Question

If $f\left(x\right)=\frac{ax+b}{x+1}$, ${lim}_{x\to \infty }f\left(x\right)=1$ and ${lim}_{x\to 0}f\left(x\right)=2$, then $f(-2)=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A

$0$

Explanation for the correct option.

Step 1: Find the value of $a$ and $b$.

It is given that ${lim}_{x\to 0}f\left(x\right)=2$

$\begin{array}{rcl}{lim}_{x\to 0}\frac{ax+b}{x+1}& =& 2\\ \Rightarrow \frac{a\left(0\right)+b}{0+1}& =& 2\\ \Rightarrow b& =& 2\end{array}$

and ${lim}_{x\to \infty }f\left(x\right)=1$

$\begin{array}{rcl}{lim}_{x\to \infty }\frac{ax+b}{x+1}& =& 1\\ & \Rightarrow & {lim}_{x\to \infty }\frac{a+{\displaystyle \frac{b}{x}}}{1+{\displaystyle \frac{1}{x}}}=1\\ \Rightarrow a& =& 1\end{array}$

Step 2: Find the value of $f(-2)$.

By substituting the value of $a$ and $b$in $f\left(x\right)=\frac{ax+b}{x+1}$, we get

$\begin{array}{rcl}f(-2)& =& \frac{\left(1\times -2\right)+2}{-2+1}\\ & =& \frac{-2+2}{-1}\\ & =& 0\end{array}$

Hence, option A is correct.