A) \[\frac{1}{n}\]
B) \[n-\frac{1}{n}\]
C) \[n+\frac{1}{n}\]
D) None
Correct Answer: C
Solution :
[c] Let \[\underset{x\to 0}{\mathop{\lim }}\,\frac{(sinnx)[(a-n)nx-tanx]}{{{x}^{2}}}=0\] \[\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\frac{\left( nx-\frac{{{n}^{3}}{{x}^{3}}}{3!} \right)\left[ n(a-n)x-\left\{ x+\frac{{{x}^{3}}}{3}+... \right\} \right]}{{{x}^{2}}}=0\](By suing expansion of sin x and tan x) \[\Rightarrow {{n}^{2}}(a-n)-n=0\Rightarrow an-{{n}^{2}}-1=0\] \[\Rightarrow a=\frac{{{n}^{2}}+1}{n}=n+\frac{1}{n}\]
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