A) \[-5<P<5\]
B) \[P<5\]
C) \[P>5\]
D) None of these
Correct Answer: C
Solution :
\[g(x)={{x}^{3}}+\tan x+\frac{{{x}^{2}}+1}{P}\] \[g(-x)={{(-x)}^{3}}+\tan (-x)+\frac{{{(-x)}^{2}}+1}{P}\] \[g(-x)=-{{x}^{3}}-\tan x+\frac{{{x}^{2}}+1}{P}\] \[g(x)+g(-x)=0\]because \[g(x)\] is a odd function \\[\left[ {{x}^{3}}+\tan x+\frac{{{x}^{2}}+1}{P} \right]\] \[+\left[ -{{x}^{3}}-\tan x+\frac{{{x}^{2}}+1}{P} \right]=0\] Þ \[\frac{2({{x}^{2}}+1)}{P}=0\] Þ \[0\le \frac{{{x}^{2}}+1}{P}<1\] because \[x\in [-2,\,2]\] Þ \[0\le \frac{5}{P}<1\Rightarrow P>5\].
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