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Question

If ∣ ∣ ∣x2+xx+1x22x2+3x13x3x3x2+2x+32x12x1∣ ∣ ∣=Ax12, then the value of A is

A
12
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B
24
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C
12
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D
24
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Solution

The correct option is B 24
Applying R3R3R1R2R22R1

∣ ∣x2+xx+1x2x1x2x+1x+3x2x+1∣ ∣

Applying R3R3R2

∣ ∣x2+xx+1x2x1x2x+1400∣ ∣

=4[(x+1)2(x2)2]

=4[2x+4x+14]

=24x12

Ax12=24x12

A=24

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