Let `f(x) = x^6 + 2x^4 + x^3 + 2x + 3, x in R`. Then the `n` natural number for which `lim_(x to 1) (x^nf(1) - f(x))/(x-1) = 44` is _______.

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to 1} \frac{x^n f(1) - f(x)}{x - 1} = 44 \]

where \( f(x) = x^6 + 2x^4 + x^3 + 2x + 3 \).

### Step 1: Calculate \( f(1) \)

First, we need to find \( f(1) \):

\[ f(1) = 1^6 + 2 \cdot 1^4 + 1^3 + 2 \cdot 1 + 3 = 1 + 2 + 1 + 2 + 3 = 9 \]

### Step 2: Substitute \( f(1) \) into the limit expression

Now substitute \( f(1) \) into the limit expression:

\[ \lim_{x \to 1} \frac{x^n \cdot 9 - f(x)}{x - 1} \]

### Step 3: Expand \( f(x) \) around \( x = 1 \)

Next, we need to expand \( f(x) \) using Taylor series around \( x = 1 \):

\[ f(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2}(x - 1)^2 + O((x - 1)^3) \]

### Step 4: Calculate \( f'(x) \)

First, we find \( f'(x) \):

\[ f'(x) = 6x^5 + 8x^3 + 3x^2 + 2 \]

Now evaluate \( f'(1) \):

\[ f'(1) = 6 \cdot 1^5 + 8 \cdot 1^3 + 3 \cdot 1^2 + 2 = 6 + 8 + 3 + 2 = 19 \]

### Step 5: Substitute \( f(1) \) and \( f'(1) \) into the expansion

Now we can write the expansion:

\[ f(x) = 9 + 19(x - 1) + O((x - 1)^2) \]

### Step 6: Substitute back into the limit expression

Substituting back into the limit gives:

\[ \lim_{x \to 1} \frac{x^n \cdot 9 - (9 + 19(x - 1) + O((x - 1)^2))}{x - 1} \]

This simplifies to:

\[ \lim_{x \to 1} \frac{x^n \cdot 9 - 9 - 19(x - 1)}{x - 1} \]

### Step 7: Factor out terms

Factoring out \( 9 \):

\[ = \lim_{x \to 1} \frac{9(x^n - 1) - 19(x - 1)}{x - 1} \]

### Step 8: Evaluate \( x^n - 1 \)

Using the fact that \( x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + ... + 1) \):

\[ = \lim_{x \to 1} \frac{9(x - 1)(x^{n-1} + x^{n-2} + ... + 1) - 19(x - 1)}{x - 1} \]

### Step 9: Cancel \( x - 1 \)

Cancelling \( x - 1 \):

\[ = \lim_{x \to 1} \left( 9(x^{n-1} + x^{n-2} + ... + 1) - 19 \right) \]

### Step 10: Evaluate the limit

As \( x \to 1 \):

\[ = 9(n) - 19 \]

Set this equal to 44:

\[ 9n - 19 = 44 \]

### Step 11: Solve for \( n \)

Solving for \( n \):

\[ 9n = 44 + 19 = 63 \\ n = \frac{63}{9} = 7 \]

Thus, the value of \( n \) is:

\[ \boxed{7} \]