A) \[\frac{a}{2}\]
B) \[\frac{na+2}{2n}\]
C) \[\frac{na-2}{2n}\]
D) None of these
Correct Answer: C
Solution :
\[I=\int_{1/n}^{\frac{an-1}{n}}{\frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}dx=\int_{1/n}^{a-\frac{1}{n}}{\frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}dx}}\] .....(i) \[=\int_{\frac{1}{n}}^{a-\frac{1}{n}}{\frac{\sqrt{\frac{1}{n}+a-\frac{1}{n}-x}\,\,\,\,\,\,\,\,\,\,\,dx}{\sqrt{a-\left( \frac{1}{n}+a-\frac{1}{n}-x \right)+}\sqrt{\frac{1}{n}+a-\frac{1}{n}-x}}}\]\[\] \[\left[ \because \int_{a}^{b}{f(x)dx=\int_{a}^{b}{f(a+b-x)\,dx}} \right]\] \[I=\int_{\frac{1}{n}}^{a-\frac{1}{n}}{\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx}\] .....(ii) Adding (i) and (ii), we get \[2I=\int_{\,1/n}^{\,a-(1/n)}{1\,dx=\left[ \,x \right]_{\,1/n}^{\,a-\frac{1}{n}}}\] \[\Rightarrow 2I=a-\frac{1}{n}-\frac{1}{n}=\frac{na-2}{n}\] \[\Rightarrow I=\frac{na-2}{2n}\].
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