A) 8
B) \[2\sqrt{2}\]
C) 2
D) \[\sqrt{2}\]
Correct Answer: B
Solution :
Let the ordinate at \[x=a\]divide the area into two equal parts Area of \[AMNB\]\[=\int_{2}^{4}{\left( 1+\frac{8}{{{x}^{2}}} \right)\text{ }dx=\left[ x-\frac{8}{x} \right]}_{2}^{4}=4\] Area of \[ACDM=\int_{2}^{a}{\left( 1+\frac{8}{{{x}^{2}}} \right)}dx=2\] On solving, we get \[a=\pm 2\sqrt{2}\];Since \[a>0\]Þ \[a=2\sqrt{2}\].You need to login to perform this action.
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