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Mathematics · Binomial Theorem

JEE Main 2025 — 22 January, Evening Shift — Question 1

Let α,β,γ\alpha, \beta, \gamma and δ\delta be the coefficients of x7,x5,x3x^{7}, x^{5}, x^{3} and xx 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 uu and vv satisfy the equations

αu+βv=18\alpha u+\beta v=18, γu+δv=20\gamma u+\delta v=20, then u+vu+v equals :

  1. Option A:

    5

    Correct
  2. Option B:

    4

  3. Option C:

    3

  4. Option D:

    8

Answer: A

Step-by-step solution

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

=2{5C0⋅x5+5C2⋅x3(x3−1)+5C4⋅x(x3−1)2}=2\left\{{ }^{5} \mathrm{C}_{0} \cdot \mathrm{x}^{5}+{ }^{5} \mathrm{C}_{2} \cdot \mathrm{x}^{3}\left(\mathrm{x}^{3}-1\right)+{ }^{5} \mathrm{C}_{4} \cdot \mathrm{x}\left(\mathrm{x}^{3}-1\right)^{2}\right\}

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

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

Now, 10u+2v=1810 \mathrm{u}+2 \mathrm{v}=18

−20u+10v=20-20 u+10 v=20

⇒u=1,v=4\Rightarrow \mathrm{u}=1, \mathrm{v}=4

u+v=5u+v=5

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
22 January, Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem
Topic
Applications of Binomial Theorem

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Let α, β, γ and δ be the coefficients of x 7 , x 5 , x 3 and x… | JEE Main 2025 PYQ with Solution · DhiX AI