Question

For the given quadratic expression $f\left(x\right)=x^2-2x+3.$ If $M,m$ are the maximum and minimum values of the $f\left(x\right)$ in $[-1,2]$. Then the value of $M+4m$ is

• A. $-7$
• B. $10$
• C. $7$
• D. $-10$

Answer 1

The correct option is B $10$

Given: $f\left(x\right)=x^2-2x+3$

On comparing with standard quadratic equation $y=a{x}^2+bx+c$ we get, $a=1,b=-2,c=3.$

Here $a>0$ which means the minimum value occure at $x=\frac{-b}{2a}.$
$x=\frac{-(-2)}{2.1}$
$\Rightarrow x=1$

$f\left(1\right)=m=(1{)}^2-2(1)+3=2$

Now we will find the value of $D$
$D=b^2-4ac=(-2{)}^2-\mathrm{4.1.3}$
$\Rightarrow D=-8$

$D<0$, this means the quadratic equation does not have any root, and the graph will look like this

Maximum value of $f\left(x\right)$ occurs at either of the end points of the interval $[-1,2].$
$\therefore f(-1)=(-1{)}^2-2(-1)+3=6$
and $f\left(2\right)=(2{)}^2-2(2)+3=3$

$\therefore$ maximum value of $f\left(x\right)$ occurs at $x=-1\Rightarrow M=6$

$\therefore M+4m=6+4\left(1\right)=10$