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Binomial Theorem question

2019 · 9 Apr · Shift 1 · Q42
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Binomial Theorem question

2019 · 9 Apr · Shift 1 · Q42

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If the fourth term in the binomial expansion of (2x+xlog⁡8x)6{\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}(x2​+xlog8​x)6 (x > 0) is 20 × 87, then a value of x is :
  1. A
    8–2
  2. B
    82
  3. C
    83
  4. D
    8
View written solutionFree

Correct answer: B

  1. Write the general term of the expansion

For (2x+xlog⁡8x)6,\left(\frac{2}{x}+x^{\log_8 x}\right)^6,(x2​+xlog8​x)6, the general term is Tr+1=(6r)(2x)6−r(xlog⁡8x)r.T_{r+1}=\binom{6}{r}\left(\frac{2}{x}\right)^{6-r}\left(x^{\log_8 x}\right)^r.Tr+1​=(r6​)(x2​)6−r(xlog8​x)r.

The fourth term corresponds to r=3r=3r=3.

So, T4=(63)(2x)3(xlog⁡8x)3.T_4=\binom{6}{3}\left(\frac{2}{x}\right)^3\left(x^{\log_8 x}\right)^3.T4​=(36​)(x2​)3(xlog8​x)3.

  1. Simplify the fourth term

We know (63)=20,\binom{6}{3}=20,(36​)=20, and (2x)3=8x3.\left(\frac{2}{x}\right)^3=\frac{8}{x^3}.(x2​)3=x38​.

Also, (xlog⁡8x)3=x3log⁡8x.\left(x^{\log_8 x}\right)^3=x^{3\log_8 x}.(xlog8​x)3=x3log8​x.

Hence,

=20\cdot 8\cdot x^{3\log_8 x-3}.$$ Given that the fourth term is $20\times 8^7$, we get $$20\cdot 8\cdot x^{3\log_8 x-3}=20\cdot 8^7.$$ Cancel $20$: $$8\,x^{3\log_8 x-3}=8^7.$$ So, $$x^{3\log_8 x-3}=8^6.$$ 3. **Test the options** Let us check the given options. ### Option B: $x=8^2$ Then $$\log_8 x=\log_8(8^2)=2.$$ So, $$3\log_8 x-3=3(2)-3=3.$$ Thus, $$x^{3\log_8 x-3}=(8^2)^3=8^6,$$ which satisfies the equation. So $x=8^2$ is a valid value. ### Option A: $x=8^{-2}$ Then $$\log_8 x=-2,$$ so $$3\log_8 x-3=-6-3=-9.$$ Then $$x^{-9}=(8^{-2})^{-9}=8^{18}\neq 8^6.$$ Not correct. ### Option C: $x=8^3$ Then $$\log_8 x=3,$$ so $$3\log_8 x-3=9-3=6.$$ Then $$x^6=(8^3)^6=8^{18}\neq 8^6.$$ Not correct. ### Option D: $x=8$ Then $$\log_8 x=1,$$ so $$3\log_8 x-3=0.$$ Then $$x^0=1\neq 8^6.$$ Not correct. 4. **Conclusion** The correct option is $$\boxed{8^2}.$$
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