A) 2 sq unit
B) 4 sq unit
C) \[log\text{ }2\]sq unit
D) \[log\text{ }4\]sq unit
E) 8 sq unit
Correct Answer: B
Solution :
Required area\[=\int_{2}^{4}{\left( 1+\frac{8}{{{x}^{2}}} \right)}\,dx\] \[=\left[ x-\frac{8}{x} \right]_{2}^{4}=4-\frac{8}{4}-2+\frac{8}{2}\] \[=4-2-2+4=4\text{ }sq\text{ }unit\]
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