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Question

If the third term in the binomial expansion of (1+xlog2x)5 equals 2560, the a possible value of x is :

A
18
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B
14
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C
22
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D
42
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Solution

The correct option is B 14
T3= 5C2(xlog2x)2=2560
5!3!2!x2log2x=2560
(xlog2x)2=256
xlog2x=16
Taking log2 both sides
log2xlog2x=4log22
(log2x)2=4
log2x=±2
log2x=2, log2x=2
x=4, x=14

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