Number of integral values of $x$ satisfying the given inequality $| 3 x - 2 | \leq \frac{1}{2}$ are?

Open in App

Solution

Given: $| 3 x - 2 | \leq \frac{1}{2}$
$\Rightarrow - \frac{1}{2} \leq 3 x - 2 \leq \frac{1}{2}$
$\Rightarrow - \frac{1}{2} + 2 \leq 3 x \leq \frac{1}{2} + 2$
$\Rightarrow \frac{3}{2} \leq 3 x \leq \frac{5}{2}$
$\Rightarrow \frac{1}{2} \leq x \leq \frac{5}{6}$
$\Rightarrow x \in [\frac{1}{2}, \frac{5}{6}]$
$∴$ the number of integral values of $x$ are $0.$

Suggest Corrections

Similar questions

x

1 \leq | 5 x + 8 | \leq 5

x

(x^{2} - x - 1) (x^{2} - x - 7) < - 5,

x, x > - 10

- 5 (x - 1) + 3 > 3 x - 4

\frac{\sqrt{x - 1}}{x - 2} \leq 0

x

\frac{(2 - x^{2})^{3} (x - 3)^{5}}{(x + 1) (x^{2} - 3 x - 4)} \geq 0

Join BYJU'S Learning Program