Question

If $\underset{x\to \infty }{lim}(\sqrt{x^4-a{x}^3-4x^2+bx+3}-\sqrt{x^4+3x^3+c{x}^2-3x+d})=3$, then

• A. $a=-9$ and $c=-3$
• B. $a=2$ and $c=0$
• C. $a=3$ and $c=10$
• D. $a=-3$ and $c=-10$

Answer 1

The correct option is D $a=-3$ and $c=-10$

$\underset{x\to \infty }{lim}\sqrt{x^4-a{x}^3-4x^2+bx+3}$
$-\sqrt{x^4+3x^3+c{x}^2-3x+d}=3$

$\underset{x\to \infty }{lim}\frac{-(a+3)x^3-(4+c)x^2+(b+3)x+3-d}{\sqrt{x^4-a{x}^3-4x^2+bx+3}+\sqrt{x^4+3x^3+c{x}^2-3x+d}}=3$
As $x\to \infty$, for limit to exist, $-(a+3)=0\Rightarrow a=-3$
When $a=-3$, we have
$\underset{x\to \infty }{lim}\frac{-(4+c)x^2+(b+3)x+3-d}{\sqrt{x^4+3x^3-4x^2+bx+3}+\sqrt{x^4+3x^3+c{x}^2-3x+d}}=3$

$\underset{x\to \infty }{lim}\frac{-(4+c)+\frac{b+3}{x}+\frac{3-d}{x^2}}{\sqrt{1+\frac{3}{x}-\frac{4}{x^2}+\frac{b}{x^3}+\frac{3}{x^4}}+\sqrt{1+\frac{3}{x}+\frac{c}{x^2}-\frac{3}{x^3}+\frac{d}{x^4}}}=3$

$\Rightarrow \frac{-(4+c)}{1+1}=3$
$\Rightarrow c=-10$ and $a=-3$