A) \[c=0,\,a=2b\]
B) \[a=b,\,c\in R\]
C) \[a=b,c=0\]
D) \[a=b,c\ne 0\]
Correct Answer: A
Solution :
\[f(x)\]is continuous at \[x=1\] \[\therefore \] \[f(1)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)\] \[\Rightarrow \] \[a+b=b+a+c\] \[\Rightarrow \] \[c=0\] Also, \[f(x)\]is difference at \[x=1\] \[Lf(x)=Rf(x)\] \[\therefore \]\[\frac{d}{dx}(a{{x}^{2}}+b)=\frac{d}{dx}(b{{x}^{2}}+ax+c)\] \[\Rightarrow \]\[2ax=2bx+a\] At \[x=1\] \[\Rightarrow \]\[2a=2b+a\] \[\Rightarrow \]\[a=2b\]You need to login to perform this action.
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