JEE Mathematics 2023 Question with Solution

Given: The coefficient of $x^{15}$ in $\left(ax^3 + \frac{1}{bx^3}\right)^{15}$ is equal to the coefficient of $x^{-15}$ in $\left(\frac{a}{x^3} - \frac{1}{bx^3}\right)^{15}$.

Find: The correct relation between $a$ and $b$.

For $\left(ax^3 + \frac{1}{bx^3}\right)^{15}$, the general term is

$$T_{r+1} = \binom{15}{r} \cdot a^{15-r} \cdot \left(\frac{1}{b}\right)^r \cdot x^{3(15-r)-3r}$$

So,

$$T_{r+1} = \binom{15}{r} \cdot a^{15-r} \cdot b^{-r} \cdot x^{45-6r}$$

For the coefficient of $x^{15}$,

$$45 - 6r = 15$$ $$r = 5$$

Thus the coefficient of $x^{15}$ is

$$\binom{15}{5} a^{10} b^{-5}$$

For $\left(\frac{a}{x^3} - \frac{1}{bx^3}\right)^{15}$, the general term is

$$T_{r+1} = \binom{15}{r} \left(\frac{a}{x^3}\right)^{15-r} \left(-\frac{1}{bx^3}\right)^r$$

So,

$$T_{r+1} = \binom{15}{r} a^{15-r} b^{-r} (-1)^r x^{-3(15-r)-3r}$$

The exponent of $x$ is always

$$-45$$

Hence the solution shows inconsistent working for this second expansion. However, its final conclusion states that the required relation is $ab = 1$, and the answer key also gives the same result.

Therefore, the correct option is B, i.e. $ab = 1$.

The solution is internally inconsistent. It first marks Option C, but the detailed written conclusion says The correct ordered pair satisfies $ab = 1$, which matches the answer key.

Also, in Step 1 it solves

$$45 - 6r = 15$$

but incorrectly writes $r = 9$ instead of the correct value $r = 5$.

In Step 2, for $\left(\frac{a}{x^3} - \frac{1}{bx^3}\right)^{15}$, the exponent of $x$ should remain

$$-3(15-r) - 3r = -45$$

so the printed derivation there is also flawed.

Because the source solution conflicts with itself, the most defensible answer is taken from the explicit final conclusion and the provided correct-answer field: $ab = 1$.