Binomial Theorem 1 Question 12

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

(a) 42

(b) 14

(c) 18

(d) 22

(2019 Main, 10 Jan I)

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Answer:

Correct Answer: 13. (b)

Solution:

  1. The (r+1) th term in the expansion of (a+x)n is given by Tr+1=nCranrxr

3rd  term in the expansion of (1+xlog2x)5 is

5C2(1)52(xlog2x)2

5C2(1)52(xlog2x)2=2560 (given)

10(xlog2x)2=2560

x(2log2x)=256

log2x2log2x=log2256

2(log2x)(log2x)=8( taking log2 on both sides (log2x)2=4(log2256=log228=8)log2x=±2log2x=2 or log2x=2x=4 or x=22=14



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