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Binomial Theorem question

2017 · 9 Apr · Shift 1 · Q25
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  5. /2017 · 9 Apr · Shift 1 · Q25

Binomial Theorem question

2017 · 9 Apr · Shift 1 · Q25

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
The coefficient of x−5 in the binomial expansion of (x+1x23−x13+1−x−1x−x12)10,{\left( {{{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}} - {{x - 1} \over {x - {x^{{1 \over 2}}}}}} \right)^{10}},(x32​−x31​+1x+1​−x−x21​x−1​)10, where x eee 0, 1, is :
  1. A
    1
  2. B
    4
  3. C
    −-− 4
  4. D
    −-− 1
View written solutionFree

Correct answer: A

  1. Simplify the expression inside the bracket

We need the coefficient of x−5x^{-5}x−5 in

(x+1x2/3−x1/3+1−x−1x−x1/2)10.\left( \frac{x+1}{x^{2/3}-x^{1/3}+1} - \frac{x-1}{x-x^{1/2}} \right)^{10}.(x2/3−x1/3+1x+1​−x−x1/2x−1​)10.

Let t=x1/3.t=x^{1/3}.t=x1/3. Then x=t3,x2/3=t2.x=t^3,\quad x^{2/3}=t^2.x=t3,x2/3=t2. So the first fraction becomes

x+1x2/3−x1/3+1=t3+1t2−t+1.\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{t^3+1}{t^2-t+1}.x2/3−x1/3+1x+1​=t2−t+1t3+1​.

Using t3+1=(t+1)(t2−t+1),t^3+1=(t+1)(t^2-t+1),t3+1=(t+1)(t2−t+1), we get

t3+1t2−t+1=t+1=x1/3+1.\frac{t^3+1}{t^2-t+1}=t+1=x^{1/3}+1.t2−t+1t3+1​=t+1=x1/3+1.

Now simplify the second fraction:

x−1x−x1/2.\frac{x-1}{x-x^{1/2}}.x−x1/2x−1​.

Let u=xu=\sqrt{x}u=x​. Then x=u2x=u^2x=u2, so

x−1x−x1/2=u2−1u2−u=(u−1)(u+1)u(u−1)=u+1u=1+u−1=1+x−1/2.\frac{x-1}{x-x^{1/2}}=\frac{u^2-1}{u^2-u}=\frac{(u-1)(u+1)}{u(u-1)}=\frac{u+1}{u}=1+u^{-1}=1+x^{-1/2}.x−x1/2x−1​=u2−uu2−1​=u(u−1)(u−1)(u+1)​=uu+1​=1+u−1=1+x−1/2.

Hence the whole inner expression is

(x1/3+1)−(1+x−1/2)=x1/3−x−1/2.(x^{1/3}+1)-(1+x^{-1/2})=x^{1/3}-x^{-1/2}.(x1/3+1)−(1+x−1/2)=x1/3−x−1/2.

So we need the coefficient of x−5x^{-5}x−5 in

(x1/3−x−1/2)10.(x^{1/3}-x^{-1/2})^{10}.(x1/3−x−1/2)10.
  1. Apply the binomial theorem

General term is

Tr+1=(10r)(x1/3)10−r(−x−1/2)rT_{r+1}=\binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^rTr+1​=(r10​)(x1/3)10−r(−x−1/2)r

for r=0,1,2,…,10r=0,1,2,\dots,10r=0,1,2,…,10.

So

Tr+1=(10r)(−1)rx10−r3−r2.T_{r+1}=\binom{10}{r}(-1)^r x^{\frac{10-r}{3}-\frac r2}.Tr+1​=(r10​)(−1)rx310−r​−2r​.

The exponent of xxx is

10−r3−r2=20−2r−3r6=20−5r6.\frac{10-r}{3}-\frac r2=\frac{20-2r-3r}{6}=\frac{20-5r}{6}.310−r​−2r​=620−2r−3r​=620−5r​.

We want this to equal −5-5−5:

20−5r6=−5.\frac{20-5r}{6}=-5.620−5r​=−5.

Thus,

20−5r=−3020-5r=-3020−5r=−30 −5r=−50-5r=-50−5r=−50 r=10.r=10.r=10.
  1. Find the corresponding coefficient

For r=10r=10r=10,

(1010)(−1)10=1.\binom{10}{10}(-1)^{10}=1.(1010​)(−1)10=1.

Therefore, the coefficient of x−5x^{-5}x−5 is

1.\boxed{1}.1​.
  1. Check options
  • A: 111 ✔
  • B: 444 ✘
  • C: −4-4−4 ✘
  • D: −1-1−1 ✘

So the correct option is A.

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