Question

If $f(x)$ is defined in $[-2, 2]$ by $f(x) = 4x^{2} - 3x +1$ and $g(x)=\frac{f(-x)-f(x)}{x^2+3}$ then $ \int_{-2}^{2} g(x)dx = $

• A. 24
• B. 0
• C. -48
• D. 64

Answer 1

Answer: (B)

Given that
$f(x) =4 x^{2}-3 x+1, g(x)=\frac{f(-x)-f(x)}{x^{2}+3} $
$\therefore g(x) =\frac{\left(4 x^{2}+3 x+1\right)-\left(4 x^{2}-3 x+1\right)}{x^{2}+3} $
$=\frac{6 x}{x^{2}+3}$
Now, $g(-x) =-\frac{6 x}{x^{2}+3} $
$=-g(x)$
Which is an odd function
$\int\limits_{-2}^{2} g(x) d x=0$