Logo
Jeeify
MCQMediumJEE 2025Binomial Expansion

JEE Mathematics 2025 Question with Solution

Let α,β,γ\alpha, \beta, \gamma and δ\delta be the coefficients of x7,x5,x3,xx^7, x^5, x^3, x respectively in the expansion of (x+x3−1)5+(x−x3−1)5,  x>1\left(x + \sqrt{x^3 - 1}\right)^5 + \left(x - \sqrt{x^3 - 1}\right)^5, \; x > 1. If αu+βv=18\alpha u + \beta v = 18, γu+δv=20\gamma u + \delta v = 20, then u+vu + v equals:

  • A

    44

  • B

    88

  • C

    33

  • D

    55

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

E=(x+x3−1)5+(x−x3−1)5E = \left(x + \sqrt{x^3 - 1}\right)^5 + \left(x - \sqrt{x^3 - 1}\right)^5

Let y=x3−1y = \sqrt{x^3 - 1}.

Find: The value of u+vu+v when α,β,γ,δ\alpha, \beta, \gamma, \delta are the coefficients of x7,x5,x3,xx^7, x^5, x^3, x respectively.

Using binomial expansion,

(x+y)5+(x−y)5=2(x5+10x3y2+5xy4)(x+y)^5 + (x-y)^5 = 2\left(x^5 + 10x^3y^2 + 5xy^4\right)

So,

E=2[x5+10x3(x3−1)+5x(x3−1)2]E = 2\left[x^5 + 10x^3(x^3-1) + 5x(x^3-1)^2\right]

Now expand:

10x3(x3−1)=10x6−10x310x^3(x^3-1) = 10x^6 - 10x^3

and

5x(x3−1)2=5x(x6−2x3+1)=5x7−10x4+5x5x(x^3-1)^2 = 5x(x^6 - 2x^3 + 1) = 5x^7 - 10x^4 + 5x

Hence,

E=2(5x7+x5+10x6−10x4−10x3+5x)E = 2\left(5x^7 + x^5 + 10x^6 - 10x^4 - 10x^3 + 5x\right)

Therefore the required coefficients are

α=10,β=2,γ=−20,δ=10\alpha = 10, \quad \beta = 2, \quad \gamma = -20, \quad \delta = 10

Now use the given equations:

10u+2v=1810u + 2v = 18 −20u+10v=20-20u + 10v = 20

From the first equation,

5u+v=95u + v = 9

From the second equation,

−2u+v=2-2u + v = 2

Subtracting,

7u=7⇒u=17u = 7 \Rightarrow u = 1

Then

v=9−5=4v = 9 - 5 = 4

So,

u+v=1+4=5u+v = 1+4 = 5

Therefore, the correct option is D.

Use cancellation of odd radical terms

Given:

(x+x3−1)5+(x−x3−1)5\left(x + \sqrt{x^3 - 1}\right)^5 + \left(x - \sqrt{x^3 - 1}\right)^5

Find: u+vu+v.

The two binomial expansions are identical except that terms containing odd powers of x3−1\sqrt{x^3-1} have opposite signs. Hence those terms cancel, and only even-power terms remain.

So the surviving terms are:

2[(50)x5+(52)x3(x3−1)+(54)x(x3−1)2]2\left[\binom{5}{0}x^5 + \binom{5}{2}x^3(x^3-1) + \binom{5}{4}x(x^3-1)^2\right]

That is,

2[x5+10x3(x3−1)+5x(x3−1)2]2\left[x^5 + 10x^3(x^3-1) + 5x(x^3-1)^2\right]

Now,

(x3−1)2=x6−2x3+1(x^3-1)^2 = x^6 - 2x^3 + 1

Therefore,

E=2[x5+10x6−10x3+5x7−10x4+5x]E = 2\left[x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x\right]

Read off coefficients:

  • coefficient of x7x^7 is 1010, so α=10\alpha = 10
  • coefficient of x5x^5 is 22, so β=2\beta = 2
  • coefficient of x3x^3 is −20-20, so γ=−20\gamma = -20
  • coefficient of xx is 1010, so δ=10\delta = 10

Hence,

10u+2v=18,−20u+10v=2010u + 2v = 18, \qquad -20u + 10v = 20

Simplifying,

5u+v=9,−2u+v=25u + v = 9, \qquad -2u + v = 2

Subtract the second from the first:

7u=7⇒u=17u = 7 \Rightarrow u = 1

Then

v=2+2u=4v = 2 + 2u = 4

Thus,

u+v=5u+v = 5

Therefore, the correct option is D.

The solution contains inconsistent intermediate work in one approach, but the algebra above matches the expression correctly and agrees with the stated final option.

Common mistakes

  • Ignoring that odd powers of x3−1\sqrt{x^3-1} cancel when adding (x+y)5\left(x+y\right)^5 and (x−y)5\left(x-y\right)^5. This is wrong because only even powers survive in the sum. Instead, keep only the y0,y2,y4y^0, y^2, y^4 terms.

  • Taking the coefficient of x3x^3 as positive. This is wrong because the term comes from 2⋅10x3(x3−1)2\cdot 10x^3(x^3-1), which contributes −20x3-20x^3. Instead, expand carefully and track the sign from −1-1.

  • Expanding (x±x3−1)5\left(x \pm \sqrt{x^3-1}\right)^5 as if x3−1=x3−1\sqrt{x^3-1} = x^3-1. This is wrong because squaring the radical changes the expression. Instead, set y=x3−1y = \sqrt{x^3-1} first and use y2=x3−1y^2 = x^3-1 only after selecting even powers.

Practice more Binomial Expansion questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Binomial Expansion

Related questions