If the tangent to the curve,y=x^{3}+ax-b at the point (1,-5) is perpendicular to the line,-x+y+4=0, then which one of the following points lies on the curve?

  • Option 1)

     \left ( -2,1 \right )

  • Option 2)

    \left ( -2,2 \right )

  • Option 3)

    \left ( 2,-1 \right )

  • Option 4)

    \left ( 2,-2 \right )

 

Answers (1)

given,

    y=x^{3}+ax-b

 If passes via \left ( 1,-5 \right )

  \Rightarrow -5=1+a-b\Rightarrow a-b=-6............(1)

Slope of the line -x+y-4=0 is 1

Slope of tier to the line -x+y+4=0 is -1

  \frac{dy}{dx}=3x^{2}+a

  \Rightarrow 3+a=-1                   at x=1

  \Rightarrow a=-4...........(2)

from (1) and (2)

 a=-4,b=+2

equation of curve become 

 y=x^{3}-4x-2

     \left ( 2,-2 \right ) satisfies the equation

 


Option 1)

 \left ( -2,1 \right )

Option 2)

\left ( -2,2 \right )

Option 3)

\left ( 2,-1 \right )

Option 4)

\left ( 2,-2 \right )

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