Question

If $x+\frac{1}{x}=99,$ then the value of $\frac{100x}{3x^2+103x+3}$ is

• A. $\frac{3}{4}$
• B. $\frac{1}{4}$
• C. $\frac{5}{4}$
• D. $\frac{7}{4}$

Answer 1

The correct option is B $\frac{1}{4}$

$Given$ X+$\frac{1}{x}=99$ multiply both sides by x

Then $x^2+1=99x\Rightarrow x^2-99x+1=0$

So $\frac{100x}{3x^2+103x+3}=\frac{100x}{3x^2-297x+3+400x}=\frac{100x}{3(x^2-99x+1)+400x}=\frac{100x}{3\times 0+400x}\frac{100x}{400x}=\frac{1}{4}$