Question

If $f\left(t\right)=\frac{1}{1+t^2+2t\cos \alpha }$ and $g\left(t\right)=t|\sin \left(2nt\right)|$ , where $\alpha \in (0,\frac{\pi }{2}]$ and $n$ is odd, then

• A. The roots of the equation $\underset{0}{\overset{1}{\int }}f\left(t\right)x^{2}dt+\underset{-\pi }{\overset{\pi }{\int }}g\left(t\right)xdt-2=0$ are $\pm 2\sqrt{\sin \alpha }$
• B.
  $\underset{0}{\overset{\frac{\pi }{4}}{\int }}g\left(t\right)dt\le \frac{{\pi }^2}{32}$
• C. $\underset{0}{\overset{\pi }{\int }}g\left(t\right)dt=\pi$
• D. The roots of the equation $\underset{0}{\overset{1}{\int }}f\left(t\right)x^{2}dt+\underset{-\pi }{\overset{\pi }{\int }}g\left(t\right)xdt-2=0$ are $\pm 2\sqrt{\frac{\sin \alpha }{\alpha }}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B
$\underset{0}{\overset{\frac{\pi }{4}}{\int }}g\left(t\right)dt\le \frac{{\pi }^2}{32}$
C $\underset{0}{\overset{\pi }{\int }}g\left(t\right)dt=\pi$
D The roots of the equation $\underset{0}{\overset{1}{\int }}f\left(t\right)x^{2}dt+\underset{-\pi }{\overset{\pi }{\int }}g\left(t\right)xdt-2=0$ are $\pm 2\sqrt{\frac{\sin \alpha }{\alpha }}$

$f\left(t\right)=\frac{1}{(t+\cos \alpha {)}^2+{\sin }^2\alpha }$
$I=\underset{-\pi }{\overset{\pi }{\int }}g\left(t\right)dt$ is an odd function
$\therefore I=0$
$J=\underset{0}{\overset{1}{\int }}f\left(t\right)dt={\left[\frac{1}{\sin \alpha }{\tan }^{-1}\frac{t+\cos \alpha }{\sin \alpha }\right]}_0^1\Rightarrow J=\frac{\alpha }{2\sin \alpha }$
$\Rightarrow J{x}^2-2=0\Rightarrow x=\pm 2\sqrt{\frac{\sin \alpha }{\alpha }}$

$g\left(t\right)=t|\sin 2nt|\le t\Rightarrow \underset{0}{\overset{\frac{\pi }{4}}{\int }}g\left(t\right)dt\le \underset{0}{\overset{\frac{\pi }{4}}{\int }}tdt\Rightarrow \underset{0}{\overset{\frac{\pi }{4}}{\int }}g\left(t\right)dt\le \frac{{\pi }^2}{32}$

$K=\underset{0}{\overset{\pi }{\int }}g\left(t\right)dt\Rightarrow K=\underset{0}{\overset{\pi }{\int }}t|\sin \left(2nt\right)|dt\Rightarrow K=\underset{0}{\overset{\pi }{\int }}(\pi -t)|\sin 2nt|dt\Rightarrow 2K=\underset{0}{\overset{\pi }{\int }}(\pi -t)|\sin 2nt|dt$
Put $2nt=x\Rightarrow dx=2ndt$
$\Rightarrow 2K=\underset{0}{\overset{2n\pi }{\int }}\frac{\pi }{2n}|\sin t|dt\Rightarrow 2K={\int }_0^{\pi }\pi |\sin t|dt\Rightarrow K=\pi$