Question Number 57633 by rahul 19 last updated on 09/Apr/19
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
![A=∣0 0 1∣ ∣(2−c) (b−c) c∣ ∣4−c^2 b^2 −c^2 c^2 ∣ =(2−c)(b−c)∣0 0 1∣ ∣1 1 c∣ ∣2+c b+c c^2 ∣ =(2−c)(b−c)(b+c−2−c) =(2−c)(b−c)(b−2) given 2,b,c in A.p 2b=2+c→ so c=2b−2 (2−2b+2)(b−2b+2)(b−2) (4−2b)(2−b)(b−2) 2(2−b)^2 ×(b−2) −2(2−b)^3 given 16 ≥−2(2−b)^3 ≥2 = 8≥−(2−b)^3 ≥1 =8≥(b−2)^3 ≥1 =2≥b−2≥1 =4≥b≥3 c=2b−2 nos 8≥2b≥6 8−2≥2b−2≥6−2 6≥c≥4 c∈[4,6]](https://www.tinkutara.com/question/Q57639.png)
$${A}=\mid\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{1}\mid \\ $$$$\:\:\:\:\:\:\:\:\mid\left(\mathrm{2}−{c}\right)\:\:\left({b}−{c}\right)\:\:\:\:{c}\mid \\ $$$$\:\:\:\:\:\:\:\:\mid\mathrm{4}−{c}^{\mathrm{2}} \:\:\:\:\:\:{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \:\:\:\:{c}^{\mathrm{2}} \mid \\ $$$$=\left(\mathrm{2}−{c}\right)\left({b}−{c}\right)\mid\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:{c}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{2}+{c}\:\:\:\:{b}+{c}\:\:\:\:\:{c}^{\mathrm{2}} \mid \\ $$$$=\left(\mathrm{2}−{c}\right)\left({b}−{c}\right)\left({b}+{c}−\mathrm{2}−{c}\right) \\ $$$$=\left(\mathrm{2}−{c}\right)\left({b}−{c}\right)\left({b}−\mathrm{2}\right) \\ $$$${given}\:\mathrm{2},{b},{c}\:{in}\:{A}.{p}\:\:\mathrm{2}{b}=\mathrm{2}+{c}\rightarrow\:\:{so}\:{c}=\mathrm{2}{b}−\mathrm{2} \\ $$$$\left(\mathrm{2}−\mathrm{2}{b}+\mathrm{2}\right)\left({b}−\mathrm{2}{b}+\mathrm{2}\right)\left({b}−\mathrm{2}\right) \\ $$$$\left(\mathrm{4}−\mathrm{2}{b}\right)\left(\mathrm{2}−{b}\right)\left({b}−\mathrm{2}\right) \\ $$$$\mathrm{2}\left(\mathrm{2}−{b}\right)^{\mathrm{2}} ×\left({b}−\mathrm{2}\right) \\ $$$$−\mathrm{2}\left(\mathrm{2}−{b}\right)^{\mathrm{3}} \\ $$$${given}\:\:\:\:\:\:\mathrm{16}\:\geqslant−\mathrm{2}\left(\mathrm{2}−{b}\right)^{\mathrm{3}} \geqslant\mathrm{2} \\ $$$$=\:\:\:\mathrm{8}\geqslant−\left(\mathrm{2}−{b}\right)^{\mathrm{3}} \geqslant\mathrm{1} \\ $$$$=\mathrm{8}\geqslant\left({b}−\mathrm{2}\right)^{\mathrm{3}} \geqslant\mathrm{1} \\ $$$$=\mathrm{2}\geqslant{b}−\mathrm{2}\geqslant\mathrm{1} \\ $$$$=\mathrm{4}\geqslant{b}\geqslant\mathrm{3} \\ $$$$\:\:{c}=\mathrm{2}{b}−\mathrm{2} \\ $$$${nos}\:\:\:\mathrm{8}\geqslant\mathrm{2}{b}\geqslant\mathrm{6} \\ $$$$\mathrm{8}−\mathrm{2}\geqslant\mathrm{2}{b}−\mathrm{2}\geqslant\mathrm{6}−\mathrm{2} \\ $$$$\mathrm{6}\geqslant{c}\geqslant\mathrm{4} \\ $$$${c}\in\left[\mathrm{4},\mathrm{6}\right] \\ $$$$ \\ $$
Commented by rahul 19 last updated on 09/Apr/19
thank you sir!