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JEE Advanced · Mathematics · 6. Binomial Theorem

Let \(a\) and \(b\) be two nonzero real numbers. If the coefficient of \(x^5\) in the expansion of \(\left(a x^2+\frac{70}{27 b x}\right)^4\) is equal to the coefficient of \(x^{-5}\) in the expansion of \(\left(a x-\frac{1}{b x^2}\right)^7\), then the value of \(2 b\) is

  1. A 3
  2. B 6
  3. C 9
  4. D 12
Verified Solution

Answer & Solution

Correct Answer

(A) 3

Step-by-step Solution

Detailed explanation

Given,
The coefficient of x5 in ax2+7027bx4 is equal to coefficient of x-5 in ax-1bx27,
Now finding the coefficient of x5 in ax2+7027bx4 we get,
Tr+1=Cr4ax24-r·7027bxr
Tr+1=Cr4a4-r·7027br·x8-2r-r
So, 8-2r-r=5r=1
So, T2=C14a3·7027b·x5
Now finding the coefficient of x-5 in ax-1bx27 we get,
Tr+1=Cr7ax7-r·1bx2r
Tr+1=Cr7a7-r·1br·x7-r-2r
So, 7-r-2r=-5r=4
So, T5=C47a3·1b4·x-5
Now equating the coefficient of x5 & x-5 we get,
C47a3·1b4=C14a3·7027b
35a3·1b4=4a3·7027b
1b3=4×227
b=322b=3
From JEE Advanced
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