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Binomial Theorem question

2025 · 22 Jan · Shift 2 · Q30
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  5. /2025 · 22 Jan · Shift 2 · Q30

Binomial Theorem question

2025 · 22 Jan · Shift 2 · Q30

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
Let α,β,γ\alpha, \beta, \gammaα,β,γ and δ\deltaδ be the coefficients of x7,x5,x3x^7, x^5, x^3x7,x5,x3 and xxx respectively in the expansion of (x+x3−1)5+(x−x3−1)5,x>1. If u and v satisfy the equations αu+βv=18,γu+δv=20,\begin{aligned} & \left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5, x\gt 1 \text {. If } u \text { and } v \text { satisfy the equations } \\\\ & \alpha u+\beta v=18, \\\\ & \gamma u+\delta v=20, \end{aligned}​(x+x3−1​)5+(x−x3−1​)5,x>1. If u and v satisfy the equations αu+βv=18,γu+δv=20,​ then u+v\mathrm{u+v}u+v equals :
  1. A
    4
  2. B
    3
  3. C
    5
  4. D
    8
View written solutionFree

Correct answer: C

  1. Let S=(x+x3−1)5+(x−x3−1)5.S=(x+\sqrt{x^3-1})^5+(x-\sqrt{x^3-1})^5.S=(x+x3−1​)5+(x−x3−1​)5.

We need the coefficients of x7,x5,x3,xx^7,x^5,x^3,xx7,x5,x3,x in SSS.

  1. Use the identity (a+b)5+(a−b)5=2(a5+10a3b2+5ab4),(a+b)^5+(a-b)^5=2\left(a^5+10a^3b^2+5ab^4\right),(a+b)5+(a−b)5=2(a5+10a3b2+5ab4), because only even powers of bbb survive.

Here, a=x,b=x3−1.a=x,\qquad b=\sqrt{x^3-1}.a=x,b=x3−1​. So S=2[x5+10x3(x3−1)+5x(x3−1)2].S=2\left[x^5+10x^3(x^3-1)+5x(x^3-1)^2\right].S=2[x5+10x3(x3−1)+5x(x3−1)2].

  1. Expand step by step:

  • First part: x5x^5x5

  • Second part: 10x3(x3−1)=10x6−10x310x^3(x^3-1)=10x^6-10x^310x3(x3−1)=10x6−10x3

  • Third part: 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+5x5x(x3−1)2=5x(x6−2x3+1)=5x7−10x4+5x

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

So S=10x7+20x6+2x5−20x4−20x3+10x.S=10x^7+20x^6+2x^5-20x^4-20x^3+10x.S=10x7+20x6+2x5−20x4−20x3+10x.

  1. Therefore the required coefficients are: α=10,β=2,γ=−20,δ=10.\alpha=10,\quad \beta=2,\quad \gamma=-20,\quad \delta=10.α=10,β=2,γ=−20,δ=10.

  2. Now use the system: αu+βv=18⇒10u+2v=18,\alpha u+\beta v=18 \Rightarrow 10u+2v=18,αu+βv=18⇒10u+2v=18, γu+δv=20⇒−20u+10v=20.\gamma u+\delta v=20 \Rightarrow -20u+10v=20.γu+δv=20⇒−20u+10v=20.

Simplify: 5u+v=9...(1)5u+v=9 \quad ...(1)5u+v=9...(1) −2u+v=2...(2)-2u+v=2 \quad ...(2)−2u+v=2...(2)

Subtract (2) from (1): 7u=7⇒u=1.7u=7 \Rightarrow u=1.7u=7⇒u=1. Then from (2): −2(1)+v=2⇒v=4.-2(1)+v=2 \Rightarrow v=4.−2(1)+v=2⇒v=4.

Hence, u+v=1+4=5.u+v=1+4=5.u+v=1+4=5.

  1. Comparing with the stored correct answer: Derived answer = 555, stored answer = C = 555. So they agree.
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