MHTCET 20 Apr 2026 Shift 2 Paper

Concept:Powers of $i$ repeat in a cycle of 4, and $\frac{1}{i} = -i$. We simplify the bracket first, then cube it.Explanation:We know $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$.First simplify $i^{18}$:$i^{18} = i^{16+2} = i^{16} \cdot i^2$.Since $i^{16} = (i^4)^4 = 1$ and $i^2 = -1$, we get:$i^{18} = 1 \cdot (-1) = -1$.Now simplify $\left(\frac{1}{i}\right)^{25}$:$\frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$.So $\left(\frac{1}{i}\right)^{25} = (-i)^{25} = -i^{25}$, because the exponent is odd.Also, $i^{25} = i^{24+1} = (i^4)^6 \cdot i = 1 \cdot i = i$.Thus, $(-i)^{25} = -i$.Therefore, the expression inside the cube is:$i^{18} + \left(\frac{1}{i}\right)^{25} = -1 + (-i) = -1 - i$.Now compute the cube:$(-1-i)^3$.First square it:$(-1-i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$.Then multiply by $(-1-i)$:$(-1-i)^3 = 2i(-1-i) = -2i - 2i^2 = -2i - 2(-1) = 2 - 2i$.Answer:$\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^{3} = 2 - 2i$.Hence, the correct option is D.