Question

If $\underset{x\to \infty }{\lim }$$\left(\sqrt{x^4+a{x}^3+3x^2+bx+2}-\sqrt{x^4+2x^3-c{x}^2+3x-d}\right)=4$,
then the values of $a$, $b$, $c$ and $d$ is

• A. $a=1$, $b=2$, $c=3$ and $d=4$
• B. $a=2$, $c=3$, $c=5$ and $d=0$
• C. $a=2$, $b=4$, $c=5$ and $d\in R$
• D. $a=2$, $c=5$ and $b$, $d\in R$

Answer 1

Answer: (D)

Given that,
$4=\underset{x\to \infty }{\lim }$$\left(\sqrt{x^4+a{x}^3+3x^2+bx+2}-\sqrt{x^4+2x^3-c{x}^2+3x-d}\right)$
$=\underset{x\to \infty }{\lim }$$\frac{\left(a-2\right)x^3+\left(3+c\right)x^2+\left(b-3\right)x+2+d}{\sqrt{x^4+a{x}^3+3x^2+bx+2}+\sqrt{x^4+2x^3-c{x}^2+3x-d}}$
Since, the limit is finite, the degree of the numerator must be at the most $2$
$\Rightarrow a-2=0$, i.e., $a=2$
Hence,
$4=\underset{x\to \infty }{\lim }$ $\frac{\left(3+c\right)+\frac{b-3}{x}+\frac{2+d}{x^2}}{\sqrt{1+\frac{a}{x}+\frac{3}{x^2}+\frac{b}{x^3}+\frac{2}{x^4}}+\sqrt{1+\frac{2}{x}-\frac{c}{x^2}-\frac{3}{x^3}-\frac{d}{x^4}}}$
$\Rightarrow 4=\frac{3+c}{2}$
$\Rightarrow c=5$
Hence, $a=2$,
$c=5$ and $b$, $d$ are any real numbers.