A) \[a=1,b=-1\]
B) \[a=-1,b=1\]
C) \[a=-1,b\text{ }=1+\sqrt{2}\]
D) \[a=-1,b=\sqrt{2}-1\]
Correct Answer: B
Solution :
Given,\[f(x)=\left\{ \begin{matrix} {{x}^{2}}/a & , & 0\le x<1 \\ a & , & 1\le x<\sqrt{2} \\ (2{{b}^{2}}-4b)/{{x}^{2}} & , & \sqrt{2}\le x<\infty \\ \end{matrix} \right.\] At\[x=1,f(x)\]is continuous. \[\therefore \] \[f(1-0)=f(1)=f(1+0)\] \[\Rightarrow \] \[\frac{1}{a}=a\] \[\Rightarrow \] \[{{a}^{2}}=1\] \[\Rightarrow \] \[a=\pm 1\] At\[x=\sqrt{2},f(x)\]is continuous. \[\therefore \] \[f(\sqrt{2}-0)=f(\sqrt{2})=f(\sqrt{2}+0)\] \[\Rightarrow \] \[a=\frac{2{{b}^{2}}-4b}{2}\] \[\Rightarrow \] \[a={{b}^{2}}-2b\] When\[a=-1,\]then \[{{b}^{2}}-2b+1=0\] \[\Rightarrow \] \[{{(b-1)}^{2}}=0\] \[\Rightarrow \] \[b=1\] \[\therefore \] \[a=-1,b=1\] When\[a=1,\]then \[{{b}^{2}}-2b-1=0\] \[\Rightarrow \] \[b=1\pm \sqrt{2}\]
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