Question

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

• A. The equation $f ( x )-3 \cos 3 x =0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
• B. The equation $f ( x )-3 \sin 3 x =-\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
• C. $\displaystyle\lim _{x \rightarrow 0} \frac{x \int\limits_{0}^{x} f(t) d t}{1-e^{x^{2}}}=-1$
• D. $\displaystyle\lim _{x \rightarrow 0} \frac{\sin x \int\limits_{0}^{x} f(t) d t}{x^{2}}=-1$

Answer 1

Answer: (A), (B), (C)

The equation $f ( x )-3 \cos 3 x =0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$

Answer 2

(A) The equation $f ( x )-3 \cos 3 x =0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
(B) The equation $f ( x )-3 \sin 3 x =-\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
(C)$\displaystyle\lim _{x \rightarrow 0} \frac{x \int\limits_{0}^{x} f(t) d t}{1-e^{x^{2}}}=-1$