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Binomial Theorem question

2019 · 10 Jan · Shift 1 · Q33
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  5. /2019 · 10 Jan · Shift 1 · Q33

Binomial Theorem question

2019 · 10 Jan · Shift 1 · Q33

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If the third term in the binomial expansion of (1+xlog⁡2x)5{\left( {1 + {x^{{{\log }_2}x}}} \right)^5}(1+xlog2​x)5 equals 2560, then a possible value of x is -
  1. A
    222\sqrt 222​
  2. B
    424\sqrt 242​
  3. C
    18{1 \over 8}81​
  4. D
    14{1 \over 4}41​
View written solutionFree

Correct answer: D

  1. Write the general term of the expansion

For (1+t)5,(1+t)^5,(1+t)5, the third term is T3=(52)t2=10t2.T_3 = \binom{5}{2} t^2 = 10t^2.T3​=(25​)t2=10t2.

Here, t=xlog⁡2x.t=x^{\log_2 x}.t=xlog2​x. So the third term is 10(xlog⁡2x)2=10x2log⁡2x.10\left(x^{\log_2 x}\right)^2=10x^{2\log_2 x}.10(xlog2​x)2=10x2log2​x.

Given that this equals 256025602560, 10x2log⁡2x=2560.10x^{2\log_2 x}=2560.10x2log2​x=2560. Hence, x2log⁡2x=256.x^{2\log_2 x}=256.x2log2​x=256.

  1. Simplify the expression

Use the identity xlog⁡2x=2(log⁡2x)2.x^{\log_2 x}=2^{(\log_2 x)^2}.xlog2​x=2(log2​x)2. Therefore, x2log⁡2x=22(log⁡2x)2.x^{2\log_2 x}=2^{2(\log_2 x)^2}.x2log2​x=22(log2​x)2.

So we get 22(log⁡2x)2=256=28.2^{2(\log_2 x)^2}=256=2^8.22(log2​x)2=256=28. Thus, 2(log⁡2x)2=8,2(\log_2 x)^2=8,2(log2​x)2=8, which gives (log⁡2x)2=4.(\log_2 x)^2=4.(log2​x)2=4.

Therefore, log⁡2x=±2.\log_2 x=\pm 2.log2​x=±2. So, x=22=4orx=2−2=14.x=2^2=4 \quad \text{or} \quad x=2^{-2}=\frac14.x=22=4orx=2−2=41​.

  1. Match with the options

The possible values are 444 and 14\frac1441​.

Checking options:

  • A: 22=23/22\sqrt2 = 2^{3/2}22​=23/2, not equal to 444 or 14\frac1441​
  • B: 42=25/24\sqrt2 = 2^{5/2}42​=25/2, not equal to 444 or 14\frac1441​
  • C: 18=2−3\frac18 = 2^{-3}81​=2−3, not equal to 444 or 14\frac1441​
  • D: 14\frac1441​, yes

So the correct option is D.\boxed{D}.D​.

  1. Comparison with stored answer

Stored correct answer: D

This matches our derived answer.

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