Solveeit Logo
Login

Question

Question: Let S be the set of all values of \(x\) for which the tangent to the curve \(y=f(x)={{x}^{3}}-{{x}^{...

Let S be the set of all values of xx for which the tangent to the curve y=f(x)=x3−x2−2xy=f(x)={{x}^{3}}-{{x}^{2}}-2x at (x,y)(x,y) is parallel to the line segment joining the points (1,f(1))(1,f(1)) and (−1,f(−1))(-1,f(-1)), then S is equal to?
(a) \left\\{ \dfrac{-1}{3},-1 \right\\}
(b) \left\\{ \dfrac{1}{3},-1 \right\\}
(c) \left\\{ \dfrac{-1}{3},1 \right\\}
(d) \left\\{ \dfrac{1}{3},1 \right\\}

Explanation

Solution

Hint: Find the derivative of the given curve to determine the slope of its tangent. Calculate the slope of the line segment joining the two points by using the formula: slope=ΔyΔxslope=\dfrac{\Delta y}{\Delta x}. Equate the derivative of the curve with the slope of the line segment and solve the obtained quadratic equation to get the values of xx.

Complete step-by-step solution -
Two lines are said to be parallel when their slopes have the same value.
Now, let us come to the question. We have been provided with the curve, y=f(x)=x3−x2−2xy=f(x)={{x}^{3}}-{{x}^{2}}-2x. Slope of the tangent of a given curve yy at any particular point is given by dydx\dfrac{dy}{dx} at that particular point. Therefore, slope of the given curve is given as,
dydx=d(x3−x2−2x)dx ∴dydx=3x2−2x−2......................(i) \begin{aligned} & \dfrac{dy}{dx}=\dfrac{d({{x}^{3}}-{{x}^{2}}-2x)}{dx} \\\ & \therefore \dfrac{dy}{dx}=3{{x}^{2}}-2x-2......................(i) \\\ \end{aligned}
Also, the slope of a line segment joining two points is ‘the ratio of change in yyto the change in xx’ or mathematically, ΔyΔx\dfrac{\Delta y}{\Delta x}.
Now, f(1)=13−12−2×1=1−1−2=−2f(1)={{1}^{3}}-{{1}^{2}}-2\times 1=1-1-2=-2, and f(−1)=(−1)3−(−1)2−2×(−1)=−1−1+2=0f(-1)={{(-1)}^{3}}-{{(-1)}^{2}}-2\times (-1)=-1-1+2=0. Therefore, the two points are: (1,−2) and (−1,0)(1,-2)\text{ and }(-1,0). So, the slope of the line segment joining these two points will be,
slope=ΔyΔx=−2−01−(−1)=−22=−1..................(ii)slope=\dfrac{\Delta y}{\Delta x}=\dfrac{-2-0}{1-(-1)}=\dfrac{-2}{2}=-1..................(ii)
Now, for the tangent of the curve and the line segment to be parallel, their slopes must be equal.
∴3x2−2x−2=−1 3x2−2x−2+1=0 3x2−2x−1=0 \begin{aligned} & \therefore 3{{x}^{2}}-2x-2=-1 \\\ & 3{{x}^{2}}-2x-2+1=0 \\\ & 3{{x}^{2}}-2x-1=0 \\\ \end{aligned}
Splitting the middle term we get,
3x2−3x+x−1=0 3x(x−1)+1(x−1)=0 (3x+1)(x−1)=0 ∴x=1 or x=−13 \begin{aligned} & 3{{x}^{2}}-3x+x-1=0 \\\ & 3x(x-1)+1(x-1)=0 \\\ & (3x+1)(x-1)=0 \\\ & \therefore x=1\text{ or }x=\dfrac{-1}{3} \\\ \end{aligned}
Hence, option (c) is the correct answer.

Note: One can note that the value of f(1) and f(−1)f(1)\text{ and }f(-1) are determined by substituting the value of x=1 and −1x=1\text{ and }-1 respectively, in the function, y=f(x)y=f(x). You should keep in mind that, when two lines are parallel then their slopes are equal and when two lines are perpendicular then the product of their slopes is −1-1.