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

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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

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Transcript

Question 13 If βˆ† = |β– 8(1&π‘Ž&π‘Ž[email protected]π‘Ž&π‘Ž2&[email protected]π‘Ž2&1&π‘Ž)| = βˆ’4 Then find the value of |β– 8(π‘Ž3βˆ’1&0&π‘Žβˆ’π‘Ž[email protected]&π‘Žβˆ’π‘Ž4&π‘Ž3βˆ’[email protected]π‘Žβˆ’π‘Ž4&π‘Ž3 βˆ’1&0)| Given |β– 8(1&π‘Ž&π‘Ž[email protected]π‘Ž&π‘Ž2&[email protected]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‘Ž[email protected]&π‘Žβˆ’π‘Ž4&π‘Ž3βˆ’[email protected]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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.