Question

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 $(a, b)$ :

• A. $a=3 b$
• B. $a=b$
• C. $ab =1$
• D. $a b=3$

Answer 1

The Correct Option is C

Coefficient Of$x^{15}\text{in}{\left(a{x}^3+\frac{1}{b{x}^{1/3}}\right)}^{15}$
$T_{r+1}={}^{15}{C}_r{\left(a{x}^3\right)}^{15-r}{\left(\frac{1}{b{x}^{1/3}}\right)}^r$
$45-3r-\frac{r}{3}=15$
$30=\frac{10r}{3}$
$r=9$
Coefficient of $x^{15}={}^{15}{C}_9{a}^6{b}^{-9}$
Coefficient of $x^{-15}\text{in}{\left(a{x}^{1/3}-\frac{1}{b{x}^3}\right)}^{15}$
$T_{r+1}={}^{15}{C}_r{\left(ax{}^{1/3}\right)}^{15-r}{\left(-\frac{1}{b{x}^3}\right)}^r$
$5-\frac{r}{3}-3r=-15$
$\frac{10r}{3}=20$
$r=6$
Coefficient $={}^{15}{C}_6{a}^9\times b^{-6}$
$\Rightarrow \frac{a^9}{b^6}=\frac{a^6}{b^9}$
$\Rightarrow a^3{b}^3=1\Rightarrow ab=1$