Here are the solutions to the questions: 1. The middle term in the expansion of $\left(\frac{x}{y} - \frac{y}{x}\right)^4$ is The power of the expansion is $n=4$. Since $n$ is an even number, there is one middle term, which is the $\left(\frac{n}{2} + 1\right)$-th term. Middle term = $\left(\frac{4}{2} + 1\right)$-th term = $(2+1)$-th term = 3rd term. The general term in the expansion of $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. Here, $a = \frac{x}{y}$, $b = -\frac{y}{x}$, and $n=4$. For the 3rd term, $r+1=3$, so $r=2$. Substitute these values into the general term formula: $$T_3 = \binom{4}{2} \left(\frac{x}{y}\right)^{4-2} \left(-\frac{y}{x}\right)^2$$ Calculate $\binom{4}{2}$: $$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{12}{2} = 6$$ Substitute the value of $\binom{4}{2}$ back into the expression for $T_3$: $$T_3 = 6 \left(\frac{x}{y}\right)^2 \left(-\frac{y}{x}\right)^2$$ $$T_3 = 6 \left(\frac{x^2}{y^2}\right) \left(\frac{y^2}{x^2}\right)$$ $$T_3 = 6 \times 1$$ $$T_3 = 6$$ The middle term is 6. The correct option is B. 2. The binomial series of $(1 + kx)^n$ is valid only when: The binomial series expansion of $(1+u)^n$ is valid for $|u| < 1$. In this expression, $u = kx$. Therefore, the series is valid when $|kx| < 1$. This inequality can be written as $-1 < kx < 1$. To isolate $x$, we divide by $|k|$ (assuming $k \neq 0$): $$-\frac{1}{|k|} < x < \frac{1}{|k|}$$ If $k$ is positive, $|k|=k$, so $-\frac{1}{k} < x < \frac{1}{k}$. Option D uses $K$ instead of $k$, but the form is consistent with the derived condition. The correct option is D. 3. The value of $\binom{n}{n-3}$ is We use the property of binomial coefficients: $\binom{n}{r} = \binom{n}{n-r}$. Applying this property to $\binom{n}{n-3}$: $$\binom{n}{n-3} = \binom{n}{n-(n-3)}$$ $$\binom{n}{n-3} = \binom{n}{3}$$ Now, expand $\binom{n}{3}$: $$\binom{n