JEE Mathematics 2023 Question with Solution

Given: The curve is $y = x^3$ and we need the area enclosed by the curve and its tangent at $(-1,-1)$.

Find: The required enclosed area.

The slope of the tangent to $y=x^3$ is obtained from

$$\frac{dy}{dx}=3x^2$$

At $x=-1$, the slope is $3$. Hence the tangent is

$$y+1=3(x+1)$$

So,

$$y=3x+2$$

Now find the other point of intersection of the tangent and the curve:

$$x^3=3x+2$$ $$x^3-3x-2=0$$

This gives the second intersection point as $(2,8)$.

Therefore, the bounded area is

$$\text{Area}=\int_{-1}^{2}(3x+2-x^3)\,dx$$

Evaluating,

$$\int_{-1}^{2}(3x+2-x^3)\,dx=\frac{27}{4}$$

Therefore, the area is $\frac{27}{4}$. So the correct option is A.

The solution labels the correct option as D, but the worked value is $\frac{27}{4}$, which matches option A.

The tangent line at $(-1,-1)$ is first found using slope $3x^2$ at $x=-1$.

$$m=3(-1)^2=3$$

Using point-slope form,

$$y+1=3(x+1)$$ $$y=3x+2$$

To find the enclosed region, solve the intersection of the line and curve:

$$x^3=3x+2$$

From the solution, the other intersection point is $(2,8)$. Over $x\in[-1,2]$, the line lies above the curve, so area is line minus curve:

$$\text{Area}=\int_{-1}^{2}(3x+2-x^3)\,dx$$

This evaluates to $\frac{27}{4}$.