Question

If $\underset{x\to \infty }{lim}x\ln \left\{\left|\begin{array}{ccc}\frac{a}{x}& 1& c\\ 0& \frac{1}{x}& b\\ 1& 0& \frac{1}{x}\end{array}\right|\right\}=-5,$ where $a,b$ and $c$ are finite real numbers, then

• A. $a=2,b=1,c\in R$
• B. $a=2,b=2,c=5$
• C. $a\in R,b=1,c\in R$
• D. $a\in R,b=1,c=5$

Answer 1

The correct option is D $a\in R,b=1,c=5$

After solving determinant, we get
$\underset{x\to \infty }{lim}x\ln (\frac{a}{x^3}+b-\frac{c}{x})=-5$
Replacing $x\to \frac{1}{h}$
$\underset{h\to 0}{lim}\frac{\ln (a{h}^3+b-ch)}{h}=-5$

For limit to exist,
$\underset{h\to 0}{lim}\ln (a{h}^3+b-ch)=0\Rightarrow \ln b=0\Rightarrow b=1$

$\underset{h\to 0}{lim}\frac{\ln (a{h}^3+1-ch)}{h}=-5$
using L-hospital $\left(\frac{0}{0}\text{form}\right)$
$\underset{h\to 0}{lim}\frac{3a{h}^2-c}{a{h}^3+1-ch}=-5\Rightarrow -c=-5\Rightarrow c=5\Rightarrow a\in R$