JEE Mathematics 2024 Question with Solution

Given: We need the sum of the coefficients of $x^{2/3}$ and $x^{-2/5}$ in the expansion of $\left(x^{2/3} + \frac{1}{2}x^{-2/5}\right)^9$.

Find: The required sum of coefficients and the correct option.

Using the general term of a binomial expansion,

$$T_{k+1} = \binom{9}{k}\left(x^{2/3}\right)^{9-k}\left(\frac{1}{2}x^{-2/5}\right)^k$$

So,

$$T_{k+1} = \binom{9}{k}\left(\frac{1}{2}\right)^k x^{\frac{2}{3}(9-k) - \frac{2}{5}k}$$

The power of $x$ is

$$\frac{2}{3}(9-k) - \frac{2}{5}k = 6 - \frac{2}{3}k - \frac{2}{5}k = 6 - \frac{16}{15}k$$

For the coefficient of $x^{2/3}$, set

$$6 - \frac{16}{15}k = \frac{2}{3}$$

Thus,

$$\frac{16}{15}k = 6 - \frac{2}{3} = \frac{16}{3}$$ $$k = \frac{16}{3} \times \frac{15}{16} = 5$$

Hence the coefficient of $x^{2/3}$ is

$$\binom{9}{5}\left(\frac{1}{2}\right)^5 = \frac{126}{32} = \frac{63}{16}$$

For the coefficient of $x^{-2/5}$, set

$$6 - \frac{16}{15}k = -\frac{2}{5}$$

Thus,

$$\frac{16}{15}k = 6 + \frac{2}{5} = \frac{32}{5}$$ $$k = \frac{32}{5} \times \frac{15}{16} = 6$$

Hence the coefficient of $x^{-2/5}$ is

$$\binom{9}{6}\left(\frac{1}{2}\right)^6 = \frac{84}{64} = \frac{21}{16}$$

Therefore, the required sum is

$$\frac{63}{16} + \frac{21}{16} = \frac{84}{16} = \frac{21}{4}$$

So, the correct option is A.

Given: The expansion is $\left(x^{2/3} + \frac{1}{2}x^{-2/5}\right)^9$.

Find: The sum of the coefficients of $x^{2/3}$ and $x^{-2/5}$.

The exponent of $x$ in the general term is

$$6 - \frac{16}{15}k$$

Now match powers directly.

For $x^{2/3}$,

$$6 - \frac{16}{15}k = \frac{2}{3} \Rightarrow k = 5$$

So coefficient

$$= \binom{9}{5}\left(\frac{1}{2}\right)^5 = \frac{63}{16}$$

For $x^{-2/5}$,

$$6 - \frac{16}{15}k = -\frac{2}{5} \Rightarrow k = 6$$

So coefficient

$$= \binom{9}{6}\left(\frac{1}{2}\right)^6 = \frac{21}{16}$$

Adding,

$$\frac{63}{16} + \frac{21}{16} = \frac{21}{4}$$

Therefore, the correct option is A.