Let p (x) be a real polynomial of degree 4 having extreme values `x=1 and x=2.if lim_(xrarr0) (p(x))/(x^2)=1`, then `p(4)` is equal to

Options

A. `0`
B. `16`
C. `32`
D. `64`

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Define the Polynomial
Let \( p(x) \) be a polynomial of degree 4. We can express it in the general form:
\[ p(x) = ax^4 + bx^3 + cx^2 + dx + e \]

### Step 2: Use the Limit Condition
We have the condition:
\[ \lim_{x \to 0} \frac{p(x)}{x^2} = 1 \]
Substituting \( p(x) \) into the limit, we get:
\[ \lim_{x \to 0} \frac{ax^4 + bx^3 + cx^2 + dx + e}{x^2} = 1 \]
This simplifies to:
\[ \lim_{x \to 0} \left( ax^2 + bx + c + \frac{d}{x} + \frac{e}{x^2} \right) = 1 \]

### Step 3: Analyze the Limit
For the limit to equal 1 as \( x \to 0 \), the terms involving \( \frac{d}{x} \) and \( \frac{e}{x^2} \) must not contribute to the limit, which means:
\[ d = 0 \quad \text{and} \quad e = 0 \]
Thus, we have:
\[ p(x) = ax^4 + bx^3 + cx^2 \]

### Step 4: Substitute Back into the Limit
Now substituting \( d = 0 \) and \( e = 0 \) into the limit:
\[ \lim_{x \to 0} \frac{ax^4 + bx^3 + cx^2}{x^2} = \lim_{x \to 0} (ax^2 + bx + c) = c \]
Setting this equal to 1 gives us:
\[ c = 1 \]
So now we have:
\[ p(x) = ax^4 + bx^3 + x^2 \]

### Step 5: Use the Extreme Values Condition
The polynomial has extreme values at \( x = 1 \) and \( x = 2 \). This means:
\[ p'(1) = 0 \quad \text{and} \quad p'(2) = 0 \]
Calculating the derivative:
\[ p'(x) = 4ax^3 + 3bx^2 + 2x \]
Setting \( p'(1) = 0 \):
\[ 4a + 3b + 2 = 0 \quad \text{(1)} \]
Setting \( p'(2) = 0 \):
\[ 32a + 12b + 4 = 0 \quad \text{(2)} \]

### Step 6: Solve the System of Equations
From equation (1):
\[ 4a + 3b = -2 \quad \text{(1)} \]
From equation (2):
\[ 32a + 12b = -4 \quad \text{(2)} \]
Multiply equation (1) by 8:
\[ 32a + 24b = -16 \quad \text{(3)} \]
Now subtract equation (2) from equation (3):
\[ (32a + 24b) - (32a + 12b) = -16 + 4 \]
This simplifies to:
\[ 12b = -12 \implies b = -1 \]
Substituting \( b = -1 \) back into equation (1):
\[ 4a + 3(-1) = -2 \implies 4a - 3 = -2 \implies 4a = 1 \implies a = \frac{1}{4} \]

### Step 7: Write the Polynomial
Now we have:
\[ a = \frac{1}{4}, \quad b = -1, \quad c = 1 \]
Thus, the polynomial is:
\[ p(x) = \frac{1}{4}x^4 - x^3 + x^2 \]

### Step 8: Calculate \( p(4) \)
Now we need to find \( p(4) \):
\[ p(4) = \frac{1}{4}(4^4) - (4^3) + (4^2) \]
Calculating each term:
\[ = \frac{1}{4}(256) - 64 + 16 \]
\[ = 64 - 64 + 16 = 16 \]

### Final Answer
Thus, \( p(4) = 16 \).

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