JEE Main 2019 — Application Of Derivatives Question with Solution

Given ƒ(x) = x³ – x² – 2x

$$\therefore$$ ƒ(1) = 1 – 1 – 2 = - 2

and ƒ(-1) = -1 – 1 + 2 = 0

So point A(1, ƒ(1)) = (1, -2)

and point B(–1, ƒ(–1)) = (-1, 0)

Slope of tangent at point (x, y) to the curve

y = ƒ(x) = x³ – x² – 2x is

$$\frac{dy}{dx}$$ = 3x² - 2x - 2

Slope of line segment joining the points (1, -2) and (–1, 0) is

= $$\frac{-2-0}{1-\left(-1\right)}$$

As tangent to the curve and line segment both are parallel then slope of them are same.

$$\therefore$$ 3x² - 2x - 2 = $$\frac{-2-0}{1-\left(-1\right)}$$

$$\Rightarrow$$ 3x² - 2x - 1 = 0

$$\Rightarrow$$ x = $$\frac{2\pm \sqrt{4-4.3\left(-1\right)}}{2.3}$$

= $$\frac{2\pm 4}{6}$$

$$\therefore$$ x = 1, $$-\frac{1}{3}$$

So, S = $$\left\{-\frac{1}{3},1\right\}$$