Question

Q14. If $\alpha ,\beta$ are the zeros of polynomial $f\left(x\right)$$=x^2-p\left(x+1\right)-c$ then $\left(\alpha +1\right)\left(\beta +1\right)$ is equal to:

• A. $c-1$
• B. $1-c$
• C. $c$
• D. $1+c$

Answer 1

The correct option is B

$1-c$

Solution:

Step 1: Relation between zeroes and Coefficient:

Given: $f\left(x\right)=x^2-p\left(x+1\right)-c$ with zeroes $\alpha and\beta .$

$f\left(x\right)=x^2-px-(p+c)$

$sumofzeroes:a+\beta =\frac{-(-p)}{1}=p\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{u}\mathbf{m}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{z}\mathbf{e}\mathbf{r}\mathbf{o}\mathbf{e}\mathbf{s}\mathbf{=}\frac{-coefficientofx}{coefficientof{x}^2}\mathbf{]}\mathbf{}$

$productofzeroes:\alpha \beta =\frac{-(p+c)}{1}=-(p+c)\mathbf{[}\mathbf{\because }\mathbf{p}\mathbf{r}\mathbf{o}\mathbf{d}\mathbf{u}\mathbf{c}\mathbf{t}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{z}\mathbf{e}\mathbf{r}\mathbf{o}\mathbf{e}\mathbf{s}\mathbf{=}\frac{\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{t}\mathbf{}\mathbf{t}\mathbf{e}\mathbf{r}\mathbf{m}}{\mathbf{c}\mathbf{o}\mathbf{e}\mathbf{f}\mathbf{f}\mathbf{i}\mathbf{c}\mathbf{i}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}{\mathbf{x}}^{\mathbf{2}}}\mathbf{]}$

Step 2: Finding the value of given expression:

$$\begin{gathered} \left(\alpha +1\right)\left(\beta +1\right)=\alpha \beta +\alpha +\beta +1 \\ =-\left(p+c\right)+p+1 \\ =-p-c+p+1 \\ =\mathbf{1}\mathbf{-}\mathit{c} \end{gathered}$$

Final answer: The correct answer is option(B).