A) 64
B) -48
C) 0
D) 24
Correct Answer: C
Solution :
Given that \[f(x)=4{{x}^{2}}-3x+1,\]\[g(x)=\frac{f(-x)-f(x)}{{{x}^{2}}+3}\] \[\therefore \]\[g(x)=\frac{(4{{x}^{2}}+3x+1)-(4{{x}^{2}}-3x+1)}{{{x}^{2}}+3}\] \[=\frac{6x}{{{x}^{2}}+3}\] Now, \[g(-x)=-\frac{6x}{{{x}^{2}}+3}\] \[-g(x)\] Which is an odd function \[\therefore \]\[\int_{-2}^{2}{g(x)\,\,dx=0}\]You need to login to perform this action.
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