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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 :
  1. 18
  2. 14
  3. 22
  4. 42


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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