JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer 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 \[\left( x,y \right)\]is parallel to the line segment joining the points \[\left( 1,\left( 1 \right) \right)\]and \[\left( 1,\left( 1 \right) \right),\]then S is equal to : [JEE Main 9-4-2019 Morning]

    A) \[\left\{ -\frac{1}{3},-1 \right\}\]                     

    B) \[\left\{ \frac{1}{3},-1 \right\}\]

    C) \[\left\{ -\frac{1}{3},1 \right\}\]

    D) \[\left\{ \frac{1}{3},1 \right\}\]

    Correct Answer: C

    Solution :

    \[\left( 1 \right)=112=2\] \[\left( 1 \right)=11+2=0\] \[m=\frac{f(1)-f(-1)}{1+1}=\frac{-2-0}{2}=-1\] \[\frac{dy}{dx}=3{{x}^{2}}-2x-2\] \[3{{x}^{2}}-2x-2=-1\] \[\Rightarrow 3{{x}^{2}}-2x-1=0\] \[\Rightarrow (x-1)(3x+1)=0\] \[\Rightarrow x=1,-\frac{1}{3}\]                   


You need to login to perform this action.
You will be redirected in 3 sec spinner