Question

If the tangent at $ x=c $ to the curve $ y={{x}^{3}}-5{{x}^{2}}-3x,\,1\le x\le 3, $ is parallel to the chord joining the points $ (1,-7) $ and $ (3,-27), $ then the value of $c$ is equal to

• A. $ 1 $
• B. $ \frac{5}{3} $
• C. $ \frac{7}{3} $
• D. $ \frac{2}{3} $

Answer 1

Answer: (A)

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 $