A) \[(2,2),\left( -\frac{2}{3},-\frac{14}{27} \right)\]
B) \[(2,-2),\left( -\frac{2}{3},-\frac{14}{37} \right)\]
C) \[(2,-2),\left( -\frac{2}{3},-\frac{14}{27} \right)\]
D) None of the above:
Correct Answer: C
Solution :
Let\[({{x}_{1}},{{y}_{1}})\]be the required point. We have the given curve \[y={{x}^{3}}-2{{x}^{2}}-x\] ...(i) \[\frac{dy}{dx}=3{{x}^{2}}-4x-1\] \[{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},{{y}_{1}})}}=3x_{1}^{2}-4{{x}_{1}}-1\] This is the slope of the tangent to the curve but the tangent is parallel to the line \[y=3x-2\] \[\therefore \] \[3x_{1}^{2}-4{{x}_{1}}-1=3\] \[\Rightarrow \] \[3x_{1}^{2}-4{{x}_{1}}-4=0\] \[\Rightarrow \] \[({{x}_{1}}-2)(3{{x}_{1}}+2)=0\] \[\Rightarrow \] \[{{x}_{1}}=2,-\frac{2}{3}\] Since, the point\[({{x}_{1}},{{y}_{1}})\]lies on the curve (i), \[\therefore \]At\[{{x}_{1}}=2{{y}_{1}}={{2}^{3}}-2{{(2)}^{2}}-2=-2\] at\[{{x}_{1}}=-\frac{2}{3},{{y}_{1}}={{\left( -\frac{2}{3} \right)}^{3}}-2{{\left( -\frac{2}{3} \right)}^{2}}+\frac{2}{3}=-\frac{14}{27}\] Hence, the required points are \[(2,-2)\]and\[\left( -\frac{2}{3},-\frac{14}{27} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
