Functions Question 656
Question: If $ g:
[-2,,2]\to R $ where $ g(x)= $ $ x^{3}+\tan x+[ \frac{x^{2}+1}{P} ] $ is a odd function then the value of parametric P is [DCE 2005]
Options:
A) $ -5<P<5 $
B) $ P<5 $
C) $ P>5 $
D) None of these
Show Answer
Answer:
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 \ $ [ x^{3}+\tan x+\frac{x^{2}+1}{P} ] $ $ +[ -x^{3}-\tan x+\frac{x^{2}+1}{P} ]=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 $ .
BETA
