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

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Binomial Theorem

If the coefficient of

x^{15}

in the expansion of

(a x^{3} + \frac{1}{b x ^{\frac{1}{3}}})^{15}

is equal to the coefficient of

x^{- 15}

in the expansion of

(a x^{\frac{1}{3}} - \frac{1}{b x ^{3}})^{15}

, where

a

and

b

are positive real numbers, then for each such ordered pair

(a, b)

Answer: (B)

(a x^{3} + \frac{1}{b x ^{1/3}})^{15}

the general term carries

x^{45 - 3 r - r /3}

; setting the exponent to

15

gives

r = 9

, with coefficient

(9 15) a^{6} b^{- 9}

. In

(a x^{1/3} - \frac{1}{b x ^{3}})^{15}

the exponent is

5 - \frac{r}{3} - 3 r

; setting it to

- 15

gives

r = 6

, with coefficient

(6 15) a^{9} b^{- 6}

. Since

(9 15) = (6 15)

, equality reduces to

a^{6} b^{- 9} = a^{9} b^{- 6}

, i.e.

a^{3} b^{3} = 1

, so for every such positive pair

ab = 1.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option