Let `f(x)` be a differentiable function satisfying `f(y)f(x/y)=f(x)` `AA`, x,y`in`R, `y!=0` and `f(1)!=0`, `f'(1)=3` then

Options

A. sgn `(f(x))` is non-differentiable at exactly one point
B. `lim _(xto0)(x ^(2) (cos x-1))/(f (x))=0`
C. `f (x) =x` has 3 solutions
D. `f (f(x))-f ^(3) (x)=0` has infinitely many solutions

Answer

To solve the problem, we start with the functional equation given:

\[ f(y) f\left(\frac{x}{y}\right) = f(x) \]

for all \( x, y \in \mathbb{R} \) and \( y \neq 0 \). We are also given that \( f(1) \neq 0 \) and \( f'(1) = 3 \).

### Step 1: Substitute specific values into the functional equation

Let’s first substitute \( y = 1 \):

\[ f(1) f(x) = f(x) \]

Since \( f(1) \neq 0 \), we can divide both sides by \( f(x) \) (assuming \( f(x) \neq 0 \)):

\[ f(1) = 1 \]

### Step 2: Explore the form of \( f(x) \)

Next, we can explore the functional equation further. Let’s substitute \( x = 1 \):

\[ f(y) f\left(\frac{1}{y}\right) = f(1) = 1 \]

This implies:

\[ f(y) f\left(\frac{1}{y}\right) = 1 \]

This suggests that \( f(y) \) and \( f\left(\frac{1}{y}\right) \) are multiplicative inverses.

### Step 3: Assume a power function form

Given the nature of the functional equation, we can assume that \( f(x) \) might be of the form \( f(x) = x^n \) for some \( n \).

Substituting \( f(x) = x^n \) into the functional equation gives:

\[ y^n \left(\frac{x}{y}\right)^n = x^n \]

This simplifies to:

\[ y^n \cdot \frac{x^n}{y^n} = x^n \]

which holds true for any \( n \).

### Step 4: Differentiate and use the derivative condition

Now, we differentiate \( f(x) = x^n \):

\[ f'(x) = n x^{n-1} \]

Given \( f'(1) = 3 \):

\[ f'(1) = n \cdot 1^{n-1} = n = 3 \]

Thus, we have \( n = 3 \) and therefore:

\[ f(x) = x^3 \]

### Step 5: Verify the options

1. **First Option**: \( \text{signum}(f(x)) \) is not differentiable at exactly one point.
- \( f(x) = x^3 \) is differentiable everywhere, but \( \text{signum}(f(x)) \) is not differentiable at \( x = 0 \). Thus, this option is correct.

2. **Second Option**: \( \lim_{x \to 0} \frac{\cos(x) - 1}{f(x)} = 0 \).
- We know \( \lim_{x \to 0} \frac{\cos(x) - 1}{x^3} = 0 \). This option is also correct.

3. **Third Option**: \( f(x) = x \) has exactly three solutions.
- Setting \( x^3 = x \) gives \( x(x^2 - 1) = 0 \), which has three solutions: \( x = 0, 1, -1 \). This option is correct.

4. **Fourth Option**: \( f(f(x)) - f^3(x) = 0 \) has infinitely many solutions.
- Since \( f(f(x)) = f(x^3) = (x^3)^3 = x^9 \) and \( f^3(x) = (x^3)^3 = x^9 \), this equation holds for all \( x \). Thus, this option is correct.

### Conclusion

All options are correct.