Step 1: Subtract the two given expectations using a difference of squares.
Rather than expanding both squares and solving them as a pair of equations, subtract them directly:
\[
E(X-1)^2 - E(X-2)^2 = E\left[(X-1)^2 - (X-2)^2\right]
\]
Using the identity \( a^2 - b^2 = (a-b)(a+b) \) with \( a = X-1 \), \( b = X-2 \):
\[
(X-1)^2 - (X-2)^2 = \big[(X-1)-(X-2)\big]\big[(X-1)+(X-2)\big] = (1)(2X-3) = 2X - 3
\]
Step 2: Take expectation of both sides.
\[
E(X-1)^2 - E(X-2)^2 = 2E(X) - 3
\]
Using the given values, \( 10 - 6 = 4 \):
\[
4 = 2E(X) - 3 \quad \Rightarrow \quad E(X) = \frac{7}{2} = \mu
\]
Step 3: Find \( E(X^2) \) using either original equation.
Expand \( E(X-2)^2 = E(X^2) - 4E(X) + 4 = 6 \):
\[
E(X^2) = 6 + 4\left(\frac{7}{2}\right) - 4 = 6 + 14 - 4 = 16
\]
Step 4: Compute the variance.
\[
\sigma^2 = E(X^2) - \mu^2 = 16 - \left(\frac{7}{2}\right)^2 = 16 - \frac{49}{4} = \frac{15}{4}
\]
So \( \mu = \dfrac{7}{2} \) and \( \sigma^2 = \dfrac{15}{4} \), which is option (D).