If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - p(x + 1) - c$, show that $(\alpha + 1)(\beta + 1) = 1 - c$.
Step-by-Step Solution
Key Concept: Rewrite $f(x) = x^2 - px - (p + c)$, extract sum $\alpha+\beta$ and product $\alpha\beta$, expand LHS.
$f(x) = x^2 - px - (p + c)$. Here coefficient of $x^2$ is $1$, coeff of $x$ is $-p$, constant is $-(p+c)$.
$\alpha + \beta = p$ and $\alpha \beta = -(p + c)$. [1.0 Mark]
$\text{LHS} = (\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1 = -(p + c) + p + 1 = -p - c + p + 1 = 1 - c = \text{RHS}$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Standard polynomial form and finding $\alpha+\beta$, $\alpha\beta$: 1.0 Mark
Expanding LHS and simplifying to $1-c$: 1.0 Mark
Correct Answer: