Question
The set of points of discontinuity of the function $$f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\,\sin \,x} \right)}^{2n}}}}{{{3^n} - {{\left( {2\,\cos \,x} \right)}^{2n}}}}$$ is given by :
A.
$$R$$
B.
$$\left\{ {n\pi \pm \frac{\pi }{3},\,n \in \,I} \right\}$$
C.
$$\left\{ {n\pi \pm \frac{\pi }{6},\,n \in \,I} \right\}$$
D.
none of these
Answer :
$$\left\{ {n\pi \pm \frac{\pi }{6},\,n \in \,I} \right\}$$
Solution :
We have,
$$\eqalign{
& f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\,\sin \,x} \right)}^{2n}}}}{{{3^n} - {{\left( {2\,\cos \,x} \right)}^{2n}}}} \cr
& = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\,\sin \,x} \right)}^{2n}}}}{{{{\left( {\sqrt 3 } \right)}^{2n}} - {{\left( {2\,\cos \,x} \right)}^{2n}}}} \cr} $$
$$f\left( x \right)$$ is discontinuous when $${\left( {\sqrt 3 } \right)^{2n}} - {\left( {2\,\cos \,x} \right)^{2n}} = 0$$
$${\text{i}}{\text{.e}}{\text{., }}\cos \,x = \pm \frac{{\sqrt 3 }}{2} \Rightarrow x = n\pi \pm \frac{\pi }{6}\left( {n\, \in \,I} \right)$$