A) \[b-a+1\]
B) \[b-a-1\]
C) \[a-b+1\]
D) \[a-b-1\]
Correct Answer: A
Solution :
| [a] Let \[p(x)={{x}^{3}}+a{{x}^{2}}+bx+c\] |
| Let \[\alpha ,\beta \] and \[\gamma \] be the zeroes of \[p(x),\] where \[\alpha =-1\] |
| Now, \[p(-1)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,{{(-1)}^{3}}+a{{(-1)}^{2}}+b(-1)+c=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,-1+a-b+c=0\Rightarrow c=1-a+b\] |
| We know that product of all zeroes |
| \[=-\frac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{3}}}}=-\frac{c}{1}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\alpha \beta \gamma =-c\Rightarrow (-1)\beta \gamma =-c\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\beta \gamma =c\Rightarrow \beta \gamma =1-a+b\] |
You need to login to perform this action.
You will be redirected in
3 sec
