question_answer

If the ordinate \[x=a\] divides the area bounded by x-axis, part of the curve \[y=1+\frac{8}{{{x}^{2}}}\] and the ordinates \[x=2,\text{ }x=4\] into two equal parts, then a is equal to

A) \[\sqrt{2}\]

B) \[2\sqrt{2}\]

C) \[3\sqrt{2}\]

D) None of these

Correct Answer: B

Solution :

[b] The area bounded by the curve

\[y=1+\frac{8}{{{x}^{2}}},\] x-axis and the ordinates \[x=2,x=4\]is

\[=\int_{2}^{4}{ydx}\]

\[=\int_{2}^{4}{\left( 1+\frac{8}{{{x}^{2}}} \right)dx}\]

\[=\left[ x-\frac{8}{x} \right]_{2}^{4}=4\].

Since, \[x=a\] divides this area into two equal parts,

\[\therefore \] Required area \[=2\int_{2}^{a}{y\,\,dx}\]

\[\therefore \,\,\,\,\,\,\,4=2\int_{2}^{a}{\left( 1+\frac{8}{{{x}^{2}}} \right)dx}\]

\[\Rightarrow 2=\left[ x-\frac{8}{x} \right]_{2}^{a}=\left( a-\frac{8}{a} \right)-(2-4)\]

\[\Rightarrow {{a}^{2}}=8\therefore a=2\sqrt{2}\]