Polynomials
NCERT Exemplar Ch 02
CBSE_NCERT_EXEMPLAR_CH02
Grade 10
Question:
If $\alpha, \beta$ are the zeroes of $p(x) = x^2 - p(x + 1) - c$, then $(\alpha + 1)(\beta + 1)$ is equal to:
$c$
$c - 1$
$1 - c$
$1 + c$
Step-by-Step Solution
Key Concept: Rewrite $p(x) = x^2 - px - (p + c)$, find $\alpha + \beta$ and $\alpha \beta$, then expand $(\alpha+1)(\beta+1)$.
Stepwise Solution:
$p(x) = x^2 - px - (p + c)$. Here $\alpha + \beta = p$ and $\alpha \beta = -(p + c)$. [0.5 Mark]
$(\alpha + 1)(\beta + 1) = \alpha \beta + (\alpha + \beta) + 1 = -(p + c) + p + 1 = -p - c + p + 1 = 1 - c$. [0.5 Mark]
Marking Scheme:
• Finding sum and product of zeroes: 0.5 Mark
• Expanding and simplifying $(\alpha+1)(\beta+1)$: 0.5 Mark
Correct Answer: $1 - c$
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