A) \[a=2,\,\,b=-2\]
B) \[a=2,\,\,b=-4\]
C) \[a=2,\,\,b=-8\]
D) \[a=4,\,\,b=-4\]
Correct Answer: C
Solution :
Given,\[y=a{{x}^{2}}+bx\] On differentiating w.r.t. x, we get \[\frac{dy}{dx}=2ax+b\] At \[(2,\,\,-8),\]\[{{\left( \frac{dy}{dx} \right)}_{(2,\,\,-8)}}=4a+b\] \[\because \]Tangent is parallel to\[X-axis\]. \[\therefore \] \[\frac{dy}{dx}=0\Rightarrow b=-4a\] ... (i) Now, point \[(2,\,\,-8)\] is on the curve\[y=a{{x}^{2}}+bx.\] \[\therefore \] \[-8=4a+2b\] ? (ii) On solving Eqs. (i) and (ii), we get \[a=2,\,\,b=-8\]
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