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Binomial Theorem question

2025 · 23 Jan · Shift 2 · Q40
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  5. /2025 · 23 Jan · Shift 2 · Q40

Binomial Theorem question

2025 · 23 Jan · Shift 2 · Q40

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If in the expansion of (1+x)p(1−x)q(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}(1+x)p(1−x)q, the coefficients of xxx and x2x^2x2 are 1 and -2 , respectively, then p2+q2\mathrm{p}^2+\mathrm{q}^2p2+q2 is equal to :
  1. A
    8
  2. B
    20
  3. C
    13
  4. D
    18
View written solutionFree

Correct answer: C

  1. Let
(1+x)p(1−x)q=1+ax+bx2+⋯(1+x)^p(1-x)^q = 1 + ax + bx^2 + \cdots(1+x)p(1−x)q=1+ax+bx2+⋯

We are given:

  • coefficient of xxx is 111
  • coefficient of x2x^2x2 is −2-2−2

So, a=1,b=−2.a=1, \qquad b=-2.a=1,b=−2.

  1. Expand each factor up to the x2x^2x2 term:
(1+x)p=1+px+p(p−1)2x2+⋯(1+x)^p = 1 + px + \frac{p(p-1)}{2}x^2 + \cdots(1+x)p=1+px+2p(p−1)​x2+⋯ (1−x)q=1−qx+q(q−1)2x2+⋯(1-x)^q = 1 - qx + \frac{q(q-1)}{2}x^2 + \cdots(1−x)q=1−qx+2q(q−1)​x2+⋯
  1. Multiply and compare coefficients.

For the coefficient of xxx:

p−q=1p - q = 1p−q=1

So,

p=q+1 .p=q+1 \, .p=q+1.
  1. Now find the coefficient of x2x^2x2. This comes from:
  • p(p−1)2\dfrac{p(p-1)}{2}2p(p−1)​ from the first factor,
  • q(q−1)2\dfrac{q(q-1)}{2}2q(q−1)​ from the second factor,
  • and the product of linear terms: (px)(−qx)=−pqx2(px)(-qx) = -pqx^2(px)(−qx)=−pqx2.

Hence,

p(p−1)2+q(q−1)2−pq=−2\frac{p(p-1)}{2} + \frac{q(q-1)}{2} - pq = -22p(p−1)​+2q(q−1)​−pq=−2

Simplify:

p2−p+q2−q−2pq2=−2\frac{p^2-p+q^2-q-2pq}{2} = -22p2−p+q2−q−2pq​=−2 (p−q)2−(p+q)2=−2\frac{(p-q)^2-(p+q)}{2} = -22(p−q)2−(p+q)​=−2

Since p−q=1p-q=1p−q=1, substitute:

1−(p+q)2=−2\frac{1-(p+q)}{2}=-221−(p+q)​=−2 1−(p+q)=−41-(p+q)=-41−(p+q)=−4 p+q=5p+q=5p+q=5
  1. Now solve the system:
p−q=1,p+q=5p-q=1, \qquad p+q=5p−q=1,p+q=5

Add the equations:

2p=6⇒p=32p=6 \Rightarrow p=32p=6⇒p=3

Then,

q=2q=2q=2
  1. Compute:
p2+q2=32+22=9+4=13p^2+q^2=3^2+2^2=9+4=13p2+q2=32+22=9+4=13

Therefore, the correct option is 13\boxed{13}13​ which is option C\boxed{\text{C}}C​.

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