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Let S be the set of all values of x for ...

Let S be the set of all values of x for which the tangent to the curve `y=f(x)=x^(3)-x^(2)-2x` at `(x, y)` is parallel to the line segment joining the points `(1, f(1))` and `(-1, f(-1))`, then S is equal to :

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The correct Answer is:
3
The point `(1, f(1))=(1,-2)` and `(-1,f(-1))=(-1,0)`
So, slope of line joining the pts `=(-2)/(2)=-1`
Again, slope of tangent to `y=x^(3)-x^(2)-2x` is `(dy)/(dx)=3x^(2)-2x-2`
`rArr 3x^(2)-2x-2=-1 rArr 3x^(2)-2x-1=0 rArr x=(2pm sqrt(4+12))/(6)=-1/3, 1`
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