A) a=2, b=3
B) a=3, b=2
C) a=-2, b=-3
D) a=-3, b=-2
Correct Answer: A
Solution :
[a] Derivative of \[f(x)=\left\{ \begin{matrix} a{{x}^{2}}+b & x<-1 \\ b{{x}^{2}}+ax+a & x\ge -1 \\ \end{matrix} \right.\] is \[f'(x)=\left\{ \begin{matrix} 2ax & x<-1 \\ 2bx+a, & x\ge -1 \\ \end{matrix} \right.\] If \[f'(x)\] is continuous everywhere then it is also continuous at \[x=-1\] \[{{\left. f'(x) \right|}_{x=-1}}=-2a=-2b+a\] or, \[3a=2b\] ? (i) From the given choice \[a=2,b=3\] satisfied this equation.
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