Question

If $m_1$ is minimum value of $x^2-6x+13$ and $m_2$ is maximum value of $-x^2-8x+4$, then $m_2-m_1=$

• A. $-16$
• B. $24$
• C. $-24$
• D. $16$

Answer 1

Answer: (D)

$16$

$\Rightarrow x^2-6x+13\to m_1=$ min value

$\Rightarrow -x^2-8x+4\to m_2=$ max value

$\Rightarrow x^2-6x+13=(x-3{)}^2-9+13=(x-3{)}^2+4\ge 4$

Therefore, $m_1=4$

and, $-x^2-8x+4=\left(\right(x+4{)}^2-16-4)=((x+4{)}^2-20)$

Max value $=20=m_2$

$\therefore m_2-m_1=20-4=16$