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(2) is (a) -15 (b) -16 (c) -17 (d) -22 (c) -17 (d) -22

Options

A. -15
B. -16
C. -17
D. -22

Explanation

To solve the problem step by step, we will find the polynomial \( P(x) \) of degree 3 that satisfies the given conditions and then evaluate \( P(2) \).

### Step 1: Understanding the conditions
We know that \( P(x) \) is a cubic polynomial, which can be expressed in the form:
\[ P(x) = ax^3 + bx^2 + cx + d \]
Given that \( P(-1) = 10 \) and \( P(1) = -6 \), and that \( P(x) \) has a maximum at \( x = -1 \) and a minimum at \( x = 1 \), we can derive the first derivative conditions.

### Step 2: Finding the first derivative
Since \( P(x) \) has a maximum at \( x = -1 \) and a minimum at \( x = 1 \), we know:
\[ P'(-1) = 0 \quad \text{and} \quad P'(1) = 0 \]
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 3: Setting up equations
1. For \( P'(-1) = 0 \):
\[ 3a(-1)^2 + 2b(-1) + c = 0 \implies 3a - 2b + c = 0 \quad \text{(Equation 1)} \]

2. For \( P'(1) = 0 \):
\[ 3a(1)^2 + 2b(1) + c = 0 \implies 3a + 2b + c = 0 \quad \text{(Equation 2)} \]

### Step 4: Solving the equations
Now we have two equations:
1. \( 3a - 2b + c = 0 \)
2. \( 3a + 2b + c = 0 \)

Subtract Equation 1 from Equation 2:
\[ (3a + 2b + c) - (3a - 2b + c) = 0 \implies 4b = 0 \implies b = 0 \]

Substituting \( b = 0 \) back into either equation gives:
\[ 3a + c = 0 \implies c = -3a \]

### Step 5: Using the polynomial values
Now we can express \( P(x) \) as:
\[ P(x) = ax^3 - 3ax + d \]

Using the conditions \( P(-1) = 10 \) and \( P(1) = -6 \):
1. For \( P(-1) = 10 \):
\[ a(-1)^3 - 3a(-1) + d = 10 \implies -a + 3a + d = 10 \implies 2a + d = 10 \quad \text{(Equation 3)} \]

2. For \( P(1) = -6 \):
\[ a(1)^3 - 3a(1) + d = -6 \implies a - 3a + d = -6 \implies -2a + d = -6 \quad \text{(Equation 4)} \]

### Step 6: Solving for \( a \) and \( d \)
Now we have:
1. \( 2a + d = 10 \) (Equation 3)
2. \( -2a + d = -6 \) (Equation 4)

Subtract Equation 4 from Equation 3:
\[ (2a + d) - (-2a + d) = 10 + 6 \implies 4a = 16 \implies a = 4 \]

Substituting \( a = 4 \) back into Equation 3:
\[ 2(4) + d = 10 \implies 8 + d = 10 \implies d = 2 \]

### Step 7: Final polynomial
Now we can find \( b \) and \( c \):
\[ b = 0, \quad c = -3(4) = -12 \]
Thus, the polynomial is:
\[ P(x) = 4x^3 - 12x + 2 \]

### Step 8: Evaluating \( P(2) \)
Now we calculate \( P(2) \):
\[ P(2) = 4(2)^3 - 12(2) + 2 = 4(8) - 24 + 2 = 32 - 24 + 2 = 10 \]

### Final Answer
The value of \( P(2) \) is \( 10 \).