A) \[1\]
B) \[-1\]
C) \[0\]
D) \[2\]
Correct Answer: D
Solution :
| [d] Let \[f(x)={{x}^{3}}-6{{x}^{2}}+3x+10\] |
| \[\therefore \] Sum of zeroes \[=-\frac{(-6)}{1}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,a+(a+b)+(a+2b)=6\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3a+3b=6\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a+b=2\] |
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