Question

If $\underset{x\to \infty }{lim}\frac{1}{\ln x-x}=\frac{a}{b}$( where $b\ne 0)$ and the equation $a{x}^3+x^2+bx+1=0$ has equal roots, then the value of $|a|+|b|$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$\frac{a}{b}=\underset{x\to \infty }{lim}\frac{1}{\ln x-x}=\underset{x\to \infty }{lim}\frac{1}{\ln x-\ln e^x}$
$=\underset{x\to \infty }{lim}\frac{1}{\ln \left(\frac{x}{e^x}\right)}$
For $x\to \infty ,\frac{x}{e^x}\to 0$
So, $\ln \left(0\right)\to -\infty$
$\therefore \frac{a}{b}=0\Rightarrow a=0$

$\therefore x^2+bx+1=0$
Equation has equal roots.
So, $D=b^2-4=0\Rightarrow b=\pm 2$
$\therefore |a|+|b|=2$