We have $v =\displaystyle \frac{1}{n} \sum_{i=1}^n (x_i - \mu)^2 $
$\qquad =\displaystyle \frac{1}{n} \sum_{i=1}^n \left[x_i^2 - 2x_i\mu+ \mu^2\right] $
$\qquad = \displaystyle \frac{1}{n} \sum_{i=1}^n x_i^2 - 2\mu\frac{1}{n}\sum_{i=1}^n x_i+ \mu^2 $
$\qquad = \displaystyle \frac{1}{n} \sum_{i=1}^n x_i^2 - 2\mu^2+ \mu^2 $
$\qquad = \displaystyle \frac{1}{n} \sum_{i=1}^n x_i^2 - \mu^2 $
Thus,
- $v_A = \displaystyle \frac{1}{n_A} \left(\sum_{i=1}^{n_A} x_i^2\right) - \mu_A^2 \quad \to (1)$
- $v_B = \displaystyle \frac{1}{n_B} \left(\sum_{i=1}^{n_B} x_i^2\right) - \mu_B^2 \quad \to (2)$
$v = \displaystyle \frac{1}{n} \left(\sum_{i=1}^n x_i^2\right) - \mu^2$
$\qquad = \displaystyle \frac{1}{n} \left(\sum_{i=1}^{n_A} x_i^2+\sum_{i=1}^{n_B} x_i^2\right) - \mu^2$
$\qquad = \displaystyle \frac{1}{n} \left(n_A( v_A + \mu_A^2)+n_B( v_B + \mu_B^2)\right) - \mu^2$ (From $(1)$ and $(2))$
$\qquad = \displaystyle \frac{n_A}{n} \left(v_A + \mu_A^2 - \mu^2\right)+\frac{n_B}{n}\left( v_B + \mu_B^2-\mu^2\right) $ (Since, $n_A+n_B = n)$
$\qquad = \displaystyle p[v_A + \mu_A^2 - \mu^2] +(1-p)[v_B +\mu_B^2 - \mu^2]$ (Since, $\frac{n_A}{n} = p)$
$\qquad = \displaystyle p[v_A + (\mu_A - \mu)^2 + 2\mu\mu_A -2\mu^2] +(1-p)[v_B +(\mu_B - \mu)^2+2\mu\mu_B-2\mu^2]$
$\qquad = \displaystyle p[v_A + (\mu_A - \mu)^2 ] +(1-p)[v_B +(\mu_B - \mu)^2] + 2p\mu\mu_A + 2(1-p)\mu\mu_B -2\mu^2$
$\qquad = \displaystyle p[v_A + (\mu_A - \mu)^2 ] +(1-p)[v_B +(\mu_B - \mu)^2] + 2\mu\left[p\mu_A + (1-p)\mu_B\right]-2\mu^2$
$\qquad = \displaystyle p[v_A + (\mu_A - \mu)^2 ] +(1-p)[v_B +(\mu_B - \mu)^2] + 2\mu^2-2\mu^2$
$\qquad = \displaystyle p[v_A + (\mu_A - \mu)^2 ] +(1-p)[v_B +(\mu_B - \mu)^2] $
So, Correct option C.