A function 
f
(
x
)
 is given below. Which of the following is true for 
f
(
x
)
?
f
(
x
)
=
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
x
x
<
0
0
x
=
0
sin
x
0
<
x
<
π
2
1
x
≥
π
2

I. 
f
(
x
)
 is continuous and differentiable at 
x
=
0
.
II. 
f
(
x
)
 is differentiable at 
x
=
π
2
.
III. 
f
(
x
)
 is continuous at 
x
=
0
 and at 
x
=
π
2
.

Let's analyze the given function and the statements about its continuity and differentiability:

The function \( f(x) \) is defined piecewise as:

\[
f(x) =
\begin{cases} 
x & \text{if } x < 0 \\
0 & \text{if } x = 0 \\
\sin(x) & \text{if } 0 < x < \frac{\pi}{2} \\
1 & \text{if } x \geq \frac{\pi}{2}
\end{cases}
\]

Now, let's assess each statement one by one:

### I. \( f(x) \) is continuous and differentiable at \( x = 0 \).

#### Continuity at \( x = 0 \):
- To check continuity at \( x = 0 \), we need to see if the left-hand limit and right-hand limit exist and match the value of the function at \( x = 0 \).
- **Left-hand limit** as \( x \to 0^- \): When \( x < 0 \), \( f(x) = x \). So, \( \lim_{x \to 0^-} f(x) = 0 \).
- **Right-hand limit** as \( x \to 0^+ \): When \( 0 < x < \frac{\pi}{2} \), \( f(x) = \sin(x) \). So, \( \lim_{x \to 0^+} f(x) = \sin(0) = 0 \).
- Since \( f(0) = 0 \), and both the left-hand and right-hand limits are equal to \( 0 \), the function is **continuous at \( x = 0 \)**.

#### Differentiability at \( x = 0 \):
- To check differentiability at \( x = 0 \), we compute the derivative from both sides.
- **Left-hand derivative**: \( \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{h - 0}{h} = 1 \).
- **Right-hand derivative**: \( \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\sin(h) - 0}{h} = \lim_{h \to 0^+} \frac{\sin(h)}{h} = 1 \).
- Since both derivatives are equal to \( 1 \), \( f(x) \) is **differentiable at \( x = 0 \)**.

Thus, **Statement I is true**.

---

### II. \( f(x) \) is differentiable at \( x = \frac{\pi}{2} \).

#### Continuity at \( x = \frac{\pi}{2} \):
- **Left-hand limit**: As \( x \to \frac{\pi}{2}^- \), \( f(x) = \sin(x) \). So, \( \lim_{x \to \frac{\pi}{2}^-} f(x) = \sin\left(\frac{\pi}{2}\right) = 1 \).
- **Right-hand limit**: As \( x \to \frac{\pi}{2}^+ \), \( f(x) = 1 \). So, \( \lim_{x \to \frac{\pi}{2}^+} f(x) = 1 \).
- Since both limits are equal to \( 1 \), the function is **continuous at \( x = \frac{\pi}{2} \)**.

#### Differentiability at \( x = \frac{\pi}{2} \):
- **Left-hand derivative**: \( \lim_{h \to 0^-} \frac{f\left(\frac{\pi}{2} + h\right) - f\left(\frac{\pi}{2}\right)}{h} = \lim_{h \to 0^-} \frac{\sin\left(\frac{\pi}{2} + h\right) - 1}{h} \). This gives \( \cos\left(\frac{\pi}{2}\right) = 0 \), so the derivative is \( 0 \).
- **Right-hand derivative**: \( \lim_{h \to 0^+} \frac{f\left(\frac{\pi}{2} + h\right) - f\left(\frac{\pi}{2}\right)}{h} = \lim_{h \to 0^+} \frac{1 - 1}{h} = 0 \).
- Since both derivatives are equal to \( 0 \), \( f(x) \) is **differentiable at \( x = \frac{\pi}{2} \)**.

Thus, **Statement II is true**.

---

### III. \( f(x) \) is continuous at \( x = 0 \) and at \( x = \frac{\pi}{2} \).

We have already checked that \( f(x) \) is continuous at both \( x = 0 \) and \( x = \frac{\pi}{2} \), so **Statement III is true**.

---

### Conclusion:
All three statements (I, II, and III) are true.

Would you like further clarification or more detailed explanations?

Here are five related questions to explore:
1. What are the conditions for a function to be continuous at a point?
2. How do we determine if a function is differentiable at a point?
3. How does the derivative of piecewise functions behave at boundary points?
4. Can a function be continuous but not differentiable? Provide examples.
5. What are the key differences between differentiability and continuity?

**Tip:** For piecewise functions, always check both continuity and differentiability separately at the transition points where the function definition changes.

A function f(x) is given below. Which of the following is true for f(x)? f(x) = { x for x < 0, 0 for x = 0, sin(x) for 0 < x < π/2, 1 for x ≥ π/2. I. f(x) is continuous and differentiable at x = 0. II. f(x) is differentiable at x = π/2. III. f(x) is continuous at x = 0 and at x = π/2.

Explore the continuity and differentiability of a piecewise function involving trigonometric and linear segments. The problem covers key concepts like limits, continuity at points, and checking differentiability using the derivative definition. Suitable for advanced high school students (Grades 11-12).

Math Problem Statement

A function f ( x ) is given below. Which of the following is true for f ( x ) ? f ( x )

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ x x < 0 0 x

I. f ( x ) is continuous and differentiable at x

0 . II. f ( x ) is differentiable at x

π 2 . III. f ( x ) is continuous at x

0 and at x

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