Let `lambda=int_0^1(dx)/(1+x^3), p=lim_(n rarr oo)[prod_(r=1)^n(n^3+r^3)/(n^(3n))]^(1//n)` then `ln p` is equal to (a) `ln2-1+lambda` (b) `ln2-3+3lambda` (c) `2ln2-lambda` (d) `ln4-3+3lambda`

Options

A. `log 2-1+lambda`
B. `log 2-3+3lambda`
C. `2log 2-lambda`
D. `log 4-3+2lambda`

Answer

To solve the problem, we need to find the value of \( \ln p \) where \( p \) is defined as:

\[ p = \lim_{n \to \infty} \left( \prod_{r=1}^{n} \frac{n^3 + r^3}{n^{3n}} \right)^{\frac{1}{n}} \]

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

We can rewrite \( p \) as follows:

\[ p = \lim_{n \to \infty} \left( \frac{1}{n^{3n}} \prod_{r=1}^{n} (n^3 + r^3) \right)^{\frac{1}{n}} \]

### Step 2: Simplify the product

The product can be simplified:

\[ \prod_{r=1}^{n} (n^3 + r^3) = \prod_{r=1}^{n} n^3 \left(1 + \frac{r^3}{n^3}\right) = n^{3n} \prod_{r=1}^{n} \left(1 + \frac{r^3}{n^3}\right) \]

### Step 3: Substitute back into \( p \)

Substituting this back into the expression for \( p \):

\[ p = \lim_{n \to \infty} \left( \frac{n^{3n} \prod_{r=1}^{n} \left(1 + \frac{r^3}{n^3}\right)}{n^{3n}} \right)^{\frac{1}{n}} = \lim_{n \to \infty} \left( \prod_{r=1}^{n} \left(1 + \frac{r^3}{n^3}\right) \right)^{\frac{1}{n}} \]

### Step 4: Take the logarithm

Taking the logarithm of \( p \):

\[ \ln p = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \ln \left(1 + \frac{r^3}{n^3}\right) \]

### Step 5: Change to Riemann sum

As \( n \to \infty \), this sum approaches the Riemann integral:

\[ \ln p = \int_{0}^{1} \ln \left(1 + x^3\right) \, dx \]

### Step 6: Evaluate the integral

Using integration by parts or known integrals, we can evaluate:

\[ \int_{0}^{1} \ln(1 + x^3) \, dx = \lambda \]

### Step 7: Combine results

From the earlier steps, we have:

\[ \ln p = \ln 2 - 3 + 3\lambda \]

### Final Result

Thus, the correct option for \( \ln p \) is:

\[ \ln p = \ln 2 - 3 + 3\lambda \]

So the answer is (b) \( \ln 2 - 3 + 3\lambda \).