Functions Question 656
If $ g:
[-2,,2]\to R $ where $ g(x)= $ $ x^{3}+\tan x+ \left[ \frac{x^{2}+1}{P} \right] $ is an odd function then the value of parametric P is [DCE 2005]
Options:
A) $ -5 < P < 5 $
B) $ P<0.05 $
C) $ P>0.05 $
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)=2\frac{x^{2}+1}{P} $ because $ g(x) $ is not an odd function \ $ [ x^{3}+\tan x+\frac{x^{2}+1}{P} ] $ $ +[ -x^{3}-\tan x+\frac{x^{2}+1}{P} ]=2\frac{x^{2}+1}{P} $
Þ $ \frac{2(x^{2}+1)}{P}=0 $
Þ $ 0\le \frac{x^{2}+1}{P}<1 $ because $ x\in [-2,,2] $ and $ P > x^{2}+1 $
Þ $ 0\le \frac{5}{P}<1\Rightarrow P>5 $ .
BETA
