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  • If the coefficient of  in the expansion of <img alt="\left(a x^3+\frac{1}{b x^{1 / 3}}\right)^{

If the coefficient of X^{15} in the expansion of \left(a x^3+\frac{1}{b x^{1 / 3}}\right)^{15}  is equal to the coefficient of X^{-15} in the expansion of \left(a x^{1 / 3}-\frac{1}{b x^3}\right)^{15} where a and b are positive real numbers, then for each such ordered pair (a, b) :

 

Option: 1

ab = 3


Option: 2

ab = 1 


Option: 3

a = b 


Option: 4

a = 3b


Answers (1)

best_answer

\left(\mathrm{ax}^3+\frac{1}{b x^{1 / 3}}\right)^{15}

general term is T_{r+1}={ }^{15} C_r\left(a x^3\right)^{15-\mathrm{r}}\left(\frac{1}{b x^{1 / 3}}\right)^r

                   \Rightarrow \mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{\mathrm{r}} \frac{\mathrm{a}^{15-\mathrm{r}}}{\mathrm{b}^{\mathrm{r}}} x^{45-3 \mathrm{r}}-\frac{\mathrm{r}}{3}

For coefficient of 

\begin{gathered} x^{15} \Rightarrow 45-3 r-\frac{r}{3}=15 \\ 30=\frac{10 r}{3} \\ r=9 \end{gathered}

\text { Coefficient of } x^{15} \text { is }={ }^{15} \mathrm{C}_9 \mathrm{a}^6 \mathrm{~b}^{-9} \quad \ldots \text { (1) }

\because$ general term of $\left(\mathrm{ax}^{1 / 3}-\frac{1}{\mathrm{bx}^3}\right)^{15}

\mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{\mathrm{r}}\left(\mathrm{ax}^{1 / 3}\right)^{15-\mathrm{r}}\left(\frac{-1}{\mathrm{bx}^3}\right)^{\mathrm{r}}

\text { For coefficient of } x^{-15} \Rightarrow \frac{15-r}{3}-3 r=-15

$$ \begin{aligned} & \Rightarrow 15-\mathrm{r}-9 \mathrm{r}=-45 \\ & \Rightarrow 60=10 \mathrm{r} \\ &\; \; \; \; \; \; \; \mathrm{r}=6 \end{aligned}

\text { Coefficient of } x \quad \text { is }=\mathrm{c}_6 \mathrm{a} b \quad \ldots(2)

\becauseboth coefficient are equal then

\begin{aligned} & { }^{15} \mathrm{C}_9 \mathrm{a}^6 \mathrm{~b}^{-9}={ }^{15} \mathrm{C}_6 \mathrm{a}^9 \mathrm{~b}^{-6} \\ & \Rightarrow \mathrm{a}^6 \mathrm{~b}^{-9}=\mathrm{a}^9 \mathrm{~b}^{-6} \\ & \Rightarrow \mathrm{a}^3 \mathrm{~b}^3=1 \\ & \Rightarrow \mathrm{ab}=1 \end{aligned}

 

Posted by

Pankaj Sanodiya

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