# Question 10 (MCQ) - Miscellaneous - Chapter 4 Class 12 Determinants

Last updated at April 16, 2024 by Teachoo

Miscellaneous

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6

Misc 7 Important

Misc 8 (MCQ)

Misc 9 (MCQ) Important

Matrices and Determinants - Formula Sheet and Summary Important

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Chapter 4 Class 12 Determinants

Serial order wise

Last updated at April 16, 2024 by Teachoo

Misc 17 (Method 1) Choose the correct answer. If a, b, c, are in A.P., then the determinant |■8(𝑥+2&𝑥+3&𝑥+2𝑎@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| is A. 0 B. 1 C. x D. 2x Since a, b & c are in A.P Then, b – a = c – b b – a – c + b = 0 2b – a – c = 0 (Common difference is equal) …(1) Solving |■8(x+2&x+3&x+2a@x+3&x+4&x+2b@x+4&x+5&x+2c)| Multiplying and dividing 2 = 2/2 |■8(x+2&x+3&x+2a@x+3&x+4&x+2b@x+4&x+5&x+2c)| Multiplying R2 by 2 = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@𝟐(𝑥+3)&𝟐(𝑥+4)&𝟐(𝑥+2𝑏)@𝑥+4&𝑥+5&𝑥+2𝑐)| = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@2𝑥+6&2𝑥+8&2𝑥+4𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| Applying R2 → R2 – R1 – R3 = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@2𝑥+6−(𝑥+2)−(𝑥+4 )&2𝑥+8−(𝑥+3)−(𝑥+5)&2𝑥+4𝑏−(𝑥+2𝑎)−(𝑥+2𝑐)@𝑥+4&𝑥+5&𝑥+2𝑐)| = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@2𝑥+6−𝑥−2−𝑥−4&2𝑥+8−𝑥−3−𝑥−5&2𝑥+4𝑏−𝑥−2𝑎−𝑥−2𝑐@𝑥+4&𝑥+5&𝑥+2𝑐)| = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@0&0&4𝑏−2𝑎−2𝑐@𝑥+4&𝑥+5&𝑥+2𝑐)| = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@0&0&2(𝟐𝒃−𝒂−𝒄)@𝑥+4&𝑥+5&𝑥+2𝑐)| = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@0&0&2(𝟎)@𝑥+4&𝑥+5&𝑥+2𝑐)| (From (1): 2b – b – c = 0) = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@0&0&0@𝑥+4&𝑥+5&𝑥+2𝑐)| If any row or column of determinant are zero, then value of determinant is also zero. = 1/2 × 0 = 0 Thus, the value of determinant is 0 Correct Answer is A Misc 17 (Method 2) Choose the correct answer. If a, b, c, are in A.P., then the determinant |■8(x+2&x+3&x+2a@x+3&x+4&x+2b@x+4&x+5&x+2c)| is A. 0 B. 1 C. x D. 2x Since a, b & c are in A.P Then a – b = c – b b + b = c + a 2b = a + c (Common difference is equal) …(1) Consider |■8(𝑥+2&𝑥+3&𝑥+2𝑎@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| Applying R1 →R1 + R3 – 2R2 = |■8((𝑥+2)+(𝑥+4)−2(𝑥+3)&(𝑥+3)+(𝑥+5)−2(𝑥+4)&(𝑥+2𝑎)+(𝑥+2𝑐)−2(𝑥+2𝑏)@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| = |■8(𝑥+2+𝑥+4−2𝑥−6&𝑥+3+𝑥+5−2𝑥−8&𝑥+2𝑎+𝑥+2𝑐−2𝑥−4𝑏@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| = |■8(2𝑥−2𝑥+6−6&2𝑥−2𝑥+8−8&2𝑥−2𝑥+2𝑎+2𝑐−4𝑏@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| = |■8(0&0&0+2(𝒂+𝒄−2𝑏)@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| = |■8(0&0&2(𝟐𝒃−2𝑏)@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| = |■8(0&0&0@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| If any row or column of determinant are zero, then value of determinant is also zero. = 0 Hence, value of determinant is 0 Correct Answer is A (From (1): 2b = a + c)