Problem 8
Question
If \(f(x)=x^{2}-3 x+1\) and \(f(2 \alpha)=2 f(\alpha)\), then \(\alpha\) is equal to (a) \(\frac{1}{\sqrt{2}}\) (b) \(-\frac{1}{\sqrt{2}}\) (c) \(\frac{1}{\sqrt{2}}\) or \(-\frac{1}{\sqrt{2}}\) (d) None of these
Step-by-Step Solution
Verified Answer
(c) \(\frac{1}{\sqrt{2}}\) or \(-\frac{1}{\sqrt{2}}\)
1Step 1: Determine f(2α) and f(α)
First, we need to calculate the function value at both \(2\alpha\) and \(\alpha\).\\[ f(2\alpha) = (2\alpha)^2 - 3(2\alpha) + 1 = 4\alpha^2 - 6\alpha + 1 \] \\[ f(\alpha) = \alpha^2 - 3\alpha + 1 \]
2Step 2: Set the Equation as Given
According to the problem, \(f(2\alpha) = 2f(\alpha)\). Substitute the expressions we calculated: \\[ 4\alpha^2 - 6\alpha + 1 = 2(\alpha^2 - 3\alpha + 1) \]
3Step 3: Simplify the Equation
Distribute and simplify the equation on the right-hand side: \\[ 4\alpha^2 - 6\alpha + 1 = 2\alpha^2 - 6\alpha + 2 \] \Now subtract the right side from the left side: \\[ 4\alpha^2 - 6\alpha + 1 - 2\alpha^2 + 6\alpha - 2 = 0 \] \This simplifies to: \\[ 2\alpha^2 - 1 = 0 \]
4Step 4: Solve for α
Rearrange \(2\alpha^2 - 1 = 0\): \\[ 2\alpha^2 = 1 \] \Now divide by 2: \\[ \alpha^2 = \frac{1}{2} \] \Take the square root of both sides: \\[ \alpha = \pm \frac{1}{\sqrt{2}} \]
5Step 5: Verify the Solutions
Both solutions \(\alpha = \frac{1}{\sqrt{2}}\) and \(\alpha = -\frac{1}{\sqrt{2}}\) satisfy the original condition. Both values of \(\alpha\) make \(f(2\alpha) = 2f(\alpha)\) true.
Key Concepts
Function EvaluationAlgebraic ManipulationSolving Equations
Function Evaluation
In mathematics, function evaluation involves calculating the value of a function for a specified input. To do this, you simply substitute the input into the function's formula in place of the variable.
For example, with a quadratic function like \( f(x) = x^2 - 3x + 1 \), evaluating the function at a specific point involves replacing \( x \) with that point's value.
This is done to determine the output or function value. In our exercise, we're tasked with finding \( f(2\alpha) \) and \( f(\alpha) \).
Here’s how:
For example, with a quadratic function like \( f(x) = x^2 - 3x + 1 \), evaluating the function at a specific point involves replacing \( x \) with that point's value.
This is done to determine the output or function value. In our exercise, we're tasked with finding \( f(2\alpha) \) and \( f(\alpha) \).
Here’s how:
- For \( f(2\alpha) \), substitute \( 2\alpha \) wherever there's an \( x \) in the function, giving \( (2\alpha)^2 - 3(2\alpha) + 1 = 4\alpha^2 - 6\alpha + 1 \).
- Similarly, for \( f(\alpha) \), substitute \( \alpha \), resulting in \( \alpha^2 - 3\alpha + 1 \).
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying expressions or equations to reveal hidden values or relationships.
The heart of solving our exercise lies in proper algebraic manipulation.
Once you have established the expressions for \( f(2\alpha) \) and \( f(\alpha) \), it's time to manipulate them. Start with: - \( f(2\alpha) = 4\alpha^2 - 6\alpha + 1 \) - Double \( f(\alpha) \): \( 2(\alpha^2 - 3\alpha + 1) = 2\alpha^2 - 6\alpha + 2 \)Setting them equal, according to the problem condition \( f(2\alpha) = 2f(\alpha) \), you form the equation:
\[ 4\alpha^2 - 6\alpha + 1 = 2\alpha^2 - 6\alpha + 2 \]
This means you’ll need to:
The heart of solving our exercise lies in proper algebraic manipulation.
Once you have established the expressions for \( f(2\alpha) \) and \( f(\alpha) \), it's time to manipulate them. Start with: - \( f(2\alpha) = 4\alpha^2 - 6\alpha + 1 \) - Double \( f(\alpha) \): \( 2(\alpha^2 - 3\alpha + 1) = 2\alpha^2 - 6\alpha + 2 \)Setting them equal, according to the problem condition \( f(2\alpha) = 2f(\alpha) \), you form the equation:
\[ 4\alpha^2 - 6\alpha + 1 = 2\alpha^2 - 6\alpha + 2 \]
This means you’ll need to:
- Subtract \( 2\alpha^2 - 6\alpha + 2 \) from \( 4\alpha^2 - 6\alpha + 1 \).
- Simplify to create a simpler quadratic equation: \( 2\alpha^2 - 1 = 0 \).
Solving Equations
Solving equations is about finding what values satisfy an equation.
In our case, solving the quadratic equation \( 2\alpha^2 - 1 = 0 \) involves finding the value or values of \( \alpha \) that make this true.
To solve for \( \alpha \), follow these logical steps:
Both values satisfy the initial condition set in the original equation, \( f(2\alpha) = 2f(\alpha) \), meaning they are valid solutions. Effective equation solving is foundational in finding correct mathematical relationships.
In our case, solving the quadratic equation \( 2\alpha^2 - 1 = 0 \) involves finding the value or values of \( \alpha \) that make this true.
To solve for \( \alpha \), follow these logical steps:
- Rearrange the equation: \( 2\alpha^2 = 1 \). This isolates the term involving \( \alpha \) on one side of the equation.
- Divide by 2 to simplify: \( \alpha^2 = \frac{1}{2} \).
- Take the square root of both sides: \( \alpha = \pm \frac{1}{\sqrt{2}} \).
Both values satisfy the initial condition set in the original equation, \( f(2\alpha) = 2f(\alpha) \), meaning they are valid solutions. Effective equation solving is foundational in finding correct mathematical relationships.
Other exercises in this chapter
Problem 6
If \(f(n+1)+f(n-1)=2 f(n)\) and \(f(0)=0\), then \(f(n), n \in N\) is (a) \(n f(1)\) (b) \((f(1))^{n}\) (c) 0 (d) None of these
View solution Problem 7
Let \(f\) be a function satisfying \(f(x+y)=f(x)+\) \(f(y)\) for all \(x, y \in R .\) If \(f(1)=K\), then \(f(n), n \in N\) is equal to (a) \(K^{n}\) (b) \(n K\
View solution Problem 10
If a function \(F\) is such that \(F(0)=2, F(1)=3\), \(F(n+2)=2 F(n)-F(n+1)\) for \(n \geq 0\), then \(F(5)\) is equal to (a) \(-7\) (b) \(-3\) (c) 7 (d) 13
View solution Problem 11
If \(f(x)=\frac{4^{x}}{4^{x}+2}\), then \(f\left(\frac{1}{1997}\right)+\) \(f\left(\frac{2}{1997}\right)+\ldots+\left(\frac{1996}{1997}\right)\) is equal to (a)
View solution