Question

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

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

Answer 1

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

(A) Let $g(x)=f(x)-3\cos 3x$
$\underset{0}{\overset{\pi /3}{\int }}g(x)dx=\underset{0}{\overset{\pi /3}{\int }}(f(x)-3\cos 3x)dx$
$=\underset{0}{\overset{\pi /3}{\int }}f(x)dx-\underset{0}{\overset{\pi /3}{\int }}3\cos 3xdx=0$
$\Rightarrow g(x)=0$ has at least one solution in $[0,\pi /3]$
(B) Let $h(x)=f(x)-3\sin 3x+6/\pi$
$\underset{0}{\overset{\pi /3}{\int }}h(x)=\underset{0}{\overset{\pi /3}{\int }}\left(f(x)-3\sin 3x+\frac{6}{\pi }\right)dx=0$
$\Rightarrow h(x)=0$ has at least one solution in $[0,\pi /3]$
(C) $\underset{x\to 0}{\lim }\frac{\underset{0}{\overset{x}{\int }}f(t)dt}{x}=\underset{x\to 0}{\lim }\frac{f(x)}{1}=f(0)=1$
$\Rightarrow \underset{x\to 0}{\lim }\frac{x\underset{0}{\overset{x}{\int }}f(t)dt}{x^2\left(\frac{1-e^{x^2}}{x^2}\right)}=\underset{x\to 0}{\lim }\frac{-\underset{0}{\overset{x}{\int }}f(t)}{x}\cdot \frac{x^2}{\left(e^{x^2}-1\right)}=-1$
(D) $\underset{x\to 0}{\lim }\frac{\sin x}{x}\cdot \frac{\underset{0}{\overset{x}{\int }}f(t)}{x}=1$