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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]

A=∣001∣∣(2−c)(b−c)c∣∣4−c2b2−c2c2∣=(2−c)(b−c)∣001∣∣11c∣∣2+cb+cc2∣=(2−c)(b−c)(b+c−2−c)=(2−c)(b−c)(b−2)given2,b,cinA.p2b=2+c→soc=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)3given16⩾−2(2−b)3⩾2=8⩾−(2−b)3⩾1=8⩾(b−2)3⩾1=2⩾b−2⩾1=4⩾b⩾3c=2b−2nos8⩾2b⩾68−2⩾2b−2⩾6−26⩾c⩾4c∈[4,6]

Commented by rahul 19 last updated on 09/Apr/19

thank you sir!

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