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SSC CGL Model Paper 18

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Question : 64 of 100

Marks: +1, -0

If

x - \frac{1}{x} = 1

, then what is the value of

\frac{1}{x}\left(\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x^2+1}-\frac{1}{x^2-1}\right)?

$\pm\sqrt{5}$
$\frac{2}{5}$
$\frac{4}{5\pm\sqrt{5}}$
$\pm\frac{\sqrt{5}}{2}$

Solution:

We have,

x - \frac{1}{x} = 1

⇒

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

⇒

x^2-1 = x

⇒

x^2 = x + 1

...(i)

⇒

x^2 - x - 1 = 0

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Here, a = 1, b = -1, c = -1

=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}

=\frac{1\pm\sqrt{1+4}}{2}=\frac{1\pm\sqrt{5}}{2}

Given expression

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

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

=\frac{1}{x}\left[\frac{2}{x^2-1}+\frac{-2}{(x^2+1)(x^2-1)}\right]

=\frac{1}{x}\left[\frac{2x^2+2-2}{(x^2+1)(x^2-1)}\right]

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

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

[From Eq. (i)]

=\frac{2}{\frac{1\pm\sqrt{5}}{2}+2}=\frac{2}{\frac{1\pm\sqrt{5}+4}{2}}

=\frac{4}{5\pm\sqrt{5}}=\frac{4}{\sqrt{5}(\sqrt{5}+1)},

=\frac{4}{\sqrt{5}(\sqrt{5}+1)}

or

\frac{4}{5+\sqrt{5}},\frac{4}{5-\sqrt{5}}

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