$\int_0^{\alpha / 3} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x=$
$\int_0^{\alpha / 3} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x=$
- $\frac{2 \alpha}{3}$
- $\frac{\alpha}{2}$
- $\frac{\alpha}{3}$
- $\frac{\alpha}{6}$
Solution
(d) Let
$
\begin{gathered}
I=\int_0^{\alpha / 3} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x \\
\Rightarrow \quad I=\int_0^{\alpha / 3} \frac{f\left(\frac{\alpha}{3}-x\right)}{f\left(\frac{\alpha}{3}-x\right)+f\left(\frac{\alpha-3\left(\frac{\alpha}{3}-x\right)}{3}\right)} d x \\
\left.\Rightarrow \quad I=\int_0^{\alpha / 3} \frac{f\left(\frac{\mathrm{i})}{3} f(x) d x=\int_0^\alpha f(a-x)\right.}{f\left(\frac{\alpha-3 x}{3}\right)+f(x)} d x\right]
\end{gathered}
$
On adding Eqs. (i) and (ii), we get
$
\begin{aligned}
& 2 I=\int_0^{\alpha / 3} 1 d x \\
& \Rightarrow \quad 2 I=[x]_0^{\alpha / 3} \\
& \Rightarrow \quad 2 I=\frac{\alpha}{3} \\
& \Rightarrow \quad I=\frac{\alpha}{6} \text {. } \\
&
\end{aligned}
$
Asked in: AP EAMCET 2018 (22 Apr Shift 1)
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