A) \[2+\sqrt{2}\]
B) \[2-\sqrt{2}\]
C) \[1+\sqrt{2}\]
D) \[1-\sqrt{2}\]
Correct Answer: B
Solution :
On solving a = 0 and b = 2 \[{{I}_{x}}=\frac{0\times 2+0\times \sqrt{8}+2\times 2}{2+2+2\sqrt{2}}\] \[{{I}_{x}}=\frac{4}{4+2\sqrt{2}}\] \[{{I}_{x}}=\frac{2}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}\] \[{{I}_{x}}=\frac{2\left( 2-\sqrt{2} \right)}{2}\]\[\Rightarrow \]\[2-\sqrt{2}\]You need to login to perform this action.
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