First slide

Algebra of real valued functions

Question

If $|x^{2} - 2 x + 2| - |2 x^{2} - 5 x + 2| = |x^{2} - 3 x|$ , then the set of values of x is

Moderate

A

$(- \infty, 0] \cup [3, \infty)$
B

$[0, \frac{1}{2}] \cup [2, 3]$
C

$(- \infty, 0] \cup [\frac{1}{2}, 2] \cup [3, \infty)$
D

$[0, 2] \cup [3, \infty)$

Solution

We have
$|2 x^{2} - 5 x + 2| + |x^{2} - 3 x| = |x^{2} - 2 x + 2|$
$= |(2 x^{2} - 5 x + 2) - (x^{2} - 3 x)|$
$\begin{aligned} ∴ (2 x^{2} - 5 x + 2) (x^{2} - 3 x) \leq 0 \\ \Rightarrow (2 x - 1) (x - 2) x (x - 3) \leq 0 \end{aligned}$
From the sign scheme, we have $x \in [0, \frac{1}{2}] \cup [2, 3]$

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