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Question: Let ƒ(x) = \(\left( \sqrt{x^{2} + 1} + \sqrt{x^{2} - 1} \right)^{6}\)+\(\left( \frac{2}{\sqrt{x^{2} ...

Let ƒ(x) = (x2+1+x2−1)6\left( \sqrt{x^{2} + 1} + \sqrt{x^{2} - 1} \right)^{6}+(2x2+1+x2–1)6\left( \frac{2}{\sqrt{x^{2} + 1} + \sqrt{x^{2}–1}} \right)^{6}. Then –

A

ƒ(x) is a polynomial of the fourth degree in x

B

ƒ(x) has exactly two terms

C

ƒ(x) is not a polynomial in x

D

Coefficient of x6 is 46

Answer

ƒ(x) has exactly two terms

Explanation

Solution

ƒ(x) = (x2+1+x2−1)6\left( \sqrt{x^{2} + 1} + \sqrt{x^{2} - 1} \right)^{6}

+(x2+1–x2−1)6\left( \sqrt{x^{2} + 1}–\sqrt{x^{2} - 1} \right)^{6}.

After expansion and simplification

ƒ(x) = 64x6 – 48x2.