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 :
Given, \[f(x)=\left\{ \begin{matrix} a{{x}^{2}}+b, & b\ne 0,\,\,x\le 1 \\ b{{x}^{2}}+ax+c, & x>1 \\ \end{matrix} \right.\] \[\Rightarrow \] \[f(x)=\left\{ \begin{matrix} 2ax, & b\ne 0,\,\,x\le 1 \\ 2bx+a, & x>1 \\ \end{matrix} \right.\] Since,\[f(x)\]is continuous at\[x=1\] \[\therefore \] \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)\] \[\Rightarrow \] \[a+b=b+a+c\] \[\Rightarrow \] \[c=0\] Also,\[f(x)\]is differentiable at\[x=1\]. \[\therefore (LHD\,\,at\,\,x=1)=(RHD\,\,at\,\,x=1\] \[\Rightarrow \] \[2a=2b(1)+a\] \[\Rightarrow \] \[a=2b\]
You need to login to perform this action.
You will be redirected in
3 sec
