If the coefficient of in the expansion of is equal to the coefficient of in the expansion of , where and are positive real numbers, then for each such ordered pair (a, b) [2023]

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Solution

Q.
If the coefficient of $x^{15}$ in the expansion of ${(a x^{3} + \frac{1}{b x^{1 / 3}})}^{15}$ is equal to the coefficient of $x^{- 15}$ in the expansion of ${(a x^{1 / 3} - \frac{1}{b x^{3}})}^{15}$ , where $a$ and $b$ are positive real numbers, then for each such ordered pair (a, b) [2023]

1 a = 3b

2 a = b

3 ab = 1

4 ab = 3

Ans.
(3)

The coefficient of $x^{15}$ in the expansion of

${(a x^{3} + \frac{1}{b x^{\frac{1}{3}}})}^{15}$ is $T_{r + 1} = {}^{15}{C_{r}} {(a x^{3})}^{15 - r} \times {(b^{- 1} x^{- \frac{1}{3}})}^{r}$

$= {}^{15}{C_{r}} a^{15 - r} \cdot x^{45 - 3 r} \cdot b^{- r} \cdot x^{- \frac{r}{3}} = {}^{15}{C_{r}} a^{15 - r} \cdot b^{- r} x^{45 - 3 r - \frac{r}{3}}$

We have to find the coefficient of $x^{15}$ , so $x^{45 - 3 r - \frac{r}{3}} = x^{15}$

$\Rightarrow 45 - 3 r - \frac{r}{3} = 15 \Rightarrow \frac{10 r}{3} = 30 \Rightarrow r = 9$

So, $T_{9 + 1} = {}^{15}{C_{9}} a^{15 - 9} \cdot b^{- 9} \cdot x^{15};$ $T_{10} = {}^{15}{C_{9}} a^{6} \cdot b^{- 9} \cdot x^{15}$

$∴$ $The coefficient of x^{15} = {}^{15}{C_{9}} a^{6} \cdot b^{- 9}$

Now, the coefficient of $x^{- 15}$ in the expansion of

${({a x}^{\frac{1}{3}} - \frac{1}{b x^{3}})}^{15}$ is, $T_{r + 1} = {{}^{15}C}_{r} {(a x^{1 / 3})}^{15 - r} \cdot {(\frac{- 1}{b x^{3}})}^{r}$

$= {{}^{15}C}_{r} a^{15 - r} x^{\frac{15 - r}{3}} \cdot b^{- r} \cdot x^{- 3 r} {(- 1)}^{r}$

$= {}^{15}{C_{r}} a^{15 - r} \cdot b^{- r} \cdot x^{\frac{15 - r}{3} - 3 r} {(- 1)}^{r}$

We need to find the coefficient of $x^{- 15}$ , so

$x^{\frac{15 - r}{3} - 3 r} = x^{- 15} \Rightarrow \frac{15 - r}{3} - 3 r = - 15 \Rightarrow r = 6$

$∴$ $The coefficient of x^{- 15} = {}^{15}{C_{6}} a^{9} b^{- 6}$

Now, according to the question, ${}^{15}{C_{9}} a^{6} b^{- 9} = {}^{15}{C_{6}} a^{9} b^{- 6}$

$\Rightarrow a^{3} b^{3} = 1 ∴ a b = 1$

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