A) \[a=2,\,b=-2\]
B) \[a=2,\,\,b=-4\]
C) \[a=2\,\,b=-8\]
D) \[a=4,\,b=-4\]
Correct Answer: C
Solution :
\[y=a{{x}^{2}}+bx\] \[\frac{dy}{dx}=2ax+b\Rightarrow {{\left( \frac{dy}{dx} \right)}_{(2,\,-8)}}=4a+b\] \[\because \] Tangent is parallel to x-axis \ \[\frac{dy}{dx}=0\Rightarrow b=-4a\] ?..(i) Now, point (2, ?8) is on the curve of \[y=a{{x}^{2}}+bx\] \ \[-8=4a+2b\] ??(ii) From (i) and (ii), we get \[a=2,\,\,b=-8\].You need to login to perform this action.
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