(a) By using $\mu = - \frac{\mathrm{\Delta}V_{p}}{\mathrm{\Delta}V_{g}} = r_{p} \times g_{m}$ $\Rightarrow 7 \times 10^{3} \times 2.5 \times 10^{- 3} = - \frac{50}{\mathrm{\Delta}V_{g}} \Rightarrow \mathrm{\Delta}V_{g} = - 2.86$
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