JEE Main 2021 — Mathematics Calculus

Given that:$\underset{x\rightarrow \infty }{lim}(\sqrt{{x}^{2}-x+1}-ax)=b$x→∞lim​(x2−x+1​−ax)=b

$$\Rightarrow \underset{x\rightarrow \infty }{lim}(\sqrt{{x}^{2}-x+1}-ax)\times \frac{\sqrt{{x}^{2}-x+1}+ax}{\sqrt{{x}^{2}-x+1}+ax}=b$$⇒x→∞lim​(x2−x+1​−ax)×x2−x+1​+axx2−x+1​+ax​=b

$$\Rightarrow \underset{x\rightarrow \infty }{lim}\frac{({x}^{2}-x+1)-(ax{)}^{2}}{\sqrt{{x}^{2}-x+1}+ax}=b$$⇒x→∞lim​x2−x+1​+ax(x2−x+1)−(ax)2​=b

$\Rightarrow \text{ Limit exist only If }{a}^{2}=1$⇒ Limit exist only If a2=1, because ${x}^{2}$x2 must not be there in the numerator

$$\therefore a=1,-1$$∴a=1,−1

$$\Rightarrow \underset{x\rightarrow \infty }{lim}\frac{-x+1}{\sqrt{{x}^{2}-x+1}+ax}=b$$⇒x→∞lim​x2−x+1​+ax−x+1​=b

$$\Rightarrow \underset{x\rightarrow \infty }{lim}\frac{-1+\frac{1}{x}}{\sqrt{1-\frac{1}{x}+\frac{1}{{x}^{2}}}+a}=b$$⇒x→∞lim​1−x1​+x21​​+a−1+x1​​=b

$$\Rightarrow \frac{-1}{1+a}=b$$⇒1+a−1​=b

Here, $a\neq -1$a=−1

So$a=1$a=1

$$b=\frac{-1}{2}$$b=2−1​

$$(a,b)\equiv (1,\frac{-1}{2})$$(a,b)≡(1,2−1​)