A) \[1\]
B) \[\frac{5}{3}\]
C) \[\frac{7}{3}\]
D) \[\frac{2}{3}\]
Correct Answer: A
Solution :
Given, curve \[y={{x}^{3}}-5{{x}^{2}}-3x\] \[\frac{dy}{dx}=3{{x}^{2}}-10x-3\] \[{{\left( \frac{dy}{dx} \right)}_{x=e}}=3{{c}^{2}}-10c-3\] Now, equation of chord joining the points \[(1,-7)\] and \[(3,-27)\] is \[(y+7)=-\frac{20}{2}\,(x-1)\] \[y+7=-10\,(x-1)\] \[y+7=-10x+10\] \[y=-10x-7+10\] \[y=-10x+3\] Since, the line \[y=-10x+3\] is parallel to the tangent to the curve \[y={{x}^{3}}-5{{x}^{2}}-3x\] Therefore, their slopes are equal \[\therefore \] \[-10=3{{c}^{2}}-10c-3\] \[\Rightarrow \] \[3{{c}^{2}}-10c+7=0\] \[\Rightarrow \] \[3{{c}^{2}}-3c-7c+7=0\] \[\Rightarrow \] \[3c\,(c-1)-7(c-1)=0\] \[\Rightarrow \] \[(3c-7)\,(c-1)=0\] \[\Rightarrow \] \[c=\frac{3}{3},1\]
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