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Question

The possible values of expression 1x2x3 is


A

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B

U (0,)

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C

U [0,)

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D

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Solution

The correct option is B

U (0,)


Take the denominator x2 - x - 3

x2 - x - 3 = x2 - 2(12).x + (12)2 - (12)2 - 3

= (x12)2 - 14 - 3

= (x12)2 - 134

Now, we know that (x12)2 ≥ 0

(x12)2 - 134134, ∀ x ∈ R

(or)

> (x12)2 - 134134
1x2x3 is in (,413] U (0,)


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