CBSE Class 12 Sample Paper for 2018 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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

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Question 13 If β = |β 8(1&π&π[email protected]π&π2&[email protected]π2&1&π)| = β4 Then find the value of |β 8(π3β1&0&πβπ[email protected]&πβπ4&π3β[email protected]πβπ4&π3 β1&0)| Given |β 8(1&π&π[email protected]π&π2&[email protected]π2&1&π)| = β4 1 (a2 Γ a β 1 Γ 1) β a(a Γ a β a2 Γ 1) + a2 (a Γ 1 β a2 Γ a2) = β 4 1 (a3 β 1) β a(a2 β a2) + a2 (a β a4) = β 4 (a3 β 1) β a(0) + a2 (a β a4) = β 4 (a3 β 1) + a2 (a β a4) = β 4 (a3 β 1) + a2 Γ a(1 β a3) = β 4 (a3 β 1) + a3 (1 β a3) = β 4 (a3 β 1) β a3 (a3 β 1) = β 4 (a3 β 1)(1 β a3) = β 4 β (a3 β 1) (a3 β 1) = β 4 (a3 β 1) (a3 β 1) = 4 (a3 β 1)2 = 4 Now, we need to find |β 8(π3β1&0&πβπ[email protected]&πβπ4&π3β[email protected]πβπ4&π3 β1&0)| = (π^3β1)[(πβπ^4 )Γ0β(π^3β1)(π^3β1)] β 0[0Γ0β(πβπ^4 )(π^3β1)] + (πβπ^4 )[0Γ(π^3β1)β(πβπ^4 )(πβπ^4 )] = (π^3β1)[0β(π^3β1)(π^3β1)] β 0 + (πβπ^4 )[0β(πβπ^4 )(πβπ^4 )] = β(π^3β1)^3β(πβπ^4 )^3 = β(π^3β1)^3βγ(π(1βπ^3)) γ^3 = β(π^3β1)^3βγπ^3 (1βπ^3)γ^3 = β(π^3β1)^3βγπ^3 Γ(β1)^3 (π^3β1)γ^3 = β(π^3β1)^3βγπ^3 Γ(β1)^3 (π^3β1)γ^3 = β(π^3β1)^3βγπ^3 Γ(β1) Γ (π^3β1)γ^3 = β(π^3β1)^3+γπ^3 (π^3β1)γ^3 = (π^3β1)^3 (β1+π^3) = (π^3β1)^3 (π^3β1) = (π^3β1)^4 = [(π^3β1)^2 ]^2 Putting (a3 β 1)2 = 4 from (1) = [4]^2 = 16 = β(π^3β1)^3βγπ^3 Γ(β1)^3 (π^3β1)γ^3 = β(π^3β1)^3βγπ^3 Γ(β1) Γ (π^3β1)γ^3 = β(π^3β1)^3+γπ^3 (π^3β1)γ^3 = (π^3β1)^3 (β1+π^3) = (π^3β1)^3 (π^3β1) = (π^3β1)^4 = [(π^3β1)^2 ]^2 Putting (a3 β 1)2 = 4 from (1) = [4]^2 = 16