A) (-2, 2)
B) (2, -2)
C) (2, -1)
D) (-2, 1)
Correct Answer: B
Solution :
\[y={{x}^{3}}+ax-b\] (1, -5) lies on the curve \[\Rightarrow -5=1+a-b\Rightarrow a-b=-6\] ... (i) Also,\[y'=3{{x}^{2}}+a\] \[y{{'}_{\left( 1,5 \right)}}=3+a\] (slope of tangent) \[\because \]this tangent is \[\bot \]to \[x+y+4=0\] \[\Rightarrow (3+a)\](1) = -1 \[\Rightarrow a=-4\] ....(ii) By (i) and (ii) : \[a=-4,b=2\] \[\therefore \]\[y={{x}^{3}}-4x-2.\] (2,-2) lies on this curve.You need to login to perform this action.
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