If `4x+3|y|=5y`, then y as a function of x is

Options

A. differentiable at x=0
B. continuous at x=0
C. `(dy)/(dx)=2"for all x"`
D. none of these

Explanation

To solve the equation \(4x + 3|y| = 5y\) for \(y\) as a function of \(x\), we will analyze the equation based on the definition of the absolute value.

### Step 1: Analyze the absolute value
The expression \( |y| \) can be defined in two cases:
1. When \( y \geq 0 \), \( |y| = y \)
2. When \( y < 0 \), \( |y| = -y \)

### Step 2: Case 1 - When \( y \geq 0 \)
Substituting \( |y| = y \) into the equation:
\[ 4x + 3y = 5y \]
Rearranging gives:
\[ 4x = 5y - 3y \implies 4x = 2y \implies y = 2x \]
This is valid for \( y \geq 0 \), which implies \( 2x \geq 0 \) or \( x \geq 0 \).

### Step 3: Case 2 - When \( y < 0 \)
Substituting \( |y| = -y \) into the equation:
\[ 4x + 3(-y) = 5y \]
This simplifies to:
\[ 4x - 3y = 5y \implies 4x = 5y + 3y \implies 4x = 8y \implies y = \frac{4x}{8} = \frac{x}{2} \]
This is valid for \( y < 0 \), which implies \( \frac{x}{2} < 0 \) or \( x < 0 \).

### Step 4: Summary of results
From the two cases, we have:
- For \( x \geq 0 \), \( y = 2x \)
- For \( x < 0 \), \( y = \frac{x}{2} \)

Thus, we can express \( y \) as a piecewise function:
\[ y(x) = \begin{cases} 2x & \text{if } x \geq 0 \\ \frac{x}{2} & \text{if } x < 0 \end{cases} \]

### Step 5: Check for continuity and differentiability
To check for continuity at \( x = 0 \):
- Left-hand limit (LHL) as \( x \to 0^- \):
\[ \lim_{x \to 0^-} y = \frac{0}{2} = 0 \]
- Right-hand limit (RHL) as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} y = 2(0) = 0 \]
- Value at \( x = 0 \):
\[ y(0) = 2(0) = 0 \]

Since LHL = RHL = \( y(0) \), the function is continuous at \( x = 0 \).

### Step 6: Check for differentiability at \( x = 0 \)
- Derivative for \( x < 0 \):
\[ \frac{dy}{dx} = \frac{1}{2} \]
- Derivative for \( x \geq 0 \):
\[ \frac{dy}{dx} = 2 \]

Since the left-hand derivative (LHD) does not equal the right-hand derivative (RHD) at \( x = 0 \) (LHD = \( \frac{1}{2} \), RHD = \( 2 \)), the function is not differentiable at \( x = 0 \).

### Final Answer
Thus, \( y \) as a function of \( x \) is:
\[ y(x) = \begin{cases} 2x & \text{if } x \geq 0 \\ \frac{x}{2} & \text{if } x < 0 \end{cases} \]