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Question

The value of ∣ ∣ ∣(a+1)(a+2)a+21(a+2)(a+3)a+31(a+3)(a+4)a+41∣ ∣ ∣ is :

A
2
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B
(a+1)(a+2)(a+3)
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C
0
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D
(a+2)(a+3)(a+4)
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Solution

The correct option is A 2
Δ=∣ ∣ ∣(a+1)(a+2)a+21(a+2)(a+3)a+31(a+3)(a+4)a+41∣ ∣ ∣

Applying C1C1C2,
Δ=∣ ∣ ∣(a+2)aa+21(a+3)(a+1)a+31(a+4)(a+2)a+41∣ ∣ ∣

Applying C2C2C3,
Δ=∣ ∣ ∣(a+2)aa+11(a+3)(a+1)a+21(a+4)(a+2)a+31∣ ∣ ∣

Applying R2R2R1 and R3R3R1,
Δ=∣ ∣a2+2aa+112a+3104a+820∣ ∣=68=2

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