Consider a function `f : R -> R; f(x^2 +yf(z)) = xf(x) + zf(y), AA x,y,z in R` If `f(x) = 0,AA x in R` is not considered a part of solution set, then

Options

A. `g(x)` is not continuous `AA xepsilonR`
B. `g(x)` is differentiable except at two points `AA x epsilonR`
C. `g(x)` is differentiable at `x=sinalpha(alpha!+npi, n epsilonI)`
D. `g(x)` is non differentiable at integers

Explanation

To solve the functional equation \( f(x^2 + yf(z)) = xf(x) + zf(y) \) for all \( x, y, z \in \mathbb{R} \) under the condition that \( f(x) = 0 \) for all \( x \in \mathbb{R} \) is not considered as a solution, we will analyze the equation step by step.

### Step 1: Analyze the functional equation

We start with the functional equation:

\[ f(x^2 + yf(z)) = xf(x) + zf(y) \]

### Step 2: Substitute specific values

1. **Let \( x = 0 \)**:
\[ f(0^2 + yf(z)) = 0 \cdot f(0) + zf(y) \implies f(yf(z)) = zf(y) \]

2. **Let \( y = 0 \)**:
\[ f(x^2 + 0 \cdot f(z)) = xf(x) + zf(0) \implies f(x^2) = xf(x) + zf(0) \]

### Step 3: Analyze the implications

From the second substitution, if we let \( z = 0 \):
\[ f(x^2) = xf(x) + 0 \implies f(x^2) = xf(x) \]

### Step 4: Explore further substitutions

3. **Let \( z = 1 \)** in the first equation:
\[ f(x^2 + yf(1)) = xf(x) + f(y) \]

### Step 5: Investigate the form of \( f \)

Now, we have two equations:
1. \( f(yf(z)) = zf(y) \)
2. \( f(x^2) = xf(x) \)

### Step 6: Consider linearity

Assuming \( f(x) \) is a linear function of the form \( f(x) = kx \), we can substitute this back into our original equation to check for consistency.

### Step 7: Substitute \( f(x) = kx \)

Substituting \( f(x) = kx \) into the original equation:
\[ f(x^2 + yf(z)) = k(x^2 + y(kz)) = kx^2 + ky(kz) \]
\[ xf(x) + zf(y) = x(kx) + z(ky) = kx^2 + kzy \]

### Step 8: Equate and solve for \( k \)

From the above, we equate:
\[ kx^2 + ky(kz) = kx^2 + kzy \]
This implies \( ky(kz) = kzy \), which is true for all \( y, z \) if \( k \neq 0 \).

### Step 9: Conclusion

Thus, the function \( f(x) = kx \) holds for any non-zero constant \( k \).

### Step 10: Check the options

Now, we check the options provided in the question:

- **Option A**: \( f(\alpha) < \alpha^4 \) for \( \alpha \in (0, 1) \) - **Incorrect**.
- **Option B**: \( f(\alpha) < \alpha^2 \) for \( \alpha \in (0, 1) \) - **Incorrect**.
- **Option C**: \( f(\alpha) = \alpha^3 \) for some \( \alpha \in \mathbb{R}^+ \) - **Correct**.
- **Option D**: \( \lim_{\alpha \to 0^+} \frac{f(\alpha)}{\alpha} < \lim_{\alpha \to 0^+} \frac{\sin(\alpha)}{\alpha} \) - **Incorrect**.

### Final Answer

The incorrect options are A, B, and D.

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