QJEEM23Ma6705: JEE Main 2023 Mathematics - Class 12 JEEM23Ma6705 Solution: The area of the region enclosed by the curve and its tangent at …

Given $y=x^3$

$\begin{aligned}& \Rightarrow \frac{d y}{d x}=3 x^2 \\\\& \left(\frac{d y}{d x}\right)_{(-1,-1)}=3\end{aligned}$

Equation of tangent at $(-1,-1)$

$\begin{aligned}& (y+1)=3(x+1) \\\\& y=3 x+2\end{aligned}$

Solving (i) and (ii)

$x^3=3 x+2$

We can rewrite this as $x^3 - 3x - 2 = 0$ . This equation will give us the x-coordinates of the points where the line $y = 3x + 2$ intersects the curve $y = x^3$ . We already know that one point of intersection is $(-1, -1)$ (as it's a point of tangency).

To find the other points of intersection, we can solve the cubic equation $x^3 - 3x - 2 = 0$ .

Let's factor this equation:

$x^3 - 3x - 2 = 0 \Rightarrow (x + 1)(x + 1)(x - 2) = 0$ .

Setting each factor equal to zero gives the solutions $x = -1, -1, 2$ .

$\therefore$ Other point is $Q(2,8)$ .

We have the required area as $\int_{-1}^{2} (3x + 2 - x^3) \, dx = \left[ \frac{3}{2}x^2 + 2x - \frac{1}{4}x^4 \right]_{-1}^{2}.$

Evaluating this at $x = 2$ and $x = -1$ gives

$\left[ \frac{3}{2} \cdot (2)^2 + 2 \cdot 2 - \frac{1}{4} \cdot (2)^4 \right] - \left[ \frac{3}{2} \cdot (-1)^2 + 2 \cdot (-1) - \frac{1}{4} \cdot (-1)^4 \right]$

$= \left[ 6 + 4 - 4 \right] - \left[ \frac{3}{2} - 2 - \frac{1}{4} \right]$

$= 6 - \left( \frac{1}{4} \right)$

$= \frac{27}{4}$