QJEEM22Ma70d5: JEE Main 2022 Mathematics - Class 12 JEEM22Ma70d5 Solution: The function defined by is continuous for all x in :

$f(x) = \mathop {\lim }\limits_{n \to \infty } {{\cos (2\pi x) - {x^{2n}}\sin (x - 1)} \over {1 + {x^{2n + 1}} - {x^{2n}}}}$

For |x| < 1,\,f(x) = \cos 2\pi x , continuous function

|x| > 1,\,f(x) = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {{x^{2n}}}}\cos 2\pi x - \sin (x - 1)} \over {{1 \over {{x^{2n}}}} + x - 1}}

$= {{ - \sin (x - 1)} \over {x - 1}}$ , continuous

For $$|x|=1,f(x)=\{\begin{array}{ccc}1& \mathrm{if}& x=1\\ -(1+\sin 2)& \mathrm{if}& x=-1\end{array}$$

Now,

$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = - 1,\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1$ , so discontinuous at $x = 1$

$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 1,\,\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = - {{\sin 2} \over 2}$ , so discontinuous at $x = - 1$

$\therefore$ $f(x)$ is continuous for all $x \in R - \{ - 1,1\}$