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

Answer: (C)

Coefficient Of$ x^{15} \text { in }\left(a x^3+\frac{1}{b x^{1 / 3}}\right)^{15} $
$T_{r+1}={ }^{15} C_r\left(a x^3\right)^{15-r}\left(\frac{1}{b x^{1 / 3}}\right)^r$
$ 45-3 r-\frac{r}{3}=15$
$ 30=\frac{10 r}{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}-3 r =-15$
$ \frac{10 r}{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 a b=1 $