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  • The expression ,  <img alt="\left ( \sqrt {2x^2 +1}+ \sqrt {2x^2-1} \right )^6 + \left ( \frac{2}{ \sqrt {2x^2 +1}+ \sqrt {2x^2-1}} \right ) ^6" src="/latex-image/?%5Cleft%20%28%20%5

The expression , 

\left ( \sqrt {2x^2 +1}+ \sqrt {2x^2-1} \right )^6 + \left ( \frac{2}{ \sqrt {2x^2 +1}+ \sqrt {2x^2-1}} \right ) ^6  is a polygon of degree 

Option: 1

5


Option: 2

6


Option: 3

7


Option: 4

8


Answers (1)

best_answer

As we have learnt,

\left ( x+a \right )^{n}+\left ( x-a \right )^{n}= 2\left ( ^{n}C_{0} \, x^{n}+ ^{n}C_{2}\, x^{n-2}\, a^{2}+---\right )

 

Now,

\frac{2}{ \sqrt {2x^2 +1}+ \sqrt {2x^2-1}} = \sqrt {2x^2 +1}- \sqrt {2x^2-1} 

(On rationalization)

\left ( \sqrt {2x^2 +1}+ \sqrt {2x^2-1} \right )^6 + \left ( \sqrt {2x^2 +1}- \sqrt {2x^2-1} \right )^6 \\\\ = 2 ( ^ 6 C_0 (2x^2 + 1 )^3+ ^ 6C_2 (2x^2+1)^2 (2x^2-1)+(^6 C_4 )(2x^2 + 1 )(2x^2 -1)^2+ ^6 C_6 (2x^2-1)^3 )

Clearly, the degree is  6 

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