Question

Let $f:[-\frac{\pi }{2},\frac{\pi }{2}]\to R$ be a continuous function such that $f\left(0\right)=1$ and $\underset{0}{\overset{\pi /3}{\int }}f\left(t\right)dt=0$

Then which of the following statements is(are) TRUE ?

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

Answer 1

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

$\underset{x\to 0}{lim}\frac{x\underset{0}{\overset{x}{\int }}f\left(t\right)dt}{1-e^{x^2}}=-1$

$f\left(0\right)=1$ and $\underset{0}{\overset{\pi /3}{\int }}f\left(t\right)dt=0$

Consider function $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt-\sin 3x$
$g\left(x\right)$ is continuous and differentiable function.
$g\left(0\right)=0$ and $g\left(\frac{\pi }{3}\right)=0$
By Rolle's theorem, $g^{\prime }\left(x\right)=0$ has at least one solution in $(0,\frac{\pi }{3})$
So, $f\left(x\right)-3\cos 3x=0$ for some $x\in (0,\frac{\pi }{3})$


Consider function $h\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt+\cos 3x+\frac{6}{\pi }x$
$h\left(x\right)$ is continuous and differentiable function and $h\left(0\right)=1,h\left(\frac{\pi }{3}\right)=1$
By Rolle's theorem, $h^{\prime }\left(x\right)=0$ for at least one $x\in (0,\frac{\pi }{3})$
So, $f\left(x\right)-3\sin 3x+\frac{6}{\pi }=0$ for some $x\in (0,\frac{\pi }{3})$


$L=\underset{x\to 0}{lim}\frac{x\underset{0}{\overset{x}{\int }}f\left(t\right)dt}{1-e^{x^2}}\left(\frac{0}{0}\text{form}\right)$
By L' Hopital rule,
$L=\underset{x\to 0}{lim}\frac{xf\left(x\right)+\underset{0}{\overset{x}{\int }}f\left(t\right)dt}{-2x{e}^{x^2}}\left(\frac{0}{0}\text{form}\right)$
Again by L' Hopital rule,
$L=\underset{x\to 0}{lim}\frac{x{f}^{\prime }\left(x\right)+f\left(x\right)+f\left(x\right)}{-4x^2{e}^{x^2}-2e^{x^2}}=\frac{0+2f\left(0\right)}{-0-2}=-1$


$\underset{x\to 0}{lim}\frac{\sin x\underset{0}{\overset{x}{\int }}f\left(t\right)dt}{x^2}\left(\frac{0}{0}\text{form}\right)=\underset{x\to 0}{lim}\frac{\sin xf\left(x\right)+\cos x\underset{0}{\overset{x}{\int }}f\left(t\right)dt}{2x}\left(\frac{0}{0}\text{form}\right)$
$=\underset{x\to 0}{lim}\frac{\cos x\cdot f\left(x\right)+\sin x\cdot f^{\prime }\left(x\right)+\cos x\cdot f\left(x\right)-\sin x\cdot {\int }_0^{x}f\left(t\right)dt}{2}=\frac{1+0+1-0}{2}=1$