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

  • Option 1)

     \left \{ \frac{1}{3}, 1\right \}     

  • Option 2)

    \left \{ -\frac{1}{3},-1 \right \}    

  • Option 3)

    \left \{ \frac{1}{3},-1 \right \}

  • Option 4)

    \left \{- \frac{1}{3},1 \right \}

 

Answers (1)

y=f\left ( x \right )=x^{3}-x^{2}-2x

\frac{dy}{dx}=3x^{2}-2x-2

Slope of line segment joining

\left ( 1,f\left ( 1 \right ) \right ) and \left ( -1,f\left ( -1 \right ) \right ) is 

m=\frac{f\left ( -1 \right )-f\left ( 1 \right )}{-1-1}

This is equal to \frac{dy}{dx}

f\left ( -1 \right )=\left ( -1 \right )^{3}-\left ( -1 \right )^{2}-2\left ( -1 \right )=0

f\left ( -1 \right )=\left ( 1 \right )^{3}-\left ( 1 \right )^{2}-2\left ( 1 \right )=-2

3x^{2}-2x-2=-1

3x^{2}-2x-1=0

x=1,x=-\frac{1}{3}

\left \{ 1,-\frac{1}{3} \right \}


Option 1)

 \left \{ \frac{1}{3}, 1\right \}     

Option 2)

\left \{ -\frac{1}{3},-1 \right \}    

Option 3)

\left \{ \frac{1}{3},-1 \right \}

Option 4)

\left \{- \frac{1}{3},1 \right \}

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