Let `f(x) = lambda + mu|x|+nu|x|^2`, where `lambda,mu, nu in R`, then `f'(0)` exists if

Options

A. `mu =0`
B. v =0
C. ` lambda=0`
D. `mu= v`

Answer

To determine the conditions under which the derivative \( f'(0) \) exists for the function \( f(x) = \lambda + \mu |x| + \nu |x|^2 \), we will analyze the left-hand limit (LHL) and right-hand limit (RHL) of the derivative at \( x = 0 \).

### Step 1: Define the function for different cases of \( x \)

The function is defined as:
\[ f(x) = \lambda + \mu |x| + \nu |x|^2 \]

We need to consider two cases based on the sign of \( x \) (i.e., whether \( x \) is less than or greater than 0).

1. **For \( x < 0 \)**:
\[ f(x) = \lambda - \mu x + \nu x^2 \]

2. **For \( x > 0 \)**:
\[ f(x) = \lambda + \mu x + \nu x^2 \]

### Step 2: Differentiate \( f(x) \) for both cases

Now, we will differentiate \( f(x) \) for both cases.

1. **For \( x < 0 \)**:
\[ f'(x) = -\mu + 2\nu x \]

2. **For \( x > 0 \)**:
\[ f'(x) = \mu + 2\nu x \]

### Step 3: Calculate the left-hand limit (LHL) as \( x \) approaches 0 from the left

To find \( f'(0) \), we calculate the left-hand limit:
\[ \text{LHL} = \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} (-\mu + 2\nu x) = -\mu \]

### Step 4: Calculate the right-hand limit (RHL) as \( x \) approaches 0 from the right

Next, we calculate the right-hand limit:
\[ \text{RHL} = \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} (\mu + 2\nu x) = \mu \]

### Step 5: Set LHL equal to RHL for the derivative to exist

For the derivative \( f'(0) \) to exist, the left-hand limit must equal the right-hand limit:
\[ -\mu = \mu \]

### Step 6: Solve for \( \mu \)

From the equation \( -\mu = \mu \), we can solve for \( \mu \):
\[ -2\mu = 0 \implies \mu = 0 \]

### Conclusion

The derivative \( f'(0) \) exists if and only if \( \mu = 0 \).