2}, frac{pi}{2}right] r | vedclass

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Question

(A) Let $g(x) = f(x) - 3 \cos 3x$. Then $\int_0^{\pi/3} g(x) dx = \int_0^{\pi/3} f(x) dx - 3 \int_0^{\pi/3} \cos 3x dx = 0 - 3 \left[ \frac{\sin 3x}{3

Vedclass · Accepted Answer

$(A), (B), (C)$

Vedclass · Answer

(A) Let $g(x) = f(x) - 3 \cos 3x$. Then $\int_0^{\pi/3} g(x) dx = \int_0^{\pi/3} f(x) dx - 3 \int_0^{\pi/3} \cos 3x dx = 0 - 3 \left[ \frac{\sin 3x}{3} \right]_0^{\pi/3} = 0 - (\sin \pi - \sin 0) = 0$. Since the integral of $g(x)$ over $[0, \pi/3]$ is $0$,by the Mean Value Theorem for integrals,there exists at least one $c \in (0, \pi/3)$ such that $g(c) = 0$. Thus,$(A)$ is $TRUE$.$(B)$ Let $h(x) = f(x) - 3 \sin 3x + \frac{6}{\pi}$. Then $\int_0^{\pi/3} h(x) dx = \int_0^{\pi/3} f(x) dx - 3 \int_0^{\pi/3} \sin 3x dx + \int_0^{\pi/3} \frac{6}{\pi} dx = 0 - 3 \left[ -\frac{\cos 3x}{3} \right]_0^{\pi/3} + \frac{6}{\pi} \cdot \frac{\pi}{3} = 0 + (\cos \pi - \cos 0) + 2 = -1 - 1 + 2 = 0$. Since the integral of $h(x)$ over $[0, \pi/3]$ is $0$,there exists at least one $c \in (0, \pi/3)$ such that $h(c) = 0$. Thus,$(B)$ is $TRUE$.$(C)$ $\lim_{x \rightarrow 0} \frac{x \int_0^x f(t) dt}{1 - e^{x^2}} = \lim_{x \rightarrow 0} \left( \frac{x^2}{1 - e^{x^2}} \right) \cdot \left( \frac{\int_0^x f(t) dt}{x} \right) = (-1) \cdot f(0) = -1 \cdot 1 = -1$. Thus,$(C)$ is $TRUE$.$(D)$ $\lim_{x \rightarrow 0} \frac{\sin x \int_0^x f(t) dt}{x^2} = \lim_{x \rightarrow 0} \left( \frac{\sin x}{x} \right) \cdot \left( \frac{\int_0^x f(t) dt}{x} \right) = 1 \cdot f(0) = 1 \cdot 1 = 1$. Thus,$(D)$ is $FALSE$.

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