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This is a question of CBSE Sample Paper - Class 12 - 2017/18.

You can download the question paper hereΒ  https://www.teachoo.com/cbse/sample-papers/


If Delta = -4, find value of | a^3 - 1 0 a - a^4 0 a-a^4 a^3 - 1

Question 13 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
Question 13 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3
Question 13 - CBSE Class 12 Sample Paper for 2018 Boards - Part 4


Transcript

Question 13 If βˆ† = |β– 8(1&π‘Ž&π‘Ž2@π‘Ž&π‘Ž2&1@π‘Ž2&1&π‘Ž)| = βˆ’4 Then find the value of |β– 8(π‘Ž3βˆ’1&0&π‘Žβˆ’π‘Ž4@0&π‘Žβˆ’π‘Ž4&π‘Ž3βˆ’1@π‘Žβˆ’π‘Ž4&π‘Ž3 βˆ’1&0)| Given |β– 8(1&π‘Ž&π‘Ž2@π‘Ž&π‘Ž2&1@π‘Ž2&1&π‘Ž)| = βˆ’4 1 (a2 Γ— a – 1 Γ— 1) – a(a Γ— a – a2 Γ— 1) + a2 (a Γ— 1 – a2 Γ— a2) = – 4 1 (a3 – 1) – a(a2 – a2) + a2 (a – a4) = – 4 (a3 – 1) – a(0) + a2 (a – a4) = – 4 (a3 – 1) + a2 (a – a4) = – 4 (a3 – 1) + a2 Γ— a(1 – a3) = – 4 (a3 – 1) + a3 (1 – a3) = – 4 (a3 – 1) – a3 (a3 – 1) = – 4 (a3 – 1)(1 – a3) = – 4 – (a3 – 1) (a3 – 1) = – 4 (a3 – 1) (a3 – 1) = 4 (a3 – 1)2 = 4 Now, we need to find |β– 8(π‘Ž3βˆ’1&0&π‘Žβˆ’π‘Ž4@0&π‘Žβˆ’π‘Ž4&π‘Ž3βˆ’1@π‘Žβˆ’π‘Ž4&π‘Ž3 βˆ’1&0)| = (π‘Ž^3βˆ’1)[(π‘Žβˆ’π‘Ž^4 )Γ—0βˆ’(π‘Ž^3βˆ’1)(π‘Ž^3βˆ’1)] – 0[0Γ—0βˆ’(π‘Žβˆ’π‘Ž^4 )(π‘Ž^3βˆ’1)] + (π‘Žβˆ’π‘Ž^4 )[0Γ—(π‘Ž^3βˆ’1)βˆ’(π‘Žβˆ’π‘Ž^4 )(π‘Žβˆ’π‘Ž^4 )] = (π‘Ž^3βˆ’1)[0βˆ’(π‘Ž^3βˆ’1)(π‘Ž^3βˆ’1)] – 0 + (π‘Žβˆ’π‘Ž^4 )[0βˆ’(π‘Žβˆ’π‘Ž^4 )(π‘Žβˆ’π‘Ž^4 )] = βˆ’(π‘Ž^3βˆ’1)^3βˆ’(π‘Žβˆ’π‘Ž^4 )^3 = βˆ’(π‘Ž^3βˆ’1)^3βˆ’γ€–(π‘Ž(1βˆ’π‘Ž^3)) γ€—^3 = βˆ’(π‘Ž^3βˆ’1)^3βˆ’γ€–π‘Ž^3 (1βˆ’π‘Ž^3)γ€—^3 = βˆ’(π‘Ž^3βˆ’1)^3βˆ’γ€–π‘Ž^3 Γ—(βˆ’1)^3 (π‘Ž^3βˆ’1)γ€—^3 = βˆ’(π‘Ž^3βˆ’1)^3βˆ’γ€–π‘Ž^3 Γ—(βˆ’1)^3 (π‘Ž^3βˆ’1)γ€—^3 = βˆ’(π‘Ž^3βˆ’1)^3βˆ’γ€–π‘Ž^3 Γ—(βˆ’1) Γ— (π‘Ž^3βˆ’1)γ€—^3 = βˆ’(π‘Ž^3βˆ’1)^3+γ€–π‘Ž^3 (π‘Ž^3βˆ’1)γ€—^3 = (π‘Ž^3βˆ’1)^3 (βˆ’1+π‘Ž^3) = (π‘Ž^3βˆ’1)^3 (π‘Ž^3βˆ’1) = (π‘Ž^3βˆ’1)^4 = [(π‘Ž^3βˆ’1)^2 ]^2 Putting (a3 – 1)2 = 4 from (1) = [4]^2 = 16 = βˆ’(π‘Ž^3βˆ’1)^3βˆ’γ€–π‘Ž^3 Γ—(βˆ’1)^3 (π‘Ž^3βˆ’1)γ€—^3 = βˆ’(π‘Ž^3βˆ’1)^3βˆ’γ€–π‘Ž^3 Γ—(βˆ’1) Γ— (π‘Ž^3βˆ’1)γ€—^3 = βˆ’(π‘Ž^3βˆ’1)^3+γ€–π‘Ž^3 (π‘Ž^3βˆ’1)γ€—^3 = (π‘Ž^3βˆ’1)^3 (βˆ’1+π‘Ž^3) = (π‘Ž^3βˆ’1)^3 (π‘Ž^3βˆ’1) = (π‘Ž^3βˆ’1)^4 = [(π‘Ž^3βˆ’1)^2 ]^2 Putting (a3 – 1)2 = 4 from (1) = [4]^2 = 16

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.