A) -2
B) 2
C) -1
D) 1
Correct Answer: C
Solution :
| Given, \[\alpha \] and \[\beta \]are the zeroes of polynomial |
| \[{{x}^{2}}-p\left( x+1 \right)+c\]which can be written as |
| \[{{x}^{2}}-px+c-p\]. |
| So, sum of zeroes, \[\alpha +\beta =p\] ... (i) |
| \[\left[ \because \,\,sum\,of\,zeroes=\frac{-\,coefficient\,of\,x}{coefficient\,\,of\,\,{{x}^{2}}} \right]\] |
| and product of zeroes, \[\alpha \beta =c-p\] (ii) |
| \[\left[ \because \,\,product\,of\,zeroes=\frac{cons\tan t\,term}{coefficient\,of\,{{x}^{2}}} \right]\] |
| Also,\[\left( \alpha +1 \right)\left( \beta +1 \right)=0\] [given] |
| \[\Rightarrow \,\,\alpha \beta +\left( \alpha +\beta \right)+1=0\] |
| \[\Rightarrow \,\,\alpha \beta +\left( \alpha +\beta \right)+1=0\]0 |
| [from Eqs. (i) and (ii)] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,c=-1\] |
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