SSC CGL Model Paper 18

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Question : 64 of 100

Marks: +1, -0

If

x -1/x= 1

, then what is the value of

1/x(1/{x-1}-1/{x+1}+1/{x^2+1}-1 /{x^2-1})?

$±√5$
$2/5$
$4/{5±√5}$
$±√5/2$

Solution:

We have,

x -1/x= 1

⇒

{x^2-1}/x=1

⇒

x^2-1 = x

⇒

x^2 =x + 1

...(i)
⇒

x^2 - x - 1 = 0

x={-b±√{(b)^2-4ac}}/{2a}

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

={-(-1)±√{(-1)^2-4(1)(-1)}}/{2(1)}

={1±√{1+4}}/2={1±√5}/2

Given expression

=1/x(1/{x-1}-1/{x+1}+1/{x^2+1}-1 /{x^2-1})

=1/x({x + 1-x + 1}/{( x - 1) ( x + 1)}+{x^2-1 - x ^2- 1}/{(x^2+1)(x^2-1)})

=1/x[2/{x^2-1}+{-2}/{(x^2+1)(x^2-1)}]

=1/x[{ 2x^2 + 2 -2}/{(x^2+1)(x^2-1)}]

=1/x×{2x^2}/{(x^2+1)x}={2x^2}/{x^2(x^2+1)}

=2/{x^2+1}=2/{x + 1 + 1}=2/{x+2}

[From Eq. (i)]

=2/{{1± √5}/2+2}=2/{{1±√5+4}/2}

=4/{ 5±√5}=4/{√5(√5+1)},

=4/{√5(√5+1)}

or

4/{5+√5},4/{5-√5}

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