Inequalities
4.0 Equations involving MOD functions
4.0 Equations involving MOD functions
\[\left| x \right| = \left\{ {\begin{array}{c}{ - x}&{{\text{when }}x \leqslant 0} \\ x&{{\text{when }}x \geqslant 0}\end{array}} \right.\]
Question 4. Solve for $x$:
(a) $\left| {x + 1} \right| \leqslant 4$
(b) ${x^2} + 5\left| x \right| + 4 = 0$
(c) $\left| {\frac{{5x - 1}}{{x + 1}}} \right| = 4$
(d) $\left| {\sin x} \right|$
Solution:
(a) $$\begin{equation} \begin{aligned} \left| {x + 1} \right| \leqslant 4 \\ - 4 \leqslant x + 1 \leqslant 4 \\ - 4 - 1 \leqslant x \leqslant 4 - 1 \\ - 5 \leqslant x \leqslant 3 \\\end{aligned} \end{equation} $$
(b) $${x^2} + 5\left| x \right| + 4 = 0$$
Case I: When $x \leqslant 0$
$$\begin{equation} \begin{aligned} {x^2} - 5x + 4 = 0 \\ {x^2} - 4x - x + 4 = 0 \\ x(x - 4) - 1(x - 4) = 0 \\ (x - 1)(x - 4) = 0 \\ x = 4{\text{ and }}x = 1 \\\end{aligned} \end{equation} $$
But initial condition was $x \leqslant 0$, so no value of $x$ is obtained using this case.
Case II: When $x \geqslant 0$
$$\begin{equation} \begin{aligned} {x^2} + 5x + 4 = 0 \\ {x^2} + 4x + x + 4 = 0 \\ x(x + 4) + 1(x + 4) = 0 \\ (x + 4)(x + 1) = 0 \\ x = - 4{\text{ and }}x = - 1 \\\end{aligned} \end{equation} $$
But initial condition was $x \geqslant 0$, so no value of $x$ is obtained using this case.
(c) $$\left| {\frac{{5x - 1}}{{x + 1}}} \right| = 4$$ $$\frac{{5x - 1}}{{x + 1}} = \pm 4$$
Case I: $$\begin{equation} \begin{aligned} \frac{{5x - 1}}{{x + 1}} = 4 \\ \frac{{5x - 1}}{{x + 1}} - 4 = 0 \\ \frac{{5x - 1 - 4x - 4}}{{x + 1}} = 0 \\ \frac{{(x - 5)(x + 1)}}{{{{(x + 1)}^2}}} = 0 \\ (x - 5)(x + 1) = 0 \\ x = 5{\text{ and }}x = - 1 \\\end{aligned} \end{equation} $$
Case II: $$\begin{equation} \begin{aligned} \frac{{5x - 1}}{{x + 1}} = - 4 \\ \frac{{5x - 1 + 4x + 4}}{{x + 1}} = 0 \\ \frac{{9x + 3}}{{x + 1}} = 0 \\ \frac{{(3x + 1)(x + 1)}}{{{{(x + 1)}^2}}} = 0 \\ (3x + 1)(x + 1) = 0 \\ x = - \frac{1}{3}{\text{ and }}x = - 1 \\\end{aligned} \end{equation} $$
(d) \[\left| {\sin x} \right| = \left\{ {\begin{array}{c}{ - \sin x}&{\sin x \leqslant 0 \Rightarrow x \in \left[ {\pi ,2\pi } \right]} \\ {\sin x}&{\sin x \geqslant 0 \Rightarrow x \in \left[ {0,\pi } \right]} \end{array}} \right.\]
