Question

For the equation ${\left({\log }_{2}x\right)}^2-4{\log }_{2}x-m^2-2m-13=0,m\in R$.
If the real roots are $x_1,x_2$ such that $x_1<x_2$, then the sum of maximum value of $x_1$ and minimum value of $x_2$ is

• A. $\frac{513}{4}$
• B. $\frac{257}{4}$
• C. $\frac{1025}{8}$
• D. $\frac{513}{8}$

Answer 1

The correct option is B $\frac{257}{4}$

${\left({\log }_{2}x\right)}^2-4\left({\log }_{2}x\right)-(m^2+2m+13)=0\cdots \left(i\right)$
It is quadratic equation in ${\log }_{2}x$,
$\Rightarrow {\log }_{2}x=\frac{4\pm \sqrt{16+4\left(1\right)(m^2+2m+13)}}{2}$

As $x_1,x_2$ are roots of the equation $\left(i\right)$,
${\log }_2{x}_1=2-\sqrt{m^2+2m+17}{\log }_2{x}_2=2+\sqrt{m^2+2m+17}$

${\log }_2{x}_2$ is minimum when $x_2$ is minimum,
${\log }_2{x}_2=2+\sqrt{m^2+2m+17}\Rightarrow {\log }_2{x}_2=2+\sqrt{(m+1{)}^2+16}$
Minimum when $m=-1$
$({\log }_2{x}_2{)}_{min.}=2+\sqrt{16}=6$
$\Rightarrow (x_2{)}_{min.}=2^6=64$

Similarly,
The maximum value of
${\log }_2{x}_1=2-\sqrt{m^2+2m+17}\Rightarrow {\log }_2{x}_1=2-\sqrt{16}=-2$
$\Rightarrow x_1=2^{-2}=\frac{1}{4}$

Hence, the required sum
$=64+\frac{1}{4}=\frac{257}{4}$