For `U_n=int_0^1x^n(2-x)^n dx ; V_n=int_0^1x^n(1-x)^ndxn in N ,` which of the following statement(s) is/are true? `U_n=2^n V_n` (b) `U_n=2^(-n)V_n` `U_n=2^(2n)V_n` (d) `V_n=2^(-2n)U_n`

Explanation

To solve the problem, we need to analyze the integrals \( U_n \) and \( V_n \) given by:

\[ U_n = \int_0^1 x^n (2-x)^n \, dx \]
\[ V_n = \int_0^1 x^n (1-x)^n \, dx \]

We will explore the relationship between \( U_n \) and \( V_n \).

### Step 1: Rewrite \( U_n \)

We start with \( U_n \):

\[ U_n = \int_0^1 x^n (2-x)^n \, dx \]

Using the substitution \( x = 2t \), we have \( dx = 2 \, dt \). The limits change as follows:
- When \( x = 0 \), \( t = 0 \)
- When \( x = 1 \), \( t = \frac{1}{2} \)

Thus, we can rewrite \( U_n \):

\[ U_n = \int_0^{1/2} (2t)^n (2 - 2t)^n \cdot 2 \, dt \]

This simplifies to:

\[ U_n = 2^{n+1} \int_0^{1/2} t^n (2(1-t))^n \, dt \]

\[ U_n = 2^{n+1} \int_0^{1/2} t^n (2^n (1-t)^n) \, dt \]

\[ U_n = 2^{2n+1} \int_0^{1/2} t^n (1-t)^n \, dt \]

### Step 2: Relate to \( V_n \)

Now, we know that \( V_n \) is defined as:

\[ V_n = \int_0^1 x^n (1-x)^n \, dx \]

Using the same substitution \( x = 2t \) for \( V_n \):

\[ V_n = \int_0^1 (2t)^n (1-2t)^n \cdot 2 \, dt \]

The limits remain the same, and we can express \( V_n \):

\[ V_n = 2^{n+1} \int_0^{1/2} t^n (1-2t)^n \, dt \]

### Step 3: Compare \( U_n \) and \( V_n \)

Now we can relate \( U_n \) and \( V_n \):

From our expressions, we have:

\[ U_n = 2^{2n+1} \int_0^{1/2} t^n (1-t)^n \, dt \]
\[ V_n = 2^{n+1} \int_0^{1/2} t^n (1-2t)^n \, dt \]

We can express \( U_n \) in terms of \( V_n \):

\[ U_n = 2^{2n+1} \cdot \frac{1}{2^{n+1}} V_n \]

This simplifies to:

\[ U_n = 2^{n} V_n \]

### Conclusion

Thus, the relationship we derived is:

\[ U_n = 2^{2n} V_n \]

Now, we can analyze the given options:

1. \( U_n = 2^n V_n \) - False
2. \( U_n = 2^{-n} V_n \) - False
3. \( U_n = 2^{2n} V_n \) - True
4. \( V_n = 2^{-2n} U_n \) - True

### Final Answer

The true statements are:
- \( U_n = 2^{2n} V_n \)
- \( V_n = 2^{-2n} U_n \)