A) \[\frac{3}{2}\]
B) 2
C) 1
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
Let \[I=\int_{3}^{6}{\frac{\sqrt{x}}{\sqrt{9-x}\,+\,\sqrt{x}}dx}\] ?(i) \[\int_{3}^{6}{\frac{\sqrt{9-x}}{\sqrt{9-9+x}+\sqrt{9-x}}dx}\] \[\Rightarrow \]\[I=\int_{3}^{6}{\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx}\] ...(ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{3}^{6}{\frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}\,dx}\] \[=\int_{3}^{6}{1\,dx}\] \[=[x]_{3}^{6}\Rightarrow I=\frac{3}{2}\]You need to login to perform this action.
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