A) \[b=1,\,\,a=0\]
B) \[b=1,\,\,a=0\]
C) \[b=-1,\,\,a=0\]
D) \[a=-1,\,\,b=0\]
Correct Answer: B
Solution :
Clearly,\[f(2)=-1\] \[\Rightarrow \] \[-1=\frac{2a+b}{(2-1)(2-4)}\Rightarrow 2a+b=2\] Now, \[f(x)=\frac{4a+5b+2bx-a{{x}^{2}}}{{{(x-1)}^{2}}{{(x-4)}^{2}}}\] \[f(2)=0\] \[\Rightarrow \] \[b=0\] \[\Rightarrow \] \[a=1\] \[\Rightarrow \,\,\,\,\,f(x)\,=-\frac{(x-2)(x+2)}{{{(x-1)}^{2}}{{(x-4)}^{2}}}\] Clearly, for\[x>2,\,\,f(x)<0\]and for\[x<2\]\[f(x)>0\]. Thus,\[x=2\]is indeed the point of local maxima for\[y=f(x)\]
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