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

\((1+x)^p(1-x)^q\) में \(x\) तथा \(x^2\) के गुणांक क्रमशः 4 तथा -5 है, तो \(2 p+3 q\) बराबर है

  1. A \(63\)
  2. B \(69\)
  3. C \(66\)
  4. D \(60\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(63\)

Step-by-step Solution

Detailed explanation

\((1+x)^p(1-x)^q\) \(\left(1+p x+\frac{p(p-1)}{2 !} x^2+\ldots\right)\) \(\left(1-q x+\frac{q(q-1)}{2 !} x^2-\ldots\right)\) \(p-q=4\) \(\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-p q=-5\) \(p^2+q^2-p-q-2 p q=-10\) \((q+4)^2+q^2-(q+4)-q-2(4+q) q=-10\)…
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