Question

If $f\left(x\right)-2\underset{0}{\overset{\pi /4}{\int }}\frac{{\sin }^{2}x}{{\cos }^{5}x}\cdot \cos t\cdot f\left(t\right)dt=\frac{{\sin }^{2}x}{{\cos }^{5}x},$ then which of the following is (are) CORRECT ?

• A. $\underset{x\to \pi /3}{lim}f\left(x\right)=72$
• B. $f\left(x\right)$ is periodic with fundamental period $\pi$
• C. The equation of normal to the graph of $y=f\left(x\right)$ at point $M$ whose abscissa is $\pi$ is given by $x-\pi =0$
• D. Number of solutions of the equation $f\left(x\right)=0$ in $(0,3)$ is one

Answer 1

Answer: (A), (C)

The equation of normal to the graph of $y=f\left(x\right)$ at point $M$ whose abscissa is $\pi$ is given by $x-\pi =0$

$f\left(x\right)-2\frac{{\sin }^{2}x}{{\cos }^{5}x}\underset{0}{\overset{\pi /4}{\int }}\cos t\cdot f\left(t\right)dt=\frac{{\sin }^{2}x}{{\cos }^{5}x}$
Let $A=\underset{0}{\overset{\pi /4}{\int }}\cos t\cdot f\left(t\right)dt$
$\therefore f\left(x\right)-2A\frac{{\sin }^{2}x}{{\cos }^{5}x}=\frac{{\sin }^{2}x}{{\cos }^{5}x}\Rightarrow f\left(x\right)=(2A+1)\frac{{\sin }^{2}x}{{\cos }^{5}x}\dots \left(1\right)$

Now, $A=\underset{0}{\overset{\pi /4}{\int }}\cos t\cdot (2A+1)\frac{{\sin }^{2}t}{{\cos }^{5}t}dt$
$=(2A+1)\underset{0}{\overset{\pi /4}{\int }}\frac{{\sin }^{2}t}{{\cos }^{4}t}dt=(2A+1)\underset{0}{\overset{\pi /4}{\int }}{\tan }^{2}t{\sec }^{2}tdt$
$\Rightarrow A=(2A+1)\frac{1}{3}\Rightarrow 3A=2A+1$
$\Rightarrow A=1.$
Hence from equation $\left(1\right),$
we get $f\left(x\right)=\frac{3{\sin }^{2}x}{{\cos }^{5}x}$

$\underset{x\to \frac{\pi }{3}}{lim}f\left(x\right)=\underset{x\to \frac{\pi }{3}}{lim}\frac{3{\sin }^{2}x}{{\cos }^{5}x}=\frac{3{\left(\frac{\sqrt{3}}{2}\right)}^2}{{\left(\frac{1}{2}\right)}^5}=72$

As $f\left(x\right)=\frac{3{\sin }^{2}x}{{\cos }^{5}x},$
so, $f\left(x\right)$ is periodic with fundamental period $2\pi$

$f\left(x\right)=\frac{3{\sin }^{2}x}{{\cos }^{5}x}\Rightarrow f^{\prime }\left(x\right)=3\left[\frac{{\cos }^{5}x\cdot (2\sin x\cdot \cos x)+{\sin }^{2}x\cdot (5{\cos }^{4}x\cdot \sin x)}{{\cos }^{10}x}\right]$
So, equation of normal to the graph of $y=f\left(x\right)$ at point $M(\pi ,0)$ is given by $x-\pi =0$

As $f\left(x\right)=0$
$\Rightarrow \frac{3{\sin }^{2}x}{{\cos }^{5}x}=0\Rightarrow \sin x=0$
$\therefore x=n\pi ,n\in Z$
So, the equation $f\left(x\right)=0$ has no root in $(0,3)$