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)

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