A) 3
B) 2
C) \[\frac{16+n}{12}\]
D) None of these
Correct Answer: A
Solution :
[a] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 2x-n\sin x}{{{x}^{3}}}=I\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2x+\frac{8{{x}^{3}}}{3!}...-nx+\frac{n{{x}^{3}}}{3!}}{{{x}^{3}}}=I\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{(2-n)x+\left( \frac{16+n}{6} \right){{x}^{3}}+....}{{{x}^{3}}}=I\] \[\Rightarrow n=2\] and, thus required value \[=\frac{16+n}{6}=3.\]
You need to login to perform this action.
You will be redirected in
3 sec
