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Question

If α1,α2,α3andα4are the roots of the equationx4+(2-3)x2+(2+3)=0, then the value of (1-α1)(1-α2)(1-α3)(1-α4) is


A

1

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B

4

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C

2+3

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D

5

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E

0

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Solution

The correct option is D

5


Find the value of(1-α1)(1-α2)(1-α3)(1-α4):

Given that α1,α2,α3andα4are the roots of the equationx4+(2-3)x2+(2+3)=0

As α1,α2,α3andα4 are the roots of the equationx4+(2-3)x2+(2+3)=0

x4+(2-3)x2+(2+3)=(x-α1)(x-α2)(x-α3)(x-α4)

putx=1in the above equation.

we get,

(1)4+(2-3)(1)2+(2+3)=(1-α1)(1-α2)(1-α3)(1-α4)(1-α1)(1-α2)(1-α3)(1-α4)=1+(2-3)+(2+3)(1-α1)(1-α2)(1-α3)(1-α4)=5

Hence, the option (D) is the correct option.


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