Question

$\text{If}\underset{x\to -\infty }{lim}(\sqrt{x^2-x+1}-ax-b)=0$ then the values of a and b are given by -

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

Answer 1

The correct option is A

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

$\text{We have,}\underset{x\to -\infty }{lim}(\sqrt{x^2-x+1}-ax-b)=0[\text{putting x=-y; x}\to -\infty ,y\to \infty ]\Rightarrow \underset{y\to \infty }{lim}(\sqrt{y^2+y+1}+ay-b)=0\Rightarrow \underset{y\to \infty }{lim}[y{(1+\frac{1}{y}+\frac{1}{y^2})}^{1/2}+ay-b]=0\Rightarrow \underset{y\to \infty }{lim}[y\{1+\frac{1}{2}(\frac{1}{y}+\frac{1}{y^2}+.........)\}+ay-b]=0\Rightarrow \underset{y\to \infty }{lim}\left[y\right(1+a)+(\frac{1}{2}-b)+\frac{1}{2y}+........]=0\Rightarrow 1+a=0and\left(1/2\right)-b=0\Rightarrow a=-1andb=1/2$