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1.1 Q-5

Question Statement

Given the function f(x)=x3βˆ’ax2+bx+1f(x) = x^3 - ax^2 + bx + 1, where f(2)=βˆ’3f(2) = -3 and f(βˆ’1)=0f(-1) = 0, find the values of aa and bb.


Background and Explanation

To solve this problem, we will use the fact that the function value is given for specific inputs, x=2x = 2 and x=βˆ’1x = -1. By substituting these values into the function, we create a system of equations that will help us solve for the unknowns aa and bb.

We also need to recall how to evaluate a polynomial at a specific point and how to solve a system of linear equations.


Solution

We are given the function: f(x)=x3βˆ’ax2+bx+1f(x) = x^3 - ax^2 + bx + 1

Step 1: Substitute x=2x = 2 into the equation.

Using the given f(2)=βˆ’3f(2) = -3, substitute x=2x = 2 into the function:

f(2)=23βˆ’a(22)+b(2)+1=βˆ’3f(2) = 2^3 - a(2^2) + b(2) + 1 = -3

Simplify:

8βˆ’4a+2b+1=βˆ’38 - 4a + 2b + 1 = -3 9βˆ’4a+2b=βˆ’39 - 4a + 2b = -3

Rearrange this equation:

βˆ’4a+2b=βˆ’12(1)-4a + 2b = -12 \tag{1}

Step 2: Substitute x=βˆ’1x = -1 into the equation.

Now, using f(βˆ’1)=0f(-1) = 0, substitute x=βˆ’1x = -1 into the function:

f(βˆ’1)=(βˆ’1)3βˆ’a(βˆ’1)2+b(βˆ’1)+1=0f(-1) = (-1)^3 - a(-1)^2 + b(-1) + 1 = 0

Simplify:

βˆ’1βˆ’aβˆ’b+1=0-1 - a - b + 1 = 0 βˆ’aβˆ’b=0β‡’a+b=0(2)-a - b = 0 \quad \Rightarrow \quad a + b = 0 \tag{2}

Step 3: Solve the system of equations.

We now have the system of equations:

  1. βˆ’4a+2b=βˆ’12-4a + 2b = -12
  2. a+b=0a + b = 0

From equation (2), solve for bb:

b=βˆ’ab = -a

Substitute b=βˆ’ab = -a into equation (1):

βˆ’4a+2(βˆ’a)=βˆ’12-4a + 2(-a) = -12

Simplify:

βˆ’4aβˆ’2a=βˆ’12-4a - 2a = -12 βˆ’6a=βˆ’12-6a = -12

Solve for aa:

a=2a = 2

Step 4: Find the value of bb.

Now that we know a=2a = 2, substitute it into a+b=0a + b = 0:

2+b=02 + b = 0 b=βˆ’2b = -2

Key Formulas or Methods Used

  • Substitution to evaluate polynomials at specific points.
  • Solving systems of linear equations.

Summary of Steps

  1. Substitute x=2x = 2 into the function to get an equation involving aa and bb.
  2. Substitute x=βˆ’1x = -1 into the function to get another equation.
  3. Solve the system of equations:
    • From a+b=0a + b = 0, find b=βˆ’ab = -a.
    • Substitute b=βˆ’ab = -a into the first equation and solve for aa.
  4. Substitute the value of aa into a+b=0a + b = 0 to find bb.
  5. The solution is a=2a = 2 and b=βˆ’2b = -2.