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  • If the magnitude of the coefficient of x7 in the expansion of <img alt="\left(a x^2+\frac{1}{b x}\right)^8" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%28a%20x%5E2&

If the magnitude of the coefficient of x7 in the expansion of \left(a x^2+\frac{1}{b x}\right)^8, where a and b are positive numbers is equal to the magnitude of the coefficient of x-7 in the expansion of \left(a x+\frac{1}{b x^2}\right)^8 \text {, then } a^2 b^2 \text { equals }

Option: 1

1.7


Option: 2

2


Option: 3

0


Option: 4

1


Answers (1)

best_answer

\left(a x^2+\frac{1}{b x}\right)^8 \Rightarrow \text { General term }={ }^8 C_r\left(a x^2\right)^{8-r}(b x)^{-r}\begin{aligned} & \therefore \quad 16-3 r=7 \Rightarrow r=3 \\ & \Rightarrow \text { Coefficient of } x^7={ }^8 C_3 \cdot a^5 \cdot b^{-3} \end{aligned}...(i)

\begin{aligned} & \left(a x+\frac{1}{b x^2}\right)^8 \Rightarrow \text { General term }={ }^8 C_r \cdot(a x)^{8-r} \cdot\left(b x^2\right)^{-r} \\ & \Rightarrow{ }^8 C_r\left(a^{8-r} x^{8-3 r} b^{-r}\right) \quad \therefore 8-3 r=-7 \Rightarrow r=5 \end{aligned}\Rightarrow \text { Coefficient of } x^{-7}={ }^8 C_5 \cdot a^3 \cdot b^{-5}

\thereforeComparing the coefficient

{ }^8 C_5 a^3 b^{-5}={ }^8 C_3 \cdot a^5 \cdot b^{-3} \quad \therefore a^2 b^2=1

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