ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 7. binomial theoram

यदि \((1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}\) के प्रसार में, \(x\) और \(x^2\) के गुणांक क्रमशः 1 और -2 हैं, तो \(\mathrm{p}^2+\mathrm{q}^2\) = __________

  1. A 18
  2. B 13
  3. C 8
  4. D 20
Verified Solution

Answer & Solution

Correct Answer

(B) 13

Step-by-step Solution

Detailed explanation

(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}+{…

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app