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If the coefficient of x^7 in \left(a x^2+\frac{1}{b x}\right)^{11} is equal to the coefficient of x^{-7} in \left(a x-\frac{1}{b x^2}\right)^{11}, then ab =
 

Option: 1

1


Option: 2

1/2


Option: 3

2


Option: 4

3


Answers (1)

best_answer

In the expansion of \left(a x^2+\frac{1}{b x}\right)^{11}, the general term is

T_{r+1}={ }^{11} C_r\left(a x^2\right)^{11-r}\left(\frac{1}{b x}\right)^r={ }^{11} C_r a^{11-r} \frac{1}{b^r} x^{22-3 r}

For x^7, we must have 22 – 3r = 7 ⇒ r = 5, and the coefficient of x^7={ }^{11} C_5 \cdot a^{11-5} \frac{1}{b^5}={ }^{11} C_5 \frac{a^6}{b^5}

Similarly, in the expansion of \left(a x-\frac{1}{b x^2}\right)^{11} the general term is T_{r+1}={ }^{11} C_r(-1)^r \frac{a^{11-r}}{b^r} \cdot x^{11-3 r}

For x^{-7} we must have, 11 – 3r = –7 ⇒ r = 6, and the coefficient of x^{-7} is { }^{11} C_6 \frac{a^5}{b^6}={ }^{11} C_5 \frac{a^5}{b^6}

As given, { }^{11} C_5 \frac{a^6}{b^5}={ }^{11} C_5 \frac{a^5}{b^6} \Rightarrow a b=1

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Sayak

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