MT-2
Grade Class 10

Question:

<p>If &alpha;, &beta; are the zeros of polynomial&nbsp;f(x) =&nbsp;x<sup>2</sup>&nbsp;&minus;&nbsp;p&nbsp;(x&nbsp;+ 1) &minus;&nbsp;c, then (&alpha; + 1) (&beta; + 1) =</p>
<p style="display:inline">c − 1</p>
<p style="display:inline">c</p>
<p style="display:inline">1 − c</p>
<p style="display:inline">1 + c</p>

Step-by-Step Solution

Key Concept: Express the polynomial in standard form to accurately determine the sum and product of zeros using Vieta's formulas.
<p>Since&nbsp;<span class="math-tex">\(\alpha\)</span>&nbsp;and&nbsp;<span class="math-tex">\(\beta\)</span>&nbsp;are the zeros of quadratic polynomial&nbsp;<span class="math-tex">\(f(x)=x^{2}-p(x+1)-c\)</span><br /> <span class="math-tex">\(=x^{2}-p x-p-c\)</span><br /> <span class="math-tex">\(\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\)</span><br /> <span class="math-tex">\(=-\left(\frac{-p}{1}\right)\)</span>&nbsp;= p<br /> <span class="math-tex">\(\alpha \times \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)</span><br /> <span class="math-tex">\(=\frac{-p-c}{1}\)</span>&nbsp;=&nbsp;<span class="math-tex">\(-p-c\)</span><br /> We have<br /> <span class="math-tex">\((\alpha+1)(\beta+1)\)</span><br /> <span class="math-tex">\(=\alpha \beta+\beta+\alpha+1\)</span><br /> <span class="math-tex">\(=\alpha \beta+(\alpha+\beta)+1\)</span><br /> <span class="math-tex">\(=-p-c+(p)+1\)</span><br /> <span class="math-tex">\(=-c+1\)</span><br /> = 1 - c<br /> The value of&nbsp;<span class="math-tex">\((\alpha+1)(\beta+1)\)</span>&nbsp;is 1 - c.</p>
Correct Answer: C

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