The value of k for which `f(x)={{:(,(x^(2^(32))-2^(32)x+4^(16)-1)/((x-1)^(2)),x ne 1),(,k,x=1):}` is continuous at x=1, is

Options

A. `2^(63)-2^(31)`
B. `2^(65)-2^(33)`
C. `2^(62)-2^(31)`
D. `2^(65)-2^(31)`

Explanation

To find the value of \( k \) for which the function

\[ f(x) = \begin{cases} \frac{x^{2^{32}} - 2^{32}x + 4^{16} - 1}{(x-1)^2} & \text{if } x \neq 1 \\ k & \text{if } x = 1 \end{cases} \]

is continuous at \( x = 1 \), we need to ensure that

\[ \lim_{x \to 1} f(x) = f(1) = k. \]

### Step 1: Calculate the limit as \( x \) approaches 1

We need to evaluate

\[ \lim_{x \to 1} \frac{x^{2^{32}} - 2^{32}x + 4^{16} - 1}{(x-1)^2}. \]

### Step 2: Substitute \( x = 1 \) into the numerator

First, we substitute \( x = 1 \) into the numerator:

\[ 1^{2^{32}} - 2^{32} \cdot 1 + 4^{16} - 1 = 1 - 2^{32} + 4^{16} - 1 = 4^{16} - 2^{32}. \]

Since \( 4^{16} = (2^2)^{16} = 2^{32} \), we have:

\[ 4^{16} - 2^{32} = 2^{32} - 2^{32} = 0. \]

### Step 3: Check the form of the limit

The limit takes the form \( \frac{0}{0} \) as \( x \to 1 \). We can apply L'Hôpital's Rule or factor the numerator.

### Step 4: Factor the numerator

We can factor \( x^{2^{32}} - 2^{32}x + 4^{16} - 1 \) using the fact that \( x = 1 \) is a root:

Let \( g(x) = x^{2^{32}} - 2^{32}x + 4^{16} - 1 \). We can find \( g'(x) \) and evaluate it at \( x = 1 \) to apply L'Hôpital's Rule.

### Step 5: Differentiate the numerator

Using the power rule, we differentiate:

\[ g'(x) = 2^{32} x^{2^{32}-1} - 2^{32}. \]

Now, substituting \( x = 1 \):

\[ g'(1) = 2^{32} \cdot 1^{2^{32}-1} - 2^{32} = 2^{32} - 2^{32} = 0. \]

### Step 6: Differentiate again

Since we still have \( \frac{0}{0} \), we differentiate again:

\[ g''(x) = 2^{32}(2^{32}-1)x^{2^{32}-2}. \]

Now substituting \( x = 1 \):

\[ g''(1) = 2^{32}(2^{32}-1) \cdot 1^{2^{32}-2} = 2^{32}(2^{32}-1). \]

### Step 7: Evaluate the limit using L'Hôpital's Rule

Now we can evaluate the limit:

\[ \lim_{x \to 1} \frac{g(x)}{(x-1)^2} = \lim_{x \to 1} \frac{g''(x)}{2} = \frac{2^{32}(2^{32}-1)}{2} = 2^{31}(2^{32}-1). \]

### Step 8: Set the limit equal to \( k \)

For continuity at \( x = 1 \):

\[ k = 2^{31}(2^{32}-1). \]

Thus, the value of \( k \) for which \( f(x) \) is continuous at \( x = 1 \) is:

\[ \boxed{2^{31}(2^{32}-1)}. \]