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Q. If the minimum value of $f ( x )= x ^2-2 kx +4 k$ is 4 for $x \in[1,4]$, then the value of $k$ is

Complex Numbers and Quadratic Equations

Solution:

$f(x)= x^2-2 k x+4 k $
$\text { vertex } x_0=k $
image
$\text { Case-I : } k \leq 1 $
$ f(1)=4 $
$1-2 k+4 k=4 $
$2 k=3$
$k=\frac{3}{2} \text { (reject) } $
$k \geq 4 $
image
$\text { Case-II: } f(4)=4 $
$ 16-8 k+4 k=4 $
$ 4 k=12$
$k = 3$ Reject
image
$\text { Case-III: } 1< k< 4 $
$ f(k)=4 $
$ k^2-2 k^2+4 k=4 $
$ k^2-4 k+4=0 $
$(k-2)^2=0 \Rightarrow k=2$