question_answer

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

A)  2a                                         

B)                  \[2\sqrt{2}\]

C)                  \[\frac{a}{2}\]                                   

D)                   none of these

Correct Answer: B

Solution :

We have, \[y=\left( 1+\frac{8}{{{x}^{2}}} \right)\] Area of curve MNBA \[=\int_{2}^{4}{\left( 1+\frac{8}{{{x}^{2}}} \right)}\,dx\] \[=\left[ x-\frac{8}{x} \right]_{2}^{4}=4\]                              ?(i) Area of curve ACDM \[=\int_{2}^{a}{\left( 1+\frac{8}{{{x}^{2}}} \right)dx}\] \[=\left[ x-\frac{8}{x} \right]_{2}^{a}=a-\frac{8}{a}-[2-4]\] \[=a-\frac{8}{a}+2\]                        ?(ii) From Eqs. (i) and (ii), we get \[a-\frac{8}{a}+2=\frac{1}{2}(4)\] \[\Rightarrow \]               \[{{a}^{2}}-8=0\Rightarrow a=2\sqrt{2}\]