Question

If 2x + 3|y| = 4y, then y as a function of x i.e. y = f(x), is -

• A. discontinuous at one point
• B. non-differentiable at one point
• C. discontinuous & non-differentiable at same point
• D. continuous & differentiable everywhere

Answer 1

Answer: (B)

non-differentiable at one point

Answer 2

To express $y$ as a function of $x$ from the equation $2x + 3|y| = 4y$, we consider two cases:

Case 1: $y \ge 0$

If $y \ge 0$, then $|y| = y$. The equation becomes:

$2x + 3y = 4y$ $2x = y$ So, $y = 2x$. This is valid when $x \ge 0$.

Case 2: $y < 0$

If $y < 0$, then $|y| = -y$. The equation becomes:

$2x + 3(-y) = 4y$ $2x - 3y = 4y$ $2x = 7y$ So, $y = \frac{2x}{7}$. This is valid when $x < 0$.

Combining these cases, the function $y = f(x)$ is defined as:

$$f(x) = \begin{cases} 2x & \text{if } x \ge 0 \\ \frac{2x}{7} & \text{if } x < 0 \end{cases}$$

Continuity Analysis:

The function $f(x)$ is a piecewise linear function. We check for continuity at $x=0$.

• $f(0) = 2(0) = 0$.

• $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{2x}{7} = 0$.

• $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 2x = 0$.

Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$, the function $f(x)$ is continuous at $x=0$ and therefore, continuous everywhere.

Differentiability Analysis:

We need to check the differentiability at $x=0$. The derivative of each piece is:

• For $x > 0$, $f'(x) = \frac{d}{dx}(2x) = 2$.

• For $x < 0$, $f'(x) = \frac{d}{dx}\left(\frac{2x}{7}\right) = \frac{2}{7}$.

Now, let's calculate the left-hand derivative (LHD) and right-hand derivative (RHD) at $x=0$.

• Left-hand derivative at $x=0$:

$f'(0^-) = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h}$

Since $h \to 0^-$, $h < 0$, so $f(h) = \frac{2h}{7}$ and $f(0)=0$.

$f'(0^-) = \lim_{h \to 0^-} \frac{\frac{2h}{7} - 0}{h} = \lim_{h \to 0^-} \frac{2}{7} = \frac{2}{7}$.

• Right-hand derivative at $x=0$:

$f'(0^+) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h}$

Since $h \to 0^+$, $h > 0$, so $f(h) = 2h$ and $f(0)=0$.

$f'(0^+) = \lim_{h \to 0^+} \frac{2h - 0}{h} = \lim_{h \to 0^+} 2 = 2$.

Since $f'(0^-) = \frac{2}{7}$ and $f'(0^+) = 2$, we have $f'(0^-) \ne f'(0^+)$. Therefore, the function $f(x)$ is not differentiable at $x=0$.

Conclusion: The function $y = f(x)$ is continuous everywhere but non-differentiable at one point ($x=0$).