JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Area Under Curves

The area of the region enclosed by the curve

y = x^{3}

and its tangent at the point

(- 1, - 1)

Answer: (A)

Solution

The tangent to

y = x^{3}

(- 1, - 1)

has slope

\frac{d y}{d x}_{x = - 1} = 3 x^{2}_{x = - 1} = 3

. So the tangent line is

y + 1 = 3 (x + 1)

, i.e.,

y = 3 x + 2

. To find where the tangent intersects the curve again, solve

x^{3} = 3 x + 2

x^{3} - 3 x - 2 = 0

. Factoring gives

(x + 1)^{2} (x - 2) = 0

, so the intersection points are

x = - 1

(tangent point) and

x = 2

. The area enclosed is

\int_{- 1}^{2} [(3 x + 2) - x^{3}] d x = [\frac{3 x ^{2}}{2} + 2 x - \frac{x ^{4}}{4}]_{- 1}^{2} = (6 + 4 - 4) - (\frac{3}{2} - 2 - \frac{1}{4}) = 6 - (- \frac{3}{4}) = \frac{27}{4} .

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Single correct option