JEE MAIN - Mathematics (2023 - 30th January Morning Shift - No. 5)
If the coefficient of $$x^{15}$$ in the expansion of $$\left(\mathrm{a} x^{3}+\frac{1}{\mathrm{~b} x^{1 / 3}}\right)^{15}$$ 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 $$(\mathrm{a}, \mathrm{b})$$ :
a = 3b
ab = 1
ab = 3
a = b
Explanation
For $$\left( {a{x^3} + {1 \over {b{x^{{1 \over 3}}}}}} \right)$$
$${T_{r + 1}} = {}^{15}{C_r}{(a{x^3})^{15 - r}}{\left( {{1 \over {b{x^{{1 \over 3}}}}}} \right)^1}$$
$$\therefore$$ $${x^{15}} \to 3(15 - r) - {r \over 3} = 15$$
$$ \Rightarrow 30 = {{10r} \over 3} \Rightarrow r = 9$$
Similarly, for $${\left( {a{x^{{1 \over 3}}} - {1 \over {b{x^3}}}} \right)^{15}}$$
$${T_{r + 1}} = {}^{15}{C_r}{\left( {a{x^{{1 \over 3}}}} \right)^{15 - r}}{\left( { - {1 \over {b{x^3}}}} \right)^2}$$
$$\therefore$$ For $${x^{ - 15}} \to {{15 - r} \over 3} - 3r = - 15 \Rightarrow r = 6$$
$$\therefore$$ $${}^{15}{C_9}{{{a^6}} \over {{b^9}}} = {}^{15}{C_6}{{{a^9}} \over {{b^6}}} \Rightarrow ab = 1$$
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