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Ex 4.2
Ex 4.2, 2 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 3 Deleted for CBSE Board 2023 Exams
Ex 4.2, 4 Deleted for CBSE Board 2023 Exams
Ex 4.2, 5 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 6 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 7 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 8 (i) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 8 (ii) Deleted for CBSE Board 2023 Exams
Ex 4.2, 9 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 10 (i) Deleted for CBSE Board 2023 Exams
Ex 4.2, 10 (ii) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 11 (i) Deleted for CBSE Board 2023 Exams
Ex 4.2, 11 (ii) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 12 Important Deleted for CBSE Board 2023 Exams You are here
Ex 4.2, 13 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 14 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 16 (MCQ) Deleted for CBSE Board 2023 Exams
Last updated at March 16, 2023 by Teachoo
Ex 4.2, 12 By using properties of determinants, show that: |■8(1&x&[email protected]&1&[email protected]&x2&1)| = (1 – x3)2 Solving L.H.S |■8(1&x&[email protected]&1&[email protected]&x2&1)| Applying R1 → R1 + R2 + R3 = |■8(𝟏+𝐱𝟐+𝐱&𝐱+𝟏+𝐱𝟐&𝐱𝟐+𝐱+𝟏@x2&1&[email protected]&x2&1)| Taking (1 + x + x2) Common from 1st row = (𝟏+𝐱+𝐱𝟐) |■8(1&1&[email protected]&1&[email protected]&x2&1)| Applying C1 → C1 − C2 = (1+x+x2) |■8(𝟏−𝟏&1&[email protected]−1&1&[email protected]−x2&x2&1)| = (1+x+x2) |■8(0&1&[email protected]−1&1&[email protected](1−x)&x2&1)| = (1+x+x2) |■8(𝟎&1&[email protected](𝐱−𝟏)(𝑥+1)&1&[email protected]−x (𝐱−𝟏)&x2&1)| Taking (x – 1) common from C1 = (x – 1) (1+x+x2) |■8(0&1&[email protected](x+1)&1&[email protected]−x&x2&1)| Applying C2 → C2 − C3 = (x – 1) (1+x+x2) |■8(0&𝟏−𝟏&[email protected]+1&1−x&[email protected]−x&x2−1&1)| = (x – 1) (1+x+x2) |■8(0&𝟎&[email protected]+1&−(𝒙−𝟏)&[email protected]−x&(x+1)(𝐱−𝟏)&1)| Taking (x – 1) common from 2nd Column = (x – 1) (1+x+x2) (x – 1) |■8(0&0&[email protected]+1&−1&[email protected]−x&𝑥+1&1)| = (x – 1)2 (1+x+x2) |■8(0&0&[email protected]+1&−1&[email protected]−x&𝑥+1&1)| Expanding Determinant along R1 = (x – 1)2 (1+x+x2) ( 0|■8(−1&𝑥@𝑥+1&1)|−0|■8(𝑥+1&[email protected]−𝑥&1)|+1|■8(𝑥+1&−[email protected]−x&𝑥+1)|) = (x – 1)2 (1+x+x2) ( 0−0+1|■8(𝑥+1&−[email protected]−x&𝑥+1)|) = (x – 1)2 (1+x+x2) (0−0+1((𝑥+1)2−𝑥)) = (x – 1)2 (1+x+x2) ((𝑥+1)^2−𝑥) = (x – 1)2 (1 + x + x2) ((x2 + 1 + 2x) – x) = (x – 1)2 (1 + x + x2) (1 + x + x2) = (x – 1)2 (1 + x + x2)2 = ((x – 1) (1 + x + x2))2 = (– (1 – x) (1 + x + x2))2 = ((1 – x) (1 + x + x2))2 = (13 – x3)2 = (1 – x3)2 = R.H.S Hence proved = (x – 1)2 (1 + x + x2)2 = ((x – 1) (1 + x + x2))2 = (– (1 – x) (1 + x + x2))2 = ((1 – x) (1 + x + x2))2 = (13 – x3)2 = (1 – x3)2 = R.H.S Hence proved We know that a3 – b3 = (a – b)(a2 + b2 + ab) Here, a = 1 , b = x