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

Answer 1

The correct option is A $\left(B\right)\to \left(P\right),\left(C\right)\to \left(S\right)$

$\underset{x\to 0}{lim}(1+ax+b{x}^2{)}^{2/x}=e^3$ $(1^{\infty }$ form$)$
$\Rightarrow e^{\underset{x\to 0}{lim}\left(\frac{2(ax+b{x}^2)}{x}\right)}=e^3\Rightarrow e^{\underset{x\to 0}{lim}(2a+2bx)}=e^3$
For existence of limit,
$a=\frac{3}{2}$ and $b\in R$

$\underset{x\to 0}{lim}\left(\frac{a{e}^x-b}{x}\right)=2$
Apply expansion of $e^x$
$\underset{x\to 0}{lim}\left(\frac{a(1+\frac{x}{1!}+\frac{x^2}{2!}+\dots )-b}{x}\right)=2$
$\Rightarrow \underset{x\to 0}{lim}\left(\frac{(a-b)+a(\frac{x}{1!}+\frac{x^2}{2!}+\dots )}{x}\right)=2$
For existence of limit,
$a-b=0\Rightarrow a=b$
$\underset{x\to 0}{lim}\left(a\right)=2$
$\Rightarrow a=2=b$