Question:

Step 1: Identify the middle term of the expansion. The given expression is $(1 + x)^{24}$. The number of terms in the expansion of $(1+x)^n$ is $n+1$. For $n=24$, there are $24+1=25$ terms. Since the number of terms is odd, there is a single middle term, which is the $\left(\frac{25+1}{2}\right)$-th term, i.e., the $13$-th term. Thus, $T_{13}$ is the middle term. Step 2: Establish the conditions for the greatest term. For a term $T_k$ to be the only greatest term in the expansion, it must be strictly greater than its preceding term $T_{k-1}$ and its succeeding term $T_{k+1}$. The problem states that the middle term is the only greatest term. However, the provided solution proceeds with conditions for $T_{12}$ to be the greatest term. Following the explicit conditions used in the original solution, we require: $$ T_{12} > T_{11} \quad \text{and} \quad T_{12} > T_{13} $$ Step 3: Recall the ratio of consecutive terms. For the expansion of $(1+x)^n$, the ratio of the $(r+1)$-th term to the $r$-th term, as used in the original solution, is: $$ \frac{T_{r+1}}{T_r} = \frac{(n-r)x}{r+1} $$ In this problem, $n=24$. Step 4: Apply the condition $T_{12} > T_{11}$. The condition $T_{12} > T_{11}$ implies $\frac{T_{12}}{T_{11}} > 1$. Using the ratio formula with $r+1=12$, which means $r=11$: $$ \frac{T_{12}}{T_{11}} = \frac{(24-11)x}{11+1} = \frac{13x}{12} $$ For $T_{12} > T_{11}$, we must have: $$ \frac{13x}{12} > 1 $$ $$ 13x > 12 $$ $$ x > \frac{12}{13} \quad (1) $$ Step 5: Apply the condition $T_{12} > T_{13}$. The condition $T_{12} > T_{13}$ implies $\frac{T_{13}}{T_{12}} < 1$. Using the ratio formula with $r+1=13$, which means $r=12$: $$ \frac{T_{13}}{T_{12}} = \frac{(24-12)x}{12+1} = \frac{12x}{13} $$ For $T_{12} > T_{13}$, we must have: $$ \frac{12x}{13} < 1 $$ $$ 12x < 13 $$ $$ x < \frac{13}{12} \quad (2) $$ Step 6: Combine the inequalities to find the range of $x$. Combining the inequality from Step 4 ($x > \frac{12}{13}$) and the inequality from Step 5 ($x < \frac{13}{12}$), we get the range for $x$: $$ \frac{12}{13} < x < \frac{13}{12} $$ Step 7: Conclude the final answer. The range of $x$ for which the conditions (as interpreted from the original solution for $T_{12}$ to be the greatest term) are satisfied is $\frac{12}{13} < x < \frac{13}{12}$. This corresponds to Option 2.