Q107 / 200 · Manipal (MET) 2014 · Mathematics
The value of $\int_{1 / n}^{(a n-1) / n} \frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}} d x$ is equal to
- A$\frac{a}{2}$
- B$\frac{n \cdot a+2}{2 n}$
- Correct answerC$\frac{n \cdot a-2}{2 n}$
- D$\frac{n \cdot a}{2}$
Solution
$\begin{aligned}
& \text { Let, } I=\int_{\frac{1}{n}}^{(a n-1) / n} \frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}} d x \\
& =\int_{\frac{1}{n}}^{a-\frac{1}{n}} \frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}} d x ...(i)\\
& =\int_{\frac{1}{n}}^{a-\frac{1}{n}} \frac{\left(\sqrt{\frac{1}{n}+a-\frac{1}{n}-x}\right)}{\sqrt{a-\left(\frac{1}{n}+a-\frac{1}{n}-x\right)}+\sqrt{\frac{1}{n}+a-\frac{1}{n}-x}} d x \\
& \Rightarrow \quad I=\int_{\frac{1}{n}}^{a-\frac{1}{n}} \frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x...(ii)
\end{aligned}$
On adding Eqs. (i) and (ii), we get $\begin{aligned} 2 I & =\int_{\frac{1}{n}}^{a-\frac{1}{n}} 1 d x=[x]_{\frac{1}{n}}^{a-\frac{1}{n}} \\ & =a-\frac{1}{n}-\frac{1}{n}=\frac{n a-2}{n} \Rightarrow I=\frac{n a-2}{2 n} \end{aligned}$
On adding Eqs. (i) and (ii), we get $\begin{aligned} 2 I & =\int_{\frac{1}{n}}^{a-\frac{1}{n}} 1 d x=[x]_{\frac{1}{n}}^{a-\frac{1}{n}} \\ & =a-\frac{1}{n}-\frac{1}{n}=\frac{n a-2}{n} \Rightarrow I=\frac{n a-2}{2 n} \end{aligned}$
Hence, the correct answer is option: 'C'.