Polynomials
RD Sharma
CBSE
Grade 10
Question:
If $\alpha, \beta$ are the zeroes of $f(x) = x^2 - p(x + 1) - c$, show that $(\alpha + 1)(\beta + 1) = 1 - c$.
Step-by-Step Solution
Key Concept: $f(x) = x^2 - px - (p + c)$. $\alpha + \beta = p, \alpha \beta = -(p + c)$.
$(\alpha + 1)(\beta + 1) = \alpha \beta + (\alpha + \beta) + 1$. [0.5 Mark]
$= -(p + c) + p + 1 = -p - c + p + 1 = 1 - c$. Proved! [1.5 Marks]
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🎯 Official CBSE Marking Scheme:
Expanding $(\alpha + 1)(\beta + 1)$: 0.5 Mark
Substituting sum and product to get $1 - c$: 1.5 Marks
Correct Answer:
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