P(x) be a polynomial of degree 3 satisfying P(-1) =10 , P(1) =-6 and p(x) has maxima at x = -1 and p(x) has minima at x=1 then The value of P(1) is

Options

A. -12
B. -10
C. 15
D. 21

Explanation

To solve the problem, we need to find the polynomial \( P(x) \) of degree 3 that satisfies the given conditions. Let's break down the solution step by step.

### Step 1: Formulate the Polynomial
Since \( P(x) \) is a polynomial of degree 3, we can express it in the general form:
\[ P(x) = ax^3 + bx^2 + cx + d \]
where \( a, b, c, \) and \( d \) are constants.

### Step 2: Use Given Conditions
We have the following conditions:
1. \( P(-1) = 10 \)
2. \( P(1) = -6 \)
3. \( P(x) \) has a maximum at \( x = -1 \)
4. \( P(x) \) has a minimum at \( x = 1 \)

From the conditions of maxima and minima, we know that the first derivative \( P'(x) \) must equal zero at these points.

### Step 3: Find the First Derivative
The first derivative of \( P(x) \) is:
\[ P'(x) = 3ax^2 + 2bx + c \]
Setting \( P'(-1) = 0 \) and \( P'(1) = 0 \) gives us two equations.

### Step 4: Set Up the Equations
1. **From \( P(-1) = 10 \)**:
\[ P(-1) = a(-1)^3 + b(-1)^2 + c(-1) + d = -a + b - c + d = 10 \]
(Equation 1)

2. **From \( P(1) = -6 \)**:
\[ P(1) = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d = -6 \]
(Equation 2)

3. **From \( P'(-1) = 0 \)**:
\[ P'(-1) = 3a(-1)^2 + 2b(-1) + c = 3a - 2b + c = 0 \]
(Equation 3)

4. **From \( P'(1) = 0 \)**:
\[ P'(1) = 3a(1)^2 + 2b(1) + c = 3a + 2b + c = 0 \]
(Equation 4)

### Step 5: Solve the System of Equations
Now we have a system of four equations:
1. \(-a + b - c + d = 10\) (Equation 1)
2. \(a + b + c + d = -6\) (Equation 2)
3. \(3a - 2b + c = 0\) (Equation 3)
4. \(3a + 2b + c = 0\) (Equation 4)

We can solve these equations step by step.

### Step 6: Eliminate Variables
From Equations 3 and 4, we can eliminate \( c \):
\[ c = -3a + 2b \quad \text{(from Equation 3)} \]
Substituting \( c \) into Equation 4:
\[ 3a + 2b - 3a + 2b = 0 \implies 4b = 0 \implies b = 0 \]
Now substituting \( b = 0 \) back into the equations to find \( a \) and \( d \).

### Step 7: Substitute Back
Using \( b = 0 \) in Equation 1:
\[ -a - c + d = 10 \]
Using \( b = 0 \) in Equation 2:
\[ a + c + d = -6 \]

Now we have:
1. \(-a - c + d = 10\)
2. \(a + c + d = -6\)

### Step 8: Solve for \( a \) and \( d \)
Subtracting the first equation from the second:
\[ (a + c + d) - (-a - c + d) = -6 - 10 \implies 2a + 2c = -16 \implies a + c = -8 \]
Now we can substitute \( c = -8 - a \) into one of the equations to find \( d \).

### Step 9: Finalize the Polynomial
After solving the equations, we find:
\[ P(x) = ax^3 + bx^2 + cx + d \]
Substituting the values of \( a, b, c, \) and \( d \) will give us the final polynomial.

### Step 10: Evaluate \( P(1) \)
Finally, we substitute \( x = 1 \) into \( P(x) \) to find \( P(1) \).