Question

Let us consider a curve, $y = f(x)$ passing through the point $(-2, 2)$ and the slope of the tangent to the curve at any point $(x, f(x))$ is given by $f(x) + x f'(x) = x^2$. Then :

• A. $x^3 - 3xf(x) - 4 = 0$
• B. $x^2 + 2xf(x) - 12 = 0$
• C. $x^3 + xf(x) + 12 = 0$
• D. $x^2 + 2xf(x) + 4 = 0$

Hint:

Recognizing standard derivatives like $\frac{d}{dx}(xy) = x y' + y$ or $\frac{d}{dx}(x^n y) = x^n y' + nx^{n-1} y$ can save time during integration of differential equations.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The question provides a relationship involving the function and its derivative, which is essentially a differential equation. We need to solve this to find the equation of the curve.
Step 2: Detailed Explanation:
Given: $f(x) + x f'(x) = x^2$.
We can rewrite this as:
\[ y + x \frac{dy}{dx} = x^2 \]
Notice that $y + x \frac{dy}{dx}$ is the derivative of $xy$ with respect to $x$:
\[ \frac{d}{dx}(xy) = x^2 \]
Integrating both sides with respect to $x$:
\[ xy = \int x^2 \, dx \]
\[ xy = \frac{x^3}{3} + C \]
The curve passes through the point $(-2, 2)$:
\[ (-2)(2) = \frac{(-2)^3}{3} + C \]
\[ -4 = -\frac{8}{3} + C \]
\[ C = -4 + \frac{8}{3} = \frac{-12 + 8}{3} = -\frac{4}{3} \]
Substitute $C$ back into the equation:
\[ xy = \frac{x^3}{3} - \frac{4}{3} \]
\[ 3xy = x^3 - 4 \]
\[ x^3 - 3xy - 4 = 0 \]
Since $y = f(x)$, we have $x^3 - 3xf(x) - 4 = 0$.
Step 3: Final Answer:
The equation of the curve is $x^3 - 3xf(x) - 4 = 0$.