Question

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

• A. $1$
• B. $2$
• C. $-1$
• D. $0$

Answer 1

The correct option is D $0$

$\underset{x\to 0}{lim}f\left(x\right)=2$

$\Rightarrow \underset{x\to 0}{lim}\frac{ax+b}{x+1}=2$

$\Rightarrow \frac{b}{1}=2$

$\therefore b=2$

Its is also given that

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

$\Rightarrow \underset{x\to \infty }{lim}\frac{ax+b}{x+1}=1$

$\Rightarrow \underset{x\to \infty }{lim}\frac{a+\frac{b}{x}}{1+\frac{1}{x}}=1$

$\Rightarrow \frac{a+0}{1+0}=1$

$\therefore a=1$

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

$f\left(x\right)=\frac{x+2}{x+1}$

$\therefore f(-2)=\frac{-2+2}{-2+1}=0$