© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Given Conditions
The polynomial P(x) is a real cubic polynomial that:
Vanishes at x = −3, i.e., P(−3) = 0.
Has local minima at x = 1 and local maxima at x = −1.
The definite integral from −1 to 1 of P(x) is 18, i.e.,
\int\limits_{-1}^1 P(x)\,dx = 18 .
We want to determine the sum of all coefficients of P(x).
Step 2: Use the Local Minima and Maxima Information
If P(x) has local minima at x = 1 and local maxima at x = −1, then P'(1) = 0 and P'(−1) = 0. For a cubic polynomial of the form
P(x) = ax^3 + bx^2 + cx + d,
its derivative P'(x) is a quadratic polynomial. If the derivative has roots at x = −1 and x = 1, we can represent it as
P'(x) = k(x + 1)(x - 1) = k(x^2 - 1),
where k is a constant to be determined.
Step 3: Integrate P'(x) to Get P(x)
Integrating P'(x) = k (x^2 - 1) with respect to x gives
P(x)
= \int k(x^2 - 1)\,dx
= k \left(\frac{x^3}{3} - x\right) + C,
where C is another constant of integration. For clarity, set a = k in the form that many steps use:
P(x) = a \left(\frac{x^3}{3} - x\right) + C.
Step 4: Use the Condition P(−3) = 0
We know P(x) vanishes at x = −3:
0 = P(-3) = a\left(\frac{(-3)^3}{3} - (-3)\right) + C.
Compute inside the parentheses:
\frac{(-3)^3}{3} = \frac{-27}{3} = -9,
and
-(-3) = +3.
So the expression becomes
0 = a(-9 + 3) + C = a(-6) + C.
Hence,
C = 6a. \quad (1)
Step 5: Use the Integral Condition
We have
\int_{-1}^{1} P(x)\,dx = 18.
Substitute P(x) = a( x^3/3 - x ) + C into the integral:
\int_{-1}^{1} \left[\,a \left(\frac{x^3}{3} - x\right) + C\right] dx = 18.
Perform the integral term by term:
\int_{-1}^{1} a \left(\frac{x^3}{3} - x\right) dx
= a \left[ \int_{-1}^{1} \frac{x^3}{3}\,dx - \int_{-1}^{1} x\,dx \right].
Note that \int_{-1}^{1} x^3\,dx = 0 because the function x³ is odd and integrates symmetrically around zero. Also \int_{-1}^{1} x\,dx = 0 for the same reason. Hence the entire contribution from the a part is zero.
The constant part contributes:
\int_{-1}^{1} C\,dx = C \int_{-1}^{1} 1\,dx = C [x]_{-1}^{1} = 2C.
Hence the integral condition becomes
2C = 18 \quad \Rightarrow \quad C = 9.
Step 6: Find a and Write P(x)
From equation (1), 6a = C, we get
6a = 9 \quad \Rightarrow \quad a = \frac{9}{6} = \frac{3}{2}.
Thus the polynomial P(x) is
P(x) = \frac{3}{2}\left(\frac{x^3}{3} - x\right) + 9
= \frac{x^3}{2} - \frac{3}{2}x + 9.
Step 7: Sum of All Coefficients
The polynomial can be written as
P(x) = \frac{1}{2}x^3 + 0\,x^2 - \frac{3}{2}x + 9.
The sum of the coefficients is
\left(\frac{1}{2}\right) + 0 + \left(-\frac{3}{2}\right) + 9 = \frac{1}{2} - \frac{3}{2} + 9 = -1 + 9 = 8.
The required sum of the coefficients is 8.
