Question

If $\underset{x\to \infty }{lim}(\frac{a{x}^2+2}{x-1}+bx+a)=3$, then which of the following option(s) is/are correct?

• A. $a$ and $b$ are additive inverse to each other
• B. $a=\frac{3}{2},b\in R$
• C. $a=1,b=-1$
• D. $a=\frac{3}{2},b=-\frac{3}{2}$

Answer 1

Answer: (A), (D)

$a=\frac{3}{2},b=-\frac{3}{2}$

$\underset{x\to \infty }{lim}(\frac{a{x}^2+2}{x-1}+bx+a)=3\Rightarrow \underset{x\to \infty }{lim}\frac{(a+b)x^2+(a-b)x+2-a}{x-1}=3$
For above limit to exist,
$a+b=0$ and $a-b=3$
Clearly, $a$ and $b$ are additive inverse to each other.
$\Rightarrow a=\frac{3}{2},b=-\frac{3}{2}$