# 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) $\left ( 2,-2 \right )$

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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