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

428 111 Complex Numbers and Quadratic Equations Report Error

A

$\frac{3}{2}$

B

2

C

3

D

4

Solution:

$f (x) = x^{2} - 2 k x + 4 k$
$vertex x_{0} = k$

$Case-I : k \leq 1$
$f (1) = 4$
$1 - 2 k + 4 k = 4$
$2 k = 3$
$k = \frac{3}{2} (reject)$
$k \geq 4$

$Case-II: f (4) = 4$
$16 - 8 k + 4 k = 4$
$4 k = 12$
$k = 3$ Reject

$Case-III: 1 < k < 4$
$f (k) = 4$
$k^{2} - 2 k^{2} + 4 k = 4$
$k^{2} - 4 k + 4 = 0$
$(k - 2)^{2} = 0 \Rightarrow k = 2$