(ii). The fith term in the expansion of \( \left(1-\frac{1}{x}\right)^{10} \) is : (a) \( \frac{210}{x^{4}} \) (b) \( \frac{x^{4}}{210} \) (c) \( \frac{1}{x^{4}} \) (d) None \( \square \times \square \) (iv). The middle term/s in the expansion of \( \left(x+\frac{1}{x}\right)^{10} \) is : (a) 504 (b) 252 (c) \( \frac{1}{x^{4}} \) (d) None (v). The value of \( i^{6} \), (where \( i=\sqrt{-1} \) ): (a) 1 (b) \( i \) (c) -1 (d) None (vi). The value of \( i^{6}+\omega^{6} \) : (a) 1 (b) 2 (c), 0 (d) None

Let's address each part of the question step-by-step. --- ### (ii). The fifth term in the expansion of \( \left(1-\frac{1}{x}\right)^{10} \) **Solution:** The general term in the binomial expansion of \( (a + b)^n \) is given by: \[ T_{k+1} = C(n, k) \cdot a^{n-k} \cdot b^k \] where \( C(n, k) \) is the binomial coefficient. Here, \( a = 1 \), \( b = -\frac{1}{x} \), and \( n = 10 \). For the **fifth term**, \( k = 4 \): \[ T_5 = C(10, 4) \cdot 1^{10-4} \cdot \left(-\frac{1}{x}\right)^4 = C(10, 4) \cdot \frac{1}{x^4} \] \[ C(10, 4) = 210 \] \[ T_5 = \frac{210}{x^4} \] **Answer:** **(a) \( \frac{210}{x^{4}} \)** --- ### (iv). The middle term/s in the expansion of \( \left(x+\frac{1}{x}\right)^{10} \) **Solution:** For an expansion of \( (a + b)^{n} \), if \( n \) is even, there are two middle terms; if odd, there's one middle term. Here, \( n = 10 \) (even), so there are two middle terms. However, considering the symmetry in \( \left(x + \frac{1}{x}\right)^{10} \), both middle terms will be the same. The general term is: \[ T_{k+1} = C(10, k) \cdot x^{10-k} \cdot \left(\frac{1}{x}\right)^k = C(10, k) \cdot x^{10-2k} \] For the middle term: \[ k = 5 \] \[ T_6 = C(10, 5) \cdot x^{0} = C(10, 5) \] \[ C(10, 5) = 252 \] **Answer:** **(b) 252** --- ### (v). The value of \( i^{6} \), where \( i = \sqrt{-1} \) **Solution:** Powers of \( i \) cycle every 4: \[ i^1 = i \\ i^2 = -1 \\ i^3 = -i \\ i^4 = 1 \\ i^5 = i \\ i^6 = -1 \] **Answer:** **(c) -1** --- ### (vi). The value of \( i^{6} + \omega^{6} \) **Assumption:** Assuming \( \omega \) is a primitive cube root of unity, i.e., \( \omega^3 = 1 \). **Solution:** From part (v): \[ i^6 = -1 \] For \( \omega^6 \): \[ \omega^6 = (\omega^3)^2 = 1^2 = 1 \] Therefore: \[ i^6 + \omega^6 = -1 + 1 = 0 \] **Answer:** **(c) 0** --- ### Summary of Answers - **(ii)** (a) - **(iv)** (b) - **(v)** (c) - **(vi)** (c)