Question

Assertion :Consider $f\left(x\right)=\underset{n\to \infty }{lim}\left\{\frac{x}{1+{\left\{3{\tan }^{2}x\right\}}^n}\right\}$
$f\left(x\right)$ is periodic function with period $\pi$. Reason: $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}0,& -\frac{\pi }{2}<x<-\frac{\pi }{6}\end{array}\\ \begin{array}{cc}\frac{x}{2},& x=-\frac{\pi }{6}\end{array}\\ \begin{array}{cc}x,& -\frac{\pi }{6}<x<\frac{\pi }{6}\end{array}\\ \begin{array}{cc}\frac{x}{2},& x=\frac{\pi }{6}\end{array}\\ \begin{array}{cc}0,& \frac{\pi }{6}<x<\frac{\pi }{2}\end{array}\end{array}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

${\tan }^{2}x>1\Rightarrow \tan x>\frac{1}{\sqrt{3}}$ or $\tan x<\frac{-1}{\sqrt{3}}$

$\Rightarrow \frac{-\pi }{2}<x<\frac{-\pi }{6}$ or $\frac{\pi }{6}<x<\frac{\pi }{2}$

$f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}0,& -\frac{\pi }{2}<x<-\frac{\pi }{6}\end{array}\\ \begin{array}{cc}\frac{x}{2},& x=-\frac{\pi }{6}\end{array}\\ \begin{array}{cc}x,& -\frac{\pi }{6}<x<\frac{\pi }{6}\end{array}\\ \begin{array}{cc}\frac{x}{2},& x=\frac{\pi }{6}\end{array}\\ \begin{array}{cc}0,& \frac{\pi }{6}<x<\frac{\pi }{2}\end{array}\end{array}$
$\because$ obviously Assertion is incorrect and Reason is correct.