Question

If $\alpha, \beta(\beta>\alpha)$ are the roots of $f(x) \equiv a x^{2}+b x+c=0, a \neq 0$ and $f(x)$ is an even function and $I=\int_{\alpha}^{\beta} \frac{e_{f}^{f\left(\frac{f(x)}{x-\alpha}\right)}}{f\left(\frac{f(x)}{x-\alpha}\right)+e^{f}\left(\frac{f(x)}{x-\beta}\right)}$, then $|I|$ is equal to

• A. $\left|\frac{b}{b}\right|$
• B. $\mid \frac{b}{2 a}$
• C. $\frac{\sqrt{b^{2}-4 a c}}{|2 a|}$
• D. None of these

Answer 1

Answer: (C)

$I=\int_{\alpha}^{\beta} \frac{e^{f\left(\frac{a(x-\alpha)(x-\beta)}{x-\alpha}\right)}}{e^{f\left(\frac{a(x-\alpha)(x-\beta)}{x-\alpha}\right)_{e}^{f}\left(\frac{a(x-\alpha)(x-\beta)}{x-\beta}\right)}} d x$
$=\int_{\alpha}^{\beta} \frac{e^{f(a(x-\beta))}}{e^{f(a(x-\beta))+f(a(x-\alpha))}} d x \ldots(1)$
$=\int_{\alpha}^{\beta} \frac{e^{f(a(\alpha+\beta-x-\beta))}}{e^{f(a(\alpha+\beta-x-\beta))}+e^{f(a(\alpha+\beta-x-\alpha))}} d x$
$I=\int_{\alpha}^{\beta} \frac{e^{f(a(x-\alpha))}}{e^{f(a(x-\alpha)}+e^{\int(a(x-\beta))}} d x \ldots(2)$
$2 I=\int_{\alpha}^{\beta} d x$
$\Rightarrow I=\frac{|\alpha-\beta|}{2}=\frac{\sqrt{b^{2}-4 a c}}{2|a|}$