Question

Assertion :The function $f\left(x\right)=\underset{n\to \infty }{lim}\frac{\cos \pi x-x^{2n}\sin (x-1)}{1+x^{2n+1}-x^{2n}}$ is discontinuous at $x=\pm 1$ Reason: $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{\cos \pi x}{1+x},& \left|x\right|<1\end{array}\\ \begin{array}{cc}-1+\sin 2,& x=-1\end{array}\\ \begin{array}{cc}-1,& x=1\end{array}\\ \begin{array}{cc}\frac{-\sin (x-1)}{x-1},& \left|x\right|>1\end{array}\end{array}$

• 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. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

We have,

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

i.e., $f\left(x\right)=\frac{\cos \left(\pi x\right)-0}{1+x-0},\left|x\right|<1$

$=\frac{\cos \left(\pi x\right)-\sin (x-1)}{1+1-1},\left|x\right|=1$

$=\underset{n\to \infty }{lim}\frac{\frac{\cos \left(\pi x\right)}{x^{2n}}-\sin (x-1)}{\frac{1}{x^{2n}}+x-1}$

$=\frac{-\sin (x-1)}{1+x},\left|x\right|>1$

i.e., $f\left(x\right)=\frac{\cos \left(\pi x\right)}{1+x},\left|x\right|<1$

$=-1+\sin 2,x=-1$

$=-1,x=1$

$=-\frac{\sin (x-1)}{x-1},\left|x\right|>1$

At $x=1,$ we have,

$\underset{h\to 0}{lim}f(-1+h)=\underset{h\to 0}{lim}\frac{\cos \pi (-1+h)}{1+(-1+h)}=\underset{h\to 0}{lim}\frac{-\cos \left(\pi h\right)}{h}=-\infty$

$f(-1)=-1+\sin 2$

implies discontinuity at $x=-1$

At $x=1,$ we have,

$\underset{h\to 0}{lim}f(1+h)=\underset{h\to 0}{lim}\frac{-\sin (1+h-1)}{1+h-1}=\underset{h\to 0}{lim}\frac{-\sin h}{1+(1-h)}=\frac{-1}{2}$

$\underset{h\to 0}{lim}f(1-h)=\underset{h\to 0}{lim}\frac{\cos \pi (1-h)}{1+(1-h)}=\frac{-1}{2}$

$f\left(x\right)=-1$

implies discontinuity at $x=1$