Question

$\text{The values of constants a and b so that}\underset{x\to \infty }{lim}(\frac{x^2+1}{x+1}-ax-b)=\frac{1}{2},are$

• A. $a=1,b=-\frac{3}{2}$
• B. $a=-1,b=\frac{3}{2}$
• C. a=0, b=0
• D. a =2, b= -1

Answer 1

The correct option is A

$a=1,b=-\frac{3}{2}$

$\underset{x\to \infty }{lim}(\frac{x^2+1}{x+1}-ax-b)=\frac{1}{2}\Rightarrow \underset{x\to \infty }{lim}\frac{(x^2+1)-(ax+b)(x+1)}{x+1}=\frac{1}{2}\Rightarrow \underset{x\to \infty }{lim}\frac{x^2(1-a)-(a+b)x-b+1}{x+1}=\frac{1}{2}\Rightarrow 1-a=0anda+b=-\frac{1}{2}\therefore a=1b=-\frac{3}{2}$