A) (2, 3)
B) (1, 2)
C) (0, 1)
D) (1, 3)
Correct Answer: C
Solution :
Let\[f(x)=a{{x}^{2}}+bx+c,\] then \[f(x)=\frac{a{{x}^{3}}}{3}+\frac{b{{x}^{2}}}{2}+cx+k\] \[=\frac{2a{{x}^{3}}+3b{{x}^{2}}+6cx+6k}{6}\] \[\Rightarrow \] \[f(1)=\frac{2a+3b+6c+6k}{6}=\frac{6k}{6}=k\] Also, \[f(0)=\frac{6k}{6}=k\] \[\therefore \] \[f(0)=f(1)=k\] \[\Rightarrow \] \[f(x)=0\]You need to login to perform this action.
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