The function `f(x) = {{:(|2x-3|[x], x ge 1),(sin((pix)/2), x lt 1):}`

Options

A. is continuous at x = 2
B. is differentiable at x = 1
C. is continuous but not differentiable at x = 1
D. none of these

Answer

To solve the problem, we need to check the continuity and differentiability of the function defined as:

\[ f(x) = \begin{cases} |2x - 3| \cdot [x], & x \geq 1 \\ \sin\left(\frac{\pi x}{2}\right), & x < 1 \end{cases} \]

### Step 1: Check Continuity at \(x = 1\)

To check continuity at \(x = 1\), we need to verify the following condition:

\[ \text{Left Hand Limit} = f(1) = \text{Right Hand Limit} \]

#### Left Hand Limit

For \(x < 1\):

\[ \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} \sin\left(\frac{\pi (1 - h)}{2}\right) \]

Calculating this limit:

\[ = \sin\left(\frac{\pi}{2} - \frac{\pi h}{2}\right) = \cos\left(\frac{\pi h}{2}\right) \]

As \(h \to 0\):

\[ \cos\left(\frac{\pi h}{2}\right) \to \cos(0) = 1 \]

So,

\[ \text{Left Hand Limit} = 1 \]

#### Right Hand Limit

For \(x \geq 1\):

\[ \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1 + h) = \lim_{h \to 0} |2(1 + h) - 3| \cdot [1 + h] \]

Calculating this limit:

\[ = |2 + 2h - 3| \cdot 1 = |2h - 1| \cdot 1 \]

As \(h \to 0\):

\[ |2h - 1| \to |0 - 1| = 1 \]

So,

\[ \text{Right Hand Limit} = 1 \]

#### Function Value at \(x = 1\)

Now, we calculate \(f(1)\):

\[ f(1) = |2(1) - 3| \cdot [1] = |2 - 3| \cdot 1 = 1 \]

### Conclusion on Continuity

Since:

\[ \text{Left Hand Limit} = 1, \quad f(1) = 1, \quad \text{Right Hand Limit} = 1 \]

Thus, \(f(x)\) is continuous at \(x = 1\).

### Step 2: Check Differentiability at \(x = 1\)

To check differentiability, we need to find the left-hand derivative and right-hand derivative at \(x = 1\).

#### Left Hand Derivative

\[ f'(1^-) = \lim_{h \to 0} \frac{f(1 - h) - f(1)}{-h} = \lim_{h \to 0} \frac{\sin\left(\frac{\pi (1 - h)}{2}\right) - 1}{-h} \]

Using L'Hôpital's Rule since it is of the form \(0/0\):

\[ = \lim_{h \to 0} \frac{\cos\left(\frac{\pi (1 - h)}{2}\right) \cdot \left(-\frac{\pi}{2}\right)}{-1} = \frac{\pi}{2} \cdot \cos(0) = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2} \]

#### Right Hand Derivative

\[ f'(1^+) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0} \frac{|2(1 + h) - 3| \cdot [1 + h] - 1}{h} \]

Calculating this limit:

\[ = \lim_{h \to 0} \frac{|2 + 2h - 3| \cdot 1 - 1}{h} = \lim_{h \to 0} \frac{|2h - 1| - 1}{h} \]

As \(h \to 0\), this approaches:

\[ = \lim_{h \to 0} \frac{-1 - 1}{h} = -2 \]

### Conclusion on Differentiability

Since:

\[ f'(1^-) = \frac{\pi}{2} \quad \text{and} \quad f'(1^+) = -2 \]

These two derivatives are not equal, hence \(f(x)\) is not differentiable at \(x = 1\).

### Final Result

The function \(f(x)\) is continuous at \(x = 1\) but not differentiable at \(x = 1\).