# NIELIT 2022 Feb Scientist D - Section D: 4

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If $x + \dfrac{1}{x} = 99,$ then the value of $\dfrac{100x}{2x^{2} + 102x + 2}$ is :

1. $1/3$
2. $1/99$
3. $99/100$
4. $3$

Given that; $\frac{100x}{2x^2+102x+2}$

Taking $2x$ as common from the denominator we get;

$\implies \frac{100x}{2x(x+51+1/x)}$

$\implies \frac{50}{99+51}=1/3$ [$\because x+\frac{1}{x}=99$]

Option (A) is correct.

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