Question

If $f\left(x\right)=\underset{t\to \infty }{lim}\frac{(1+\cos \frac{\pi x}{2}{)}^t-1}{(1+\cos \frac{\pi x}{2}{)}^t+1}$, then $f\left(x\right)$ is continuous at $x=$

• A. $1$
  
• B. $2$
  
• C. $3$
  
• D. $4$
  

Answer 1

Answer: (B), (D)

The correct options are
B $2$

D $4$


At $x=1$
$f\left(1\right)=\underset{t\to \infty }{lim}\frac{{(1+\cos \frac{\pi }{2})}^t-1}{{(1+\cos \frac{\pi }{2})}^t+1}=\frac{1-1}{1+1}=0$

$x\to 1^{+}\to \frac{{\pi }^{+}}{2}\Rightarrow 2\text{nd quadrant}\Rightarrow \cos \left(\frac{{\pi }^{+}}{2}\right)\to -ve$

$\text{RHL}=f\left(1^{+}\right)=\underset{t\to \infty }{lim}\frac{{(1+\cos \frac{{\pi }^{+}}{2})}^t-1}{{(1+\cos \frac{{\pi }^{+}}{2})}^t+1}=\frac{0-1}{0+1}=-1$

$x\to 1^{-}\to \frac{{\pi }^{-}}{2}\Rightarrow 1\text{st quadrant}\Rightarrow \cos \left(\frac{{\pi }^{-}}{2}\right)\to +ve\Rightarrow \underset{t\to \infty }{lim}{[1+\cos (\frac{{\pi }^{-}}{2})]}^t\to \infty$

$\text{LHL}=f\left(1^{-}\right)=\frac{{(1+\cos \frac{{\pi }^{-}}{2})}^t-1}{{(1+\cos \frac{{\pi }^{-}}{2})}^t+1}=\frac{1-\frac{1}{{(1+\cos \frac{{\pi }^{-}}{2})}^t}}{1+\frac{1}{{(1+\cos \frac{{\pi }^{-}}{2})}^t}}=\frac{1-0}{1+0}=1$

$\text{LHL}\ne \text{RHL}\ne f\left(1\right)$
$\therefore f\left(x\right)$ is not continuous at $x=1$.

Similarly, $f\left(x\right)$ is not continuous at $x=3$.


At $x=2$
$f\left(2\right)=\underset{t\to \infty }{lim}\frac{(1+\cos \pi {)}^t-1}{(1+\cos \pi {)}^t+1}=\frac{0-1}{0+1}=-1$

$x\to 2^{+}\to {\pi }^{+}\Rightarrow 3\text{rd quadrant}\Rightarrow \cos {\pi }^{+}\to -ve$

$\Rightarrow f\left(2^{+}\right)=\frac{(1+\cos {\pi }^{+}{)}^t-1}{(1+\cos {\pi }^{+}{)}^t+1}=\frac{0-1}{0+1}=-1$

$x\to 2^{-}\to {\pi }^{-}\Rightarrow 2\text{nd quadrant}\Rightarrow \cos {\pi }^{-}\to -ve$

$\Rightarrow f\left(2^{-}\right)=\frac{(1+\cos {\pi }^{-}{)}^t-1}{(1+\cos {\pi }^{-}{)}^t+1}=\frac{0-1}{0+1}=-1$

$\text{LHL}=\text{RHL}=f\left(2\right)$
$\therefore f\left(x\right)$ is continuous at $x=2$.

Similarly, $f\left(x\right)$ is continuous at $x=4$.