JEE Mathematics 2023 Question with Solution

Given: The term without $x$ in the expansion of $\left( x^{\frac{2}{3}} + \frac{\alpha}{x^3} \right)^{22}$ is $7315$.

Find: $|\alpha|$.

For the expansion of $\left( x^{\frac{2}{3}} + \frac{\alpha}{x^3} \right)^{22}$, the general term is

$$T_{r+1} = \binom{22}{r}\left(x^{\frac{2}{3}}\right)^{22-r}\left(\frac{\alpha}{x^3}\right)^r$$

So,

$$T_{r+1} = \binom{22}{r}\alpha^r x^{\frac{2}{3}(22-r)-3r}$$

For the term independent of $x$, the exponent of $x$ must be zero:

$$\frac{2}{3}(22-r)-3r = 0$$ $$44 - 11r = 0$$ $$r = 4$$

Hence the constant term is

$$T_5 = \binom{22}{4}\alpha^4$$

Now,

$$\binom{22}{4} = \frac{22\times21\times20\times19}{4\times3\times2\times1} = 7315$$

Therefore,

$$7315\alpha^4 = 7315$$ $$\alpha^4 = 1$$

So,

$$|\alpha| = 1$$

Therefore, the required value is $1$.

Given: The constant term in $\left( x^{\frac{2}{3}} + \frac{\alpha}{x^3} \right)^{22}$ is $7315$.

Find: $|\alpha|$.

A common source of confusion is the form of the general term. If the first term is chosen $22-r$ times and the second term is chosen $r$ times, then

$$T_{r+1} = \binom{22}{r}\left(x^{\frac{2}{3}}\right)^{22-r}\left(\frac{\alpha}{x^3}\right)^r$$

Now simplify the power of $x$:

$$T_{r+1} = \binom{22}{r}\alpha^r x^{\frac{44-2r}{3}-3r}$$ $$= \binom{22}{r}\alpha^r x^{\frac{44-11r}{3}}$$

For the constant term,

$$\frac{44-11r}{3} = 0$$ $$44-11r=0$$ $$r=4$$

Thus,

$$T_5 = \binom{22}{4}\alpha^4$$

Given constant term $=7315$, we get

$$\binom{22}{4}\alpha^4 = 7315$$

Since

$$\binom{22}{4} = 7315$$

we have

$$\alpha^4 = 1$$

Hence,

$$|\alpha| = 1$$

Therefore, the answer is $1$.

The solution used $a$ in place of $\alpha$; this notation change does not affect the result.