Question

What is the solution set to the inequality

Answer 1

$(4x - 3)(2x - 1) \geqslant 0 \\ \Leftrightarrow \begin{cases}4x - 3 \geqslant 0 \\ 2x - 1 \geqslant 0\end{cases}\: \vee \:\begin{cases}4x - 3 \leqslant 0 \\2x - 1 \leqslant 0 \end{cases} \\ \Leftrightarrow \begin{cases}x \geqslant \frac{3}{4} \\ x \geqslant \frac{1}{2} \end{cases}\: \vee \:\begin{cases}x\leqslant \frac{3}{4} \\x \leqslant \frac{1}{2} \end{cases} \\ \Leftrightarrow x \geqslant \frac{3}{4} \: \vee \: x \leqslant \frac{1}{2} \\ \Rightarrow The\: third\: option$

Answer 2

Answer:

Option C.

Step-by-step explanation:

The given inequality is

$\left(4x-3\right)\left(2x-1\right)\geq 0$

We know that if $ab\geq 0$, then either both a and b are positive or both are negative.

Case 1: If both factors are positive.

$\left(4x-3\right)\geq 0\Rightarrow x\geq \frac{3}{4}$

$\left(2x-1\right)\geq 0\Rightarrow x\geq \frac{1}{2}$

Using these two equation we get

$x\geq \frac{3}{4}$

Case 2: If both factors are negative.

$\left(4x-3\right)\leq 0\Rightarrow x\leq \frac{3}{4}$

$\left(2x-1\right)\leq 0\Rightarrow x\leq \frac{1}{2}$

Using these two equation we get

$x\leq \frac{1}{2}$

From case (1) and (2) we get the solution set for the given inequality.

$\{x|x\leq \frac{1}{2}\text{ or }x\geq \frac{3}{4}\}$

Therefore, the correct option is C.