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Question Number 112280 by bemath last updated on 07/Sep/20

If 15 ≤ ∣10−(1/3)a∣ < 20 find a ∈ R

$$\:\:\mathrm{If}\:\mathrm{15}\:\leqslant\:\mid\mathrm{10}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{a}\mid\:<\:\mathrm{20}\: \\ $$ $$\mathrm{find}\:\mathrm{a}\:\in\:\mathbb{R} \\ $$

Answered by john santu last updated on 07/Sep/20

⇒ ∣10−(1/3)a∣ ≥15 ∧ ∣10−(1/3)a∣ < 20 (1) ∣10−(1/3)a∣ ≥ 15 10−(1/3)a ≤ −15 ∪ 10−(1/3)a ≥15 −(1/3)a ≤ −25 ∪ −(1/3)a ≥ 5 a ≥ 75 ∪ a ≤−15 (2) ∣10−(1/3)a∣ < 20 −20 < 10−(1/3)a < 20 −30 < −(1/3)a < 10 −30<a<90 solution (1)∩(2) −30< a≤ −15 ∪ 75 ≤ a < 90

$$\Rightarrow\:\mid\mathrm{10}−\frac{\mathrm{1}}{\mathrm{3}}{a}\mid\:\geqslant\mathrm{15}\:\wedge\:\mid\mathrm{10}−\frac{\mathrm{1}}{\mathrm{3}}{a}\mid\:<\:\mathrm{20} \\ $$ $$\left(\mathrm{1}\right)\:\mid\mathrm{10}−\frac{\mathrm{1}}{\mathrm{3}}{a}\mid\:\geqslant\:\mathrm{15}\: \\ $$ $$\:\mathrm{10}−\frac{\mathrm{1}}{\mathrm{3}}{a}\:\leqslant\:−\mathrm{15}\:\cup\:\mathrm{10}−\frac{\mathrm{1}}{\mathrm{3}}{a}\:\geqslant\mathrm{15} \\ $$ $$−\frac{\mathrm{1}}{\mathrm{3}}{a}\:\leqslant\:−\mathrm{25}\:\cup\:−\frac{\mathrm{1}}{\mathrm{3}}{a}\:\geqslant\:\mathrm{5} \\ $$ $$\:\:{a}\:\geqslant\:\mathrm{75}\:\cup\:{a}\:\leqslant−\mathrm{15} \\ $$ $$\left(\mathrm{2}\right)\:\mid\mathrm{10}−\frac{\mathrm{1}}{\mathrm{3}}{a}\mid\:<\:\mathrm{20}\: \\ $$ $$\:−\mathrm{20}\:<\:\mathrm{10}−\frac{\mathrm{1}}{\mathrm{3}}{a}\:<\:\mathrm{20} \\ $$ $$\:−\mathrm{30}\:<\:−\frac{\mathrm{1}}{\mathrm{3}}{a}\:<\:\mathrm{10}\: \\ $$ $$\:\:−\mathrm{30}<{a}<\mathrm{90} \\ $$ $${solution}\:\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right) \\ $$ $$−\mathrm{30}<\:{a}\leqslant\:−\mathrm{15}\:\cup\:\mathrm{75}\:\leqslant\:{a}\:<\:\mathrm{90} \\ $$

Answered by ajfour last updated on 07/Sep/20

(a−30)^2 ≥45^2 & (a−30)^2 <60^2 ⇒ 45≤ a−30 < 60 or −60 < a−30 ≤ −45

$$\left({a}−\mathrm{30}\right)^{\mathrm{2}} \geqslant\mathrm{45}^{\mathrm{2}} \:\:\:\:\&\:\:\:\left({a}−\mathrm{30}\right)^{\mathrm{2}} <\mathrm{60}^{\mathrm{2}} \\ $$ $$\Rightarrow\:\:\:\:\:\:\:\:\mathrm{45}\leqslant\:{a}−\mathrm{30}\:<\:\mathrm{60}\:\:\: \\ $$ $${or}\:\:−\mathrm{60}\:<\:{a}−\mathrm{30}\:\leqslant\:−\mathrm{45} \\ $$

Commented bybemath last updated on 07/Sep/20

(a−30)^2 ≥ 45^2

$$\left(\mathrm{a}−\mathrm{30}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{45}^{\mathrm{2}} \: \\ $$

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