Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}\underset{x\to \infty }{lim}(\frac{x^2+1}{x+1}-ax-b)=0,\text{then}& \text{(P)}& a=\frac{3}{2},b\in R\\ \text{(B)}& \text{If}\underset{x\to 0}{lim}(1+ax+b{x}^2{)}^{2/x}=e^3,\text{then}& \text{(Q)}& a=1,b=-\frac{1}{2}\\ \text{(C)}& \text{If}\underset{x\to 0}{lim}\left(\frac{a{e}^x-b}{x}\right)=2,\text{then}& \text{(R)}& a=1,b=-1\\ \text{(D)}& \text{If}\underset{x\to \infty }{lim}\{\sqrt{(x^2-x+1)}-ax-b\}=0,\text{then}& \text{(S)}& a=2,b=2\end{array}$

Which of the following is the only CORRECT combination?

• A. $\left(A\right)\to \left(Q\right),\left(D\right)\to \left(Q\right)$
• B. $\left(A\right)\to \left(R\right),\left(D\right)\to \left(S\right)$
• C. $\left(A\right)\to \left(R\right),\left(D\right)\to \left(Q\right)$
• D. $\left(A\right)\to \left(S\right),\left(D\right)\to \left(R\right)$

Answer 1

The correct option is C $\left(A\right)\to \left(R\right),\left(D\right)\to \left(Q\right)$

$\underset{x\to \infty }{lim}(\frac{x^2+1}{x+1}-ax-b)=0\Rightarrow \underset{x\to \infty }{lim}\left(\frac{1+x^2-a{x}^2-bx-ax-b}{x+1}\right)=0\Rightarrow \underset{x\to \infty }{lim}\left(\frac{(1-a)x^2-(a+b)x+1-b}{x+1}\right)=0$
For existence of limit,
$1-a=0\Rightarrow a=1-(a+b)=0\Rightarrow b=-1$

$\underset{x\to \infty }{lim}\{\sqrt{(x^2-x+1)}-ax-b\}=0\Rightarrow \underset{x\to \infty }{lim}\{\sqrt{(x^2-x+1)}-ax-b\}\cdot \frac{\sqrt{(x^2-x+1)}+ax+b}{\sqrt{(x^2-x+1)}+ax+b}=0\Rightarrow \underset{x\to \infty }{lim}\frac{x^2-x+1-(ax+b{)}^2}{\sqrt{(x^2-x+1)}+ax+b}=0\Rightarrow \underset{x\to \infty }{lim}\frac{x^2(1-a^2)-x(2ab+1)+(1-b^2)}{x(\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}+a+\frac{b}{x})}=0\Rightarrow \underset{x\to \infty }{lim}\frac{x(1-a^2)-(2ab+1)+\frac{1-b^2}{x}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}+a+\frac{b}{x}}=0$

For existence of limit,
$1-a^2=0\Rightarrow a=\pm 1$
But at $a=-1,$ denominator becomes $0$
So, $a=1$
$\frac{-2b-1}{2}=0\Rightarrow b=-\frac{1}{2}$