A) \[a=b=1\]
B) \[a=b=0\]
C) \[a=1,b=0\]
D) \[a=b=2\]
Correct Answer: C
Solution :
[c] \[y=\frac{ax+b}{(x-1)(x-4)}=\frac{ax+b}{{{x}^{2}}-5x+4}\] has turning point at \[P(2,-1)\]. Thus, \[P(2,-1)\] lies on the curve. Therefore, \[2a+b=2...(1)\] Also, \[\frac{dy}{dx}=0\] at \[P(2,-1)\]. Now, \[\frac{dy}{dx}=\frac{a({{x}^{2}}-5x+4)-(2x-5)(ax+b)}{{{({{x}^{2}}-5x+4)}^{2}}}\]. At \[P(2,-1),\frac{dy}{dx}=\frac{-2a+2a+b}{4}=0\] or \[b=0\] or \[a=1\] [From equation (1)].
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