Let `f (x) = {{:(min (x"," x ^(2)), x ge 0),( max (2x "," x-1), x lt 0):},` then which of the following is not true ?

Options

A. f (x) is not differentiable at `x=0`
B. f (x) is not differentiable are exactly two points
C. f (x) is continous everywhere
D. f (x) is strictly increasing `AA x in R`

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} \min(x, x^2) & \text{if } x \geq 0 \\ \max(2x, x-1) & \text{if } x < 0 \end{cases} \]

### Step 1: Analyze \( f(x) \) for \( x \geq 0 \)

For \( x \geq 0 \):
- We have two functions: \( y = x \) and \( y = x^2 \).
- To find the minimum, we need to determine where these two functions intersect:
\[ x = x^2 \implies x^2 - x = 0 \implies x(x - 1) = 0 \]
This gives us the points \( x = 0 \) and \( x = 1 \).

- Now we evaluate the functions at these points:
- At \( x = 0 \): \( f(0) = \min(0, 0^2) = 0 \)
- At \( x = 1 \): \( f(1) = \min(1, 1^2) = 1 \)

- For \( 0 \leq x < 1 \), \( x^2 < x \), so \( f(x) = x^2 \).
- For \( x > 1 \), \( x^2 > x \), so \( f(x) = x \).

Thus, for \( x \geq 0 \):
\[ f(x) = \begin{cases} x^2 & \text{if } 0 \leq x < 1 \\ x & \text{if } x \geq 1 \end{cases} \]

### Step 2: Analyze \( f(x) \) for \( x < 0 \)

For \( x < 0 \):
- We have two functions: \( y = 2x \) and \( y = x - 1 \).
- To find the maximum, we need to determine where these two functions intersect:
\[ 2x = x - 1 \implies x = -1 \]

- Now we evaluate the functions at this point:
- At \( x = -1 \): \( f(-1) = \max(2(-1), -1 - 1) = \max(-2, -2) = -2 \)

- For \( x < -1 \), \( 2x > x - 1 \), so \( f(x) = 2x \).
- For \( -1 < x < 0 \), \( 2x < x - 1 \), so \( f(x) = x - 1 \).

Thus, for \( x < 0 \):
\[ f(x) = \begin{cases} 2x & \text{if } x < -1 \\ x - 1 & \text{if } -1 \leq x < 0 \end{cases} \]

### Step 3: Check Continuity and Differentiability

1. **Continuity**: We check continuity at the transition points \( x = 0 \) and \( x = -1 \).
- At \( x = 0 \):
- \( \lim_{x \to 0^-} f(x) = f(0) = 0 \)
- \( \lim_{x \to 0^+} f(x) = f(0) = 0 \)
- At \( x = -1 \):
- \( \lim_{x \to -1^-} f(x) = 2(-1) = -2 \)
- \( \lim_{x \to -1^+} f(x) = f(-1) = -2 \)

Thus, \( f(x) \) is continuous everywhere.

2. **Differentiability**:
- At \( x = 0 \): The left-hand derivative is \( 0 \) (from \( x^2 \)) and the right-hand derivative is \( 1 \) (from \( x \)). Since they are not equal, \( f(x) \) is not differentiable at \( x = 0 \).
- At \( x = -1 \): The left-hand derivative is \( 2 \) (from \( 2x \)) and the right-hand derivative is \( 1 \) (from \( x - 1 \)). Since they are not equal, \( f(x) \) is not differentiable at \( x = -1 \).

### Conclusion

- The function \( f(x) \) is not differentiable at two points: \( x = 0 \) and \( x = -1 \).
- Therefore, the statement that \( f(x) \) is not differentiable at exactly two points is **false**.

### Final Answer
The option that is **not true** is the one stating that \( f(x) \) is not differentiable at exactly two points.