Question

Let $y = f(x)$ be the real valued function defined on the interval $(0, \infty)$, satisfying $f(1) = 0$ and the differential equation \[ x \frac{dy}{dx} = y - x^3. \] Then which of the following statements is (are) TRUE?

• A. The function $f$ has a local minimum at $x = \frac{1}{\sqrt{3}}$
• B. The function $f$ has a local maximum at $x = \frac{1}{\sqrt{3}}$
• C. The function $f$ is increasing in the interval $(1, 2)$
• D. If $g(x) = 4x^3 - 5x^2 + \frac{3}{2}x$ for $x > 0$, then the number of elements in the set $\{ x \in (0, \infty) : f(x) = g(x) \}$ is 2}

Hint:

For linear differential equations of the form $x y' \pm y = \dots$, usually the Integrating Factor is simple ($x$ or $1/x$). Always substitute the initial condition immediately to keep calculations manageable.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We first need to solve the linear differential equation to find $f(x)$, then use calculus to analyze its extrema and monotonicity, and finally find intersection points with another function $g(x)$.

Step 2: Key Formula or Approach:

• Solve $x y' - y = -x^3$ using the Integrating Factor (I.F.) method.

• $f'(x) = 0$ and $f''(x)$ to determine local maxima/minima.

• Solve $f(x) = g(x)$ for positive real $x$.

Step 3: Detailed Explanation:

• Rewrite as $y' - \frac{1}{x}y = -x^2$. I.F. $= e^{\int -1/x dx} = 1/x$.

• Solution: $y \cdot \frac{1}{x} = \int (-x^2) \cdot \frac{1}{x} dx = -\frac{x^2}{2} + C \implies y = -\frac{x^3}{2} + Cx$.

• Given $f(1) = 0 \implies 0 = -1/2 + C \implies C = 1/2$.

• So, $f(x) = \frac{x - x^3}{2}$.

• Checking extrema: $f'(x) = \frac{1 - 3x^2}{2}$. $f'(x) = 0 \implies x = \frac{1}{\sqrt{3}}$ (since $x > 0$).
$f''(x) = -3x$. At $x = 1/\sqrt{3}$, $f'' < 0$, so it's a local maximum. Statement (B) is TRUE, (A) is FALSE.

• Checking (C): For $x \in (1, 2)$, $x^2 > 1 \implies 1 - 3x^2 < -2$, so $f'(x) < 0$. The function is decreasing. Statement (C) is FALSE.

• Checking (D): $f(x) = g(x) \implies 0.5x - 0.5x^3 = 4x^3 - 5x^2 + 1.5x$.
\[ 4.5x^3 - 5x^2 + x = 0 \implies x(4.5x^2 - 5x + 1) = 0. \] Since $x > 0$, we check the quadratic $4.5x^2 - 5x + 1 = 0 \implies 9x^2 - 10x + 2 = 0$.
Discriminant $D = 100 - 4(9)(2) = 100 - 72 = 28 > 0$. Both roots are positive ($\frac{10 \pm \sqrt{28}}{18}$).
Thus, there are 2 elements in the set. Statement (D) is TRUE.

Step 4: Final Answer:
The true statements are (B) and (D).