A) \[1\]
B) \[0\]
C) \[-1\]
D) \[-\frac{1}{2}\]
Correct Answer: C
Solution :
[c] Given \[a+c=b\] |
Let \[f(x)=a{{x}^{2}}+bx+c\] |
put \[x=-1,\] |
\[\therefore \,\,\,\,\,\,\,\,\,f(-1)=a{{(-1)}^{2}}+b(-1)+c\] |
\[=a-b+c=(a+c)-b=b-b=0\] |
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