Question

If $e^{\underset{x\to 0}{lim}\frac{a{x}^2+bx+c}{c{x}^2+ax+b}}=\underset{x\to 0}{lim}(b{x}^2+ax+1{)}^{\frac{1}{{\log }_e(a{x}^2+cx+1)}},$ then which among the following option(s) is/are correct

• A. $a,c,b$ are in G.P
• B. $c^2<4a$ and $a>0$
• C. $a,b,c$ are in G.P
• D. $c^2<4a$ and $a<0$

Answer 1

Answer: (A), (B)

$c^2<4a$ and $a>0$

$\text{L.H.S}=e^{\underset{x\to 0}{lim}\frac{a{x}^2+bx+c}{c{x}^2+ax+b}}=e^{\frac{c}{b}}$

$\text{R.H.S}=\underset{x\to 0}{lim}(1+b{x}^2+ax{)}^{\frac{1}{{\log }_e(a{x}^2+cx+1)}}$
$=e^{\underset{x\to 0}{lim}\frac{b{x}^2+ax}{{\log }_e(1+a{x}^2+cx)}}$
$=e^{\underset{x\to 0}{lim}\frac{x(bx+a)}{x(ax+c)}\cdot \frac{x(ax+c)}{{\log }_e(1+x(ax+c\left)\right)}}$
$=e^{\frac{a}{c}}[\because \underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1]$

$\therefore e^{\frac{c}{b}}=e^{\frac{a}{c}}\Rightarrow \frac{c}{b}=\frac{a}{c}\Rightarrow c^2=ab$

For ${\log }_e(a{x}^2+cx+1)$ to be defined
$a{x}^2+cx+1>0$
$\Rightarrow \Delta <0$ and $a>0$
$\Rightarrow c^2-4a<0$ and $a>0$