Find the continuous function `f` where `(x^4-4x^2)lt=f(x)lt=(2x^2-x^3)` such that the area bounded by `y=f(x),y=x^4-4x^2dot` then y-axis, and the line `x=t ,` where `(0lt=tlt=2)` is `k` times the area bounded by `y=f(x),y=2x^2-x^3,y-a xi s ,` and line `x=t(w h e r e0lt=tlt=2)dot`

Explanation

To find the continuous function \( f \) that satisfies the given conditions, we will follow these steps:

### Step 1: Understand the Area Conditions
We need to find a function \( f(x) \) such that:
\[ x^4 - 4x^2 \leq f(x) \leq 2x^2 - x^3 \]
We will also need to calculate the areas bounded by \( f(x) \), \( x^4 - 4x^2 \), and \( 2x^2 - x^3 \) between the y-axis and the line \( x = t \) for \( 0 < t < 2 \).

### Step 2: Set Up the Area Equations
The area \( A_1 \) bounded by \( f(x) \) and \( x^4 - 4x^2 \) from \( 0 \) to \( t \) is given by:
\[ A_1 = \int_0^t (f(x) - (x^4 - 4x^2)) \, dx \]
The area \( A_2 \) bounded by \( f(x) \) and \( 2x^2 - x^3 \) from \( 0 \) to \( t \) is given by:
\[ A_2 = \int_0^t ((2x^2 - x^3) - f(x)) \, dx \]

### Step 3: Relate the Areas
According to the problem, we have:
\[ A_1 = k \cdot A_2 \]

### Step 4: Differentiate with Respect to \( t \)
Differentiating both sides with respect to \( t \):
\[ \frac{d}{dt} A_1 = \frac{d}{dt} A_2 \cdot k \]
Using the Fundamental Theorem of Calculus:
\[ f(t) - (t^4 - 4t^2) = k \left( (2t^2 - t^3) - f(t) \right) \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ f(t) + kf(t) = k(2t^2 - t^3) + t^4 - 4t^2 \]
\[ f(t)(1 + k) = kt^2(2 - t) + t^4 - 4t^2 \]
\[ f(t) = \frac{kt^2(2 - t) + t^4 - 4t^2}{1 + k} \]

### Step 6: Replace \( t \) with \( x \)
Thus, we can express \( f(x) \) as:
\[ f(x) = \frac{kx^2(2 - x) + x^4 - 4x^2}{1 + k} \]

### Final Function
So the required continuous function \( f(x) \) is:
\[ f(x) = \frac{x^4 + kx^2(2 - x) - 4x^2}{1 + k} \]