Question

If the tangent at $x=c$ to the curve $y=x^3-5x^2-3x,1\le x\le 3,$ 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 $y=x^3-5x^2-3x$
$\frac{dy}{dx}=3x^2-10x-3$
${\left(\frac{dy}{dx}\right)}_{x=e}=3c^2-10c-3$
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
$y=x^3-5x^2-3x$
Therefore, their slopes are equal $\therefore$
$-10=3c^2-10c-3$
$\Rightarrow$ $3c^2-10c+7=0$
$\Rightarrow$ $3c^2-3c-7c+7=0$
$\Rightarrow$ $3c(c-1)-7(c-1)=0$
$\Rightarrow$ $(3c-7)(c-1)=0$
$\Rightarrow$ $c=\frac{3}{3},1$