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JEE Main 202523 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Binomial Theorem

If in the expansion of (1+x)^ p (1-x)^ q , the coefficients of x...

If in the expansion of (1+x)p(1−x)q(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}, the coefficients of xx and x2x^2 are 1 and -2 , respectively, then p2+q2\mathrm{p}^2+\mathrm{q}^2 is equal to :

Options

  1. A18
  2. B13
  3. C8
  4. D20

Correct answer

B. 13

Step-by-step solution

(1+x)p(1−x)q=(pC0+pC1x+pC2x2+…)(qC0−qC1x+qC2x2+…) coff of  x≡pC0qC1+pC1qC0=1p−q=1 coff of x2≡pC0qC2−pC1qC1+pC2qC0=−2q(q−1)2−pq+p(p−1)2=−2q2−q−2pq+p2−p=−41−(p+q)=−4p+q=5p=3q=2 so p2+q2=13\begin{aligned} & (1+\mathrm{x})^{\mathrm{p}}(1-\mathrm{x})^{\mathrm{q}}=\left({ }^{\mathrm{p}} \mathrm{C}_0+{ }^{\mathrm{p}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{p}} \mathrm{C}_2 \mathrm{x}^2+\ldots\right)\left({ }^{\mathrm{q}} \mathrm{C}_0-{ }^{\mathrm{q}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{q}} \mathrm{C}_2 \mathrm{x}^2+\ldots\right) \\ & \text{ coff\text{ of } } \mathrm{x} \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^{\mathrm{q}} \mathrm{C}_1+{ }^{\mathrm{p}} \mathrm{C}_1{ }^{\mathrm{q}} \mathrm{C}_0=1 \\ & \mathrm{p}-\mathrm{q}=1 \\ & \text{ coff of } \mathrm{x}^2 \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^{\mathrm{q}} \mathrm{C}_2-{ }^{\mathrm{p}} \mathrm{C}_1{ }^{\mathrm{q}} \mathrm{C}_1+{ }^{\mathrm{p}} \mathrm{C}_2{ }^{\mathrm{q}} \mathrm{C}_0=-2 \\ & \frac{\mathrm{q}(\mathrm{q}-1)}{2}-\mathrm{pq}+\frac{\mathrm{p}(\mathrm{p}-1)}{2}=-2 \\ & \mathrm{q}^2-\mathrm{q}-2 \mathrm{pq}+\mathrm{p}^2-\mathrm{p}=-4 \\ & 1-(\mathrm{p}+\mathrm{q})=-4 \\ & \mathrm{p}+\mathrm{q}=5 \\ & \mathrm{p}=3 \\ & \mathrm{q}=2 \\ & \text{ so } \mathrm{p}^2+\mathrm{q}^2=13\end{aligned}
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