`f^(prime)(x)=varphi^(prime)(x)=f(x)` for all `xdot` Also, `f(3)=5a n df^(prime)(3)=4.` Then the value of `[f(10)]^2` -`[ varphi(10)]^2` is _____

Options

A. 0
B. 9
C. 41
D. none of these

Explanation

To solve the problem, we start with the given information:

1. \( f'(x) = \varphi'(x) = f(x) \) for all \( x \).
2. \( f(3) = 5 \) and \( f'(3) = 4 \).

We need to find the value of \( [f(10)]^2 - [\varphi(10)]^2 \).

### Step-by-Step Solution:

**Step 1: Differentiate the expression \( f(x)^2 - \varphi(x)^2 \)**

Using the chain rule, we differentiate \( f(x)^2 - \varphi(x)^2 \):

\[ \frac{d}{dx}(f(x)^2 - \varphi(x)^2) = 2f(x)f'(x) - 2\varphi(x)\varphi'(x) \]

**Step 2: Substitute the derivatives**

Since \( f'(x) = f(x) \) and \( \varphi'(x) = f(x) \):

\[ = 2f(x)f(x) - 2\varphi(x)f(x) = 2f(x)^2 - 2\varphi(x)f(x) \]

**Step 3: Set the derivative to zero**

For the expression to be constant, the derivative must equal zero:

\[ 2f(x)^2 - 2\varphi(x)f(x) = 0 \]

This implies:

\[ f(x)^2 = \varphi(x)f(x) \]

**Step 4: Rearranging the equation**

We can rearrange this to:

\[ f(x)^2 - \varphi(x)f(x) = 0 \]

Factoring out \( f(x) \):

\[ f(x)(f(x) - \varphi(x)) = 0 \]

Since \( f(x) \) cannot be zero for all \( x \) (as it is given that \( f(3) = 5 \)), we have:

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

**Step 5: Conclusion about the constant**

Since \( f(x) = \varphi(x) \), we can conclude that:

\[ f(x)^2 - \varphi(x)^2 = 0 \]

This means:

\[ f(x)^2 - \varphi(x)^2 = C \]

for some constant \( C \).

**Step 6: Evaluate at specific points**

Using the values given:

At \( x = 3 \):

\[ f(3) = 5 \quad \text{and} \quad \varphi(3) = f'(3) = 4 \]

Thus:

\[ f(3)^2 - \varphi(3)^2 = 5^2 - 4^2 = 25 - 16 = 9 \]

Since \( f(x)^2 - \varphi(x)^2 \) is constant, it holds for \( x = 10 \) as well:

\[ f(10)^2 - \varphi(10)^2 = 9 \]

### Final Answer:

The value of \( [f(10)]^2 - [\varphi(10)]^2 \) is \( \boxed{9} \).