Problem 2
Question
If \(b^{2}-4 a c=0\) and \(a>0\), then the domain of the function \(f(x)=\log \left(a x^{3}+(2 a+b) x^{2}+(2 b+c) x+2 c\right)\) is (A) \((-2, \infty) \backslash\left\\{-\frac{b}{2 a}\right\\}\) (B) \([-2, \infty) \backslash\left\\{-\frac{b}{2 a}\right\\}\) (C) \((-\infty,-2) \backslash\left\\{-\frac{b}{2 a}\right\\}\) (D) None of these
Step-by-Step Solution
Verified Answer
The domain is \((-2, \, \infty) \setminus \{-\frac{b}{2a}\}\). (Option A)
1Step 1: Understanding the Problem
We need to find the domain of the logarithmic function \(f(x) = \log(a x^3 + (2a+b)x^2 + (2b+c)x + 2c)\). The expression inside the logarithm must be greater than zero. Additionally, given \(b^2 - 4ac = 0\), we know the quadratic equation has one unique real root.
2Step 2: Substitution and Simplification
Since \(b^2 - 4ac = 0\), the quadratic equation \(ax^2 + bx + c = 0\) has a double root at \(x = -\frac{b}{2a}\). Substitute this into our expression: \(x = -\frac{b}{2a}\). Thus, the original cubic polynomial at this point becomes zero or a turning point.
3Step 3: Analysis of Polynomial Behavior
Analyze the polynomial's behavior using the double root \(x = -\frac{b}{2a}\). Given this is a cubic expression due to higher powers of \(x\), test intervals around \(x = -\frac{b}{2a}\) and \(x = -2\) to determine where the expression is positive.
4Step 4: Interval Testing for Domain
1. Choose test points in intervals: \((-\infty, -2)\), \((-2, -\frac{b}{2a})\), and \((-\frac{b}{2a}, \infty)\).2. Calculate the sign of the expression in these intervals; the function is defined where it is positive.3. Since \(a > 0\), test intervals suggest positivity from \(-2\) to \(\infty\), but excluding \(-\frac{b}{2a}\) due to zero at this point.
5Step 5: Conclusion
Hence, the expression inside the log function is greater than zero for \(( -2, \, \infty)\), but undefined at \(-\frac{b}{2a}\), which must be excluded from the domain.
Key Concepts
Domain of a FunctionLogarithmic FunctionQuadratic EquationsPolynomial Analysis
Domain of a Function
When working with functions, especially in mathematics, the domain of a function is crucial. It represents all the possible input values that a function can accept. For the function \( f(x) = \log(a x^3 + (2a+b)x^2 + (2b+c)x + 2c) \), the domain is determined by the expression inside the logarithm.
This specific expression must always be positive, as logarithms of zero or negative numbers are undefined.
Thus, we must ensure that the cubic polynomial inside the log is greater than zero. This involves solving inequalities to find where the expression is positive, and checking critical points where it may turn zero.
This specific expression must always be positive, as logarithms of zero or negative numbers are undefined.
Thus, we must ensure that the cubic polynomial inside the log is greater than zero. This involves solving inequalities to find where the expression is positive, and checking critical points where it may turn zero.
Logarithmic Function
Logarithmic functions are vital in mathematics, especially in complex equations. They are the inverses of exponential functions. The natural logarithm, \( \log(x) \), only accepts positive argument values.
For \( f(x) = \log(a x^3 + (2a+b)x^2 + (2b+c)x + 2c) \), solely positive values of the polynomial are valid.
For \( f(x) = \log(a x^3 + (2a+b)x^2 + (2b+c)x + 2c) \), solely positive values of the polynomial are valid.
- Zero or negative arguments result in undefined outputs.
- Thus, finding where the polynomial is positive specifies the domain.
Quadratic Equations
Quadratic equations, often represented as \( ax^2 + bx + c = 0 \), play a role in this exercise. They help identify key properties of the expression inside the logarithm. Specifically, \( b^2 - 4ac = 0 \) reveals that the quadratic has one unique root: \( x = -\frac{b}{2a} \).
This unique solution means the quadratic expression is a perfect square, reaching zero only at this point. It also acts as a turning point in the polynomial behavior around this x-value. The double root property is crucial in segmenting the domain, helping to exclude points where the expression goes to zero.
This unique solution means the quadratic expression is a perfect square, reaching zero only at this point. It also acts as a turning point in the polynomial behavior around this x-value. The double root property is crucial in segmenting the domain, helping to exclude points where the expression goes to zero.
Polynomial Analysis
Polynomial analysis involves studying the properties and behaviors of polynomial functions like the cubic polynomial in this task.
By breaking it into intervals, we explore where the function is positive or negative, which dictates the domain.
If \( a > 0 \), the graph tends to rise from left to right, impacting positivity intervals respectively.
Critical points, like \( -\frac{b}{2a} \), create transitions in sign or behavior, requiring analytical checks step-by-step.
This methodical analysis helps in determining where the logarithmic function is valid.
By breaking it into intervals, we explore where the function is positive or negative, which dictates the domain.
- Identify roots and their multiplicities.
- Test intervals between roots to see sign changes.
If \( a > 0 \), the graph tends to rise from left to right, impacting positivity intervals respectively.
Critical points, like \( -\frac{b}{2a} \), create transitions in sign or behavior, requiring analytical checks step-by-step.
This methodical analysis helps in determining where the logarithmic function is valid.
Other exercises in this chapter
Problem 1
Let \(f(x)=x^{3}+x^{2}+100 x+7 \sin x\), then the equation \(\frac{1}{y-f(1)}+\frac{2}{y-f(2)}+\frac{3}{y-f(3)}=0\) has (A) one real root (B) two real roots (C)
View solution Problem 3
If \(e^{x}+e^{f(x)}=e\), then range of the function \(f\) is (A) \((-\infty, 1]\) (B) \((-\infty, 1)\) (C) \((1, \infty)\) (D) \([1, \infty)\)
View solution Problem 4
Which of the following functions is are injective \(\operatorname{map}(\mathrm{s}) ?\) (A) \(f(x)=x^{2}+2, x \in(-\infty, \infty)\) (B) \(f(x)=|x+2|, x \in[-2,
View solution Problem 5
The graph of the function \(\cos x \cos (x+2)-\cos ^{2}(x+1)\) is (A) a straight line passing through \(\left(0,-\sin ^{2} 1\right)\) with slope 2 (B) a straigh
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