Consider the graph of the function `f(x) = x+ sqrt|x|` Statement-1: The graph of `y=f(x)` has only one critical point Statement-2: `fprime (x)` vanishes only at one point

Options

A. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 2 is a correct explanation for Statement 1.
B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
C. Statement 1 is true, Statement 2 is false.
D. Statement 1 is false, Statement 2 is true.

Explanation

To solve the problem, we will analyze the function \( f(x) = x + \sqrt{|x|} \) and determine the critical points by finding the derivative and identifying where it vanishes.

### Step 1: Define the function
The function is given by:
\[ f(x) = x + \sqrt{|x|} \]

### Step 2: Differentiate the function
We need to find the derivative \( f'(x) \). Since the function involves the absolute value, we will consider two cases: when \( x \geq 0 \) and when \( x < 0 \).

**Case 1: \( x \geq 0 \)**
\[ f(x) = x + \sqrt{x} \]
Differentiating:
\[ f'(x) = 1 + \frac{1}{2\sqrt{x}} \]

**Case 2: \( x < 0 \)**
\[ f(x) = x + \sqrt{-x} \]
Differentiating:
\[ f'(x) = 1 - \frac{1}{2\sqrt{-x}} \]

### Step 3: Set the derivative to zero
We need to find where \( f'(x) = 0 \) in both cases.

**For \( x \geq 0 \):**
\[ 1 + \frac{1}{2\sqrt{x}} = 0 \]
This equation does not have any solutions since \( \frac{1}{2\sqrt{x}} \) is always positive for \( x \geq 0 \).

**For \( x < 0 \):**
\[ 1 - \frac{1}{2\sqrt{-x}} = 0 \]
Solving for \( x \):
\[ \frac{1}{2\sqrt{-x}} = 1 \implies 2\sqrt{-x} = 1 \implies \sqrt{-x} = \frac{1}{2} \implies -x = \frac{1}{4} \implies x = -\frac{1}{4} \]

### Step 4: Identify critical points
The only critical point we found is \( x = -\frac{1}{4} \).

### Conclusion
- **Statement 1**: The graph of \( y = f(x) \) has only one critical point, which is true.
- **Statement 2**: \( f'(x) \) vanishes only at one point, which is also true.

Thus, both statements are correct.