What is the solution set to the inequality
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[tex](4x - 3)(2x - 1) \geqslant 0 \\ \Leftrightarrow \begin{cases}4x - 3 \geqslant 0 \\ 2x - 1 \geqslant 0\end{cases}\: \vee \:\begin{cases}4x - 3 \leqslant 0 \\2x - 1 \leqslant 0 \end{cases} \\ \Leftrightarrow \begin{cases}x \geqslant \frac{3}{4} \\ x \geqslant \frac{1}{2} \end{cases}\: \vee \:\begin{cases}x\leqslant \frac{3}{4} \\x \leqslant \frac{1}{2} \end{cases} \\ \Leftrightarrow x \geqslant \frac{3}{4} \: \vee \: x \leqslant \frac{1}{2} \\ \Rightarrow The\: third\: option[/tex]
Answer:
Option C.
Step-by-step explanation:
The given inequality is
[tex]\left(4x-3\right)\left(2x-1\right)\geq 0[/tex]
We know that if [tex]ab\geq 0[/tex], then either both a and b are positive or both are negative.
Case 1: If both factors are positive.
[tex]\left(4x-3\right)\geq 0\Rightarrow x\geq \frac{3}{4}[/tex]
[tex]\left(2x-1\right)\geq 0\Rightarrow x\geq \frac{1}{2}[/tex]
Using these two equation we get
[tex]x\geq \frac{3}{4}[/tex]
Case 2: If both factors are negative.
[tex]\left(4x-3\right)\leq 0\Rightarrow x\leq \frac{3}{4}[/tex]
[tex]\left(2x-1\right)\leq 0\Rightarrow x\leq \frac{1}{2}[/tex]
Using these two equation we get
[tex]x\leq \frac{1}{2}[/tex]
From case (1) and (2) we get the solution set for the given inequality.
[tex]\{x|x\leq \frac{1}{2}\text{ or }x\geq \frac{3}{4}\}[/tex]
Therefore, the correct option is C.