Question

If $\displaystyle \lim_{x \to \infty}$$\left(\sqrt{x^{4}+ax^{3}+3x^{2}+bx+2}-\sqrt{x^{4}+2x^{3}-cx^{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=\displaystyle \lim_{x \to \infty}$$\left(\sqrt{x^{4}+ax^{3}+3x^{2}+bx+2}-\sqrt{x^{4}+2x^{3}-cx^{2}+3x-d}\right)$
$=\displaystyle \lim_{x \to \infty}$$\frac{\left(a-2\right)x^{3}+\left(3+c\right)x^{2}+\left(b-3\right)x+2+d}{\sqrt{x^{4}+ax^{3}+3x^{2}+bx+2}+\sqrt{x^{4}+2x^{3}-cx^{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,
$4=\displaystyle \lim_{x \to \infty}$ $\frac{\left(3+c\right)+\frac{b-3}{x}+\frac{2+d}{x^{2}}}{\sqrt{1+\frac{a}{x}+\frac{3}{x^{2}}+\frac{b}{x^{3}}+\frac{2}{x^{4}}}+\sqrt{1+\frac{2}{x}-\frac{c}{x^{2}}-\frac{3}{x^{3}}-\frac{d}{x^{4}}}}$
$\Rightarrow 4=\frac{3+c}{2}$
$\Rightarrow c=5$
Hence, $a = 2$,
$c = 5$ and $b$, $d$ are any real numbers.