First slide

Binomial theorem for positive integral Index

Question

The expression ${(\sqrt{2 x^{2} + 1} + \sqrt{2 x^{2} - 1})}^{6} + {(\frac{2}{\sqrt{2 x^{2} + 1} + \sqrt{2 x^{2} - 1}})}^{6}$ is a polynomial of degree

Moderate

A

6
B

8
C

10
D

12

Solution

We have,
$\begin{matrix} \frac{2}{\sqrt{2 x^{2} + 1} + \sqrt{2 x^{2} - 1}} = \frac{2 (\sqrt{2 x^{2} + 1} - \sqrt{2 x^{2} - 1})}{(2 x^{2} + 1) - (2 x^{2} - 1)} \\ = \sqrt{2 x^{2} + 1} - \sqrt{2 x^{2} - 1} \end{matrix}$
Thus, the given expression can be written as
${(\sqrt{2 x^{2} + 1} + \sqrt{2 x^{2} - 1})}^{6} + {(\sqrt{2 x^{2} + 1} - \sqrt{2 x^{2} - 1})}^{6}$
But $(a + b)^{6} + (a - b)^{6} = 2 [a^{6} +^{6} C_{2} a^{4} b^{2} +^{6} C_{4} a^{2} b^{4} + b^{6}]$
Therefore,
${(\sqrt{2 x^{2} + 1} + \sqrt{2 x^{2} - 1})}^{6} + {(\sqrt{2 x^{2} + 1} - \sqrt{2 x^{2} - 1})}^{6} = 2 [{(2 x^{2} + 1)}^{3} + 15 {(2 x^{2} + 1)}^{2} (2 x^{2} - 1) + 15 (2 x^{2} + 1) \times {(2 x^{2} - 1)}^{2} + {(2 x^{2} - 1)}^{3}]$
which is a polynomial of degree 6.

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