Question:

From $p(x) = x^2 - 2x + 3$, $\alpha + \beta = 2$ and $\alpha \beta = 3$. [0.5 Mark]
(i) For zeroes $\alpha + 2$ and $\beta + 2$:
New sum $S_1 = (\alpha + 2) + (\beta + 2) = (\alpha + \beta) + 4 = 2 + 4 = 6$.
New product $P_1 = (\alpha + 2)(\beta + 2) = \alpha \beta + 2(\alpha + \beta) + 4 = 3 + 2(2) + 4 = 11$.
Polynomial $q_1(x) = x^2 - 6x + 11$. [2.0 Marks]
(ii) For zeroes $\gamma = \dfrac{\alpha - 1}{\alpha + 1}$ and $\delta = \dfrac{\beta - 1}{\beta + 1}$:
New sum $S_2 = \dfrac{\alpha - 1}{\alpha + 1} + \dfrac{\beta - 1}{\beta + 1} = \dfrac{(\alpha - 1)(\beta + 1) + (\beta - 1)(\alpha + 1)}{(\alpha + 1)(\beta + 1)} = \dfrac{2\alpha \beta - 2}{\alpha \beta + (\alpha + \beta) + 1} = \dfrac{2(3) - 2}{3 + 2 + 1} = \dfrac{4}{6} = \dfrac{2}{3}$.
New product $P_2 = \dfrac{(\alpha - 1)(\beta - 1)}{(\alpha + 1)(\beta + 1)} = \dfrac{\alpha \beta - (\alpha + \beta) + 1}{\alpha \beta + (\alpha + \beta) + 1} = \dfrac{3 - 2 + 1}{3 + 2 + 1} = \dfrac{2}{6} = \dfrac{1}{3}$.
Polynomial $q_2(x) = k\left(x^2 - \dfrac{2}{3}x + \dfrac{1}{3}\right) \Rightarrow 3x^2 - 2x + 1$. [2.5 Marks]

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🎯 Official CBSE Marking Scheme:
Original sum and product ($\alpha+\beta=2, \alpha\beta=3$): 0.5 Mark
Part (i) New sum, product, and polynomial $x^2 - 6x + 11$: 2.0 Marks
Part (ii) New sum $S_2 = 2/3$: 1.0 Mark
Part (ii) New product $P_2 = 1/3$: 1.0 Mark
Part (ii) Final polynomial $3x^2 - 2x + 1$: 0.5 Mark