IIFT 2018 Question 113

Question 113

The square root of $$1 + x^2 + \sqrt{1 + x^2 + x^4}$$is

Solution

$$1+x^2+\sqrt{\ 1+x^2+x^4}$$

=$$\frac{1}{2}\left(2+2x^2+2\sqrt{\ 1+x^2+x^4}\right)$$

=$$\frac{1}{2}\left(1+x+x^2+1-x+x^2+2\sqrt{\left(1+x+x^2\right)\left(1-x+x^2\right)}\right)$$

=$$\frac{1}{2}\left(\sqrt{\ 1+x+x^2}+\sqrt{\ 1-x+x^2}\right)^2$$

$$\sqrt{\ 1+x^2+\sqrt{\ 1+x^2+x^4}}=\sqrt{\frac{1}{2}\left(\sqrt{\ 1+x+x^2}+\sqrt{\ 1-x+x^2}\right)^2\ }$$

$$\sqrt{\ 1+x^2+\sqrt{\ 1+x^2+x^4}}=\frac{1}{\sqrt{\ 2}}\left(\sqrt{1+x+x^2}+\sqrt{\ 1-x+x^2}\ \right)$$

The answer is option A.



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