Let `p(x)` be a polynomial of degree 4 having extremum at `x = 1,2` and `lim_(x->0)(1+(p(x))/x^2)=2.` Then find the value of `p(2).`

Answer

To solve the problem, we will follow these steps:

1. **Understanding the Polynomial**:
Since \( p(x) \) is a polynomial of degree 4 with extremum points at \( x = 1 \) and \( x = 2 \), we can express it in the form:
\[ p(x) = a x^4 + b x^3 + c x^2 + d x + e \]

2. **Using the Limit Condition**:
We are given that:
\[ \lim_{x \to 0} \left( 1 + \frac{p(x)}{x^2} \right) = 2 \]
This implies:
\[ \lim_{x \to 0} \frac{p(x)}{x^2} = 1 \]
Therefore, \( p(x) \) must behave like \( x^2 \) as \( x \to 0 \). This means that the constant term \( e \) must be 0, and the coefficient of \( x^2 \) must be 1. Thus, we can write:
\[ p(x) = a x^4 + b x^3 + x^2 \]

3. **Finding the Derivative**:
The derivative of \( p(x) \) is:
\[ p'(x) = 4a x^3 + 3b x^2 + 2x \]
Since \( p(x) \) has extrema at \( x = 1 \) and \( x = 2 \), we set \( p'(1) = 0 \) and \( p'(2) = 0 \).

4. **Setting Up the Equations**:
For \( x = 1 \):
\[ p'(1) = 4a(1)^3 + 3b(1)^2 + 2(1) = 4a + 3b + 2 = 0 \quad \text{(Equation 1)} \]
For \( x = 2 \):
\[ p'(2) = 4a(2)^3 + 3b(2)^2 + 2(2) = 32a + 12b + 4 = 0 \quad \text{(Equation 2)} \]

5. **Solving the Equations**:
From Equation 1:
\[ 4a + 3b + 2 = 0 \implies 4a + 3b = -2 \quad \text{(1)} \]
From Equation 2:
\[ 32a + 12b + 4 = 0 \implies 32a + 12b = -4 \quad \text{(2)} \]
Now, we can multiply Equation (1) by 8:
\[ 32a + 24b = -16 \quad \text{(3)} \]
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):
\[ 4a + 3(-1) = -2 \implies 4a - 3 = -2 \implies 4a = 1 \implies a = \frac{1}{4} \]

6. **Constructing the Polynomial**:
Now we have:
\[ p(x) = \frac{1}{4} x^4 - x^3 + x^2 \]

7. **Finding \( p(2) \)**:
Now we compute \( p(2) \):
\[ p(2) = \frac{1}{4}(2^4) - (2^3) + (2^2) = \frac{1}{4}(16) - 8 + 4 \]
\[ = 4 - 8 + 4 = 0 \]

Thus, the value of \( p(2) \) is \( \boxed{0} \).