Question

If $\underset{x\to 1}{lim}(1+ax+b{x}^2{)}^{\frac{c}{(x-1)}}=e^3$, then the value of $bc$ is ...........

• A. $6$
• B. $-1$
• C. $3$
• D. $-3$

Answer 1

The correct option is C $3$

$\underset{x\to 1}{lim}(1+ax+b{x}^2{)}^{\frac{c}{x-1}}=e^3$
or $e^{\underset{x\to 1}{lim}(1+ax+b{x}^2-1{)}^{\frac{c}{x-1}}}=e^3$
or $e^{\underset{x\to 1}{lim}\frac{c(ax+b{x}^2)}{x-1}}=e^3$
or $\underset{x\to 1}{lim}\frac{c(ax+b{x}^2)}{x-1}=3$
or $\underset{h\to 0}{lim}\frac{c\left(a\right(1+h)+b(1+h{)}^2)}{1+h-1}=3$
or $\underset{h\to 0}{lim}\frac{(ca+b)+(ac+2b)h+b{h}^2}{h}=3$
or $ca+b=0$ and $ac+2b=3$
or $b=3$ and $ac=-3$
Also, the form must be $1^{\infty }$ for which $a+b=0$, i.e., $a=-3$ and $c=1$
$\Rightarrow bc=3$
Hence, option 'C' is correct.