In the given circuit, charge $$Q_2$$ on the 2 $$\mu$$F capacitor changes as C is varied from 1 $$\mu$$F to 3 $$\mu$$F. $$Q_2$$ as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale)

$$C_p = 1\ \mu\text{F} + 2\ \mu\text{F} = 3\ \mu\text{F}$$

$$C_{eq} = \frac{C \cdot C_p}{C + C_p} = \frac{3C}{C + 3}\ \mu\text{F}$$

$$Q_{total} = C_{eq} \cdot E = \frac{3CE}{C + 3}$$

$$Q_2 = \left( \frac{C_2}{C_1 + C_2} \right) Q_{total} = \left( \frac{2}{1 + 2} \right) Q_{total} = \frac{2}{3} Q_{total}$$

$$Q_2(C) = \frac{2}{3} \left( \frac{3CE}{C + 3} \right) = \frac{2CE}{C + 3}$$

$$Q_2 = 2E \left( \frac{C}{C + 3} \right) = 2E \left( 1 - \frac{3}{C + 3} \right)$$

$$\frac{dQ_2}{dC} = \frac{6E}{(C + 3)^2}$$ (always positive)

$$\frac{d^2Q_2}{dC^2} = -\frac{12E}{(C + 3)^3}$$ (always negative because capacitance is always positive)

Graph C correctly shows an increasing curve that is concave downwards.