

$[\text{Cr(H}_2\text{O)}_6]^{3+} \text{ contains } \text{Cr}^{3+}: [\text{Ar}]3d^3 \cdot t_{2g}^3 e_g^0$
$[\text{Fe(H}_2\text{O)}_6]^{3+} \text{ contains } \text{Fe}^{3+}: [\text{Ar}]3d^5 \cdot t_{2g}^3 e_g^2$
$[\text{Ni(H}_2\text{O)}_6]^{2+} \text{ contains } \text{Ni}^{2+}: [\text{Ar}]3d^8 \cdot t_{2g}^6 e_g^2$
$[\text{V(H}_2\text{O)}_6]^{3+} \text{ contains } \text{V}^{3+}: [\text{Ar}]3d^2 \cdot t_{2g}^2 e_g^0$
In the circuit shown, assuming the threshold voltage of the diode is negligibly small, then the voltage \( V_{AB} \) is correctly represented by:
 
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: