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SSC CGL 23 Aug 2021 Shift 2 Paper

Question : 69 of 100

Marks: +1, -0

x+\frac{1}{x}=2\sqrt{5}

\frac{x^{4}+\frac{1}{x^{2}}}{x^{2}+1}?

Solution:

Given,

x+\frac{1}{x}=2\sqrt{5}

\Rightarrow (x^{2}+1)=2\sqrt{5}x

Now,

\frac{x^{4}+\frac{1}{x^{2}}}{x^{2}+1}=\frac{x^{8}+1}{x^{2}(x^{2}+1)}

We know that,

(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)

Here,

a

x^{2}, b=1

(x^{2}+1)^{3}=x^{8}+1+3\times x^{2}\times 1(x^{2}+1)

\Rightarrow (2\sqrt{5}x)^{3}=x^{8}+1+3x^{2}(2\sqrt{5}x)

[\because x^{2}+1=2\sqrt{5}x]

\Rightarrow 40\sqrt{5}x^{3}=x^{8}+1+6\sqrt{5}x^{3}

\Rightarrow (x^{8}+1)=40\sqrt{5}x^{3}-6\sqrt{5}x^{3}

=34\sqrt{5}x^{3}

\therefore \frac{x^{8}+1}{x^{2}(x^{2}+1)}=\frac{34\sqrt{5}x^{3}}{x^{2}\times 2\sqrt{5}x}

=\frac{34\sqrt{5}x^{3}}{2\sqrt{5}x^{3}}=17

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