If a2+b2+c2=−2 and f(x)=∣∣
∣
∣∣1+a2x(1+b2)x(1+c2)x(1+a2)c1+b2x(1+c2)x(1+a2)x(1+b2)x1+c2x∣∣
∣
∣∣, then f(x) is a polynomial of degree.
A
3
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B
2
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C
1
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D
0
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Solution
The correct option is D2 Applying C1→C1+C2+C3, we get f(x)=∣∣
∣
∣∣1(1+b2)x(1+c2)x1(1+b2x)(1+c2)x1(1+b2)x1+c2x∣∣
∣
∣∣, (a2+b2+c2+2=0) Again, applying R2→R2−R1,R3→R3−R1, we get =∣∣
∣
∣∣1(1+b2)x(1+c2)x01−x0001−x∣∣
∣
∣∣=(1−x)2 Hence, degree of f(x)=2.