If the domain of the function f(x) = loge((2x+3)/(4x²+x-3)) + cos⁻¹((2x-1)/(x+2)) is (α, β], then th

Q: Wait, why can't I just cross-multiply the inequality $\frac{2x-1}{x+2} \le 1$ to get $2x-1 \le x+2$?

That is the most dangerous trap in inequality algebra! When you cross-multiply, you are multiplying by $(x+2)$. Since $x$ is a variable, we don't know if $(x+2)$ is positive or negative. If it's negative, the inequality sign must flip, but if it's positive, it stays the same. Because we don't know the sign, we cannot safely cross-multiply. Always bring all terms to one side to form a single rational expression and use the Wavy Curve Method instead.

Q: How do I know if the brackets should be open or closed at the boundary points?

Look at the original inequality sign. If the inequality is strict ($>$ or $<$), the boundary point makes the expression zero or undefined, so we use an open bracket (e.g., $()$). If the inequality is non-strict ($\ge$ or $\le$), and the point makes the expression zero, we include it with a closed bracket (e.g., $[]$). However, if a point makes the denominator zero, it must always be an open bracket, regardless of the inequality sign, because division by zero is undefined!

Q: Why do we take the intersection of the two domains instead of the union?

Think of the function $f(x)$ as a machine that requires two inputs to work. If the logarithm condition is met but the inverse cosine condition fails, the machine breaks. If the inverse cosine condition is met but the logarithm condition fails, the machine still breaks. For the function to exist, both conditions must be satisfied simultaneously. In set theory, 'simultaneously' translates to the intersection of the two sets.

Question

If the domain of the function $f(x) = \log_e (\frac{2x+3}{4x^2+x-3}) + \cos^{-1}(\frac{2x-1}{x+2})$ is $(\alpha, \beta]$, then the value of $5\beta - 4\alpha$ is equal to

Options:

(A) 10

(B) 12

(C) 11

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the domain of the function $f(x) = \log_e \left(\frac{2x+3}{4x^2+x-3}ight) + \cos^{-1}\left(\frac{2x-1}{x+2}ight)$, we must ensure that both components of the function are simultaneously defined.

The function is defined only where the logarithmic argument is strictly positive and the inverse cosine argument lies within the closed interval $[-1, 1]$.

### The Logarithm's Strict Demand

The natural logarithm $\log_e(u)$ requires $u > 0$. Thus, we require:

$$\frac{2x+3}{4x^2+x-3} > 0$$

First, we factor the denominator $4x^2+x-3$. By splitting the middle term, we obtain:

$$4x^2 + 4x - 3x - 3 = 4x(x+1) - 3(x+1) = (4x-3)(x+1)$$

The inequality becomes:

$$\frac{2x+3}{(4x-3)(x+1)} > 0$$

Using the Wavy Curve Method with critical points at $x = -\frac{3}{2}$, $x = -1$, and $x = \frac{3}{4}$, we determine the solution set for the logarithm:

$$x \in \left( -\frac{3}{2}, -1 ight) \cup \left( \frac{3}{4}, \infty ight)$$

### The Inverse Cosine's Boundary

The inverse cosine function $\cos^{-1}(v)$ is defined for $v \in [-1, 1]$. This implies:

$$-1 \le \frac{2x-1}{x+2} \le 1$$

We solve this as two separate inequalities. First, $\frac{2x-1}{x+2} \ge -1$:

$$\frac{2x-1}{x+2} + 1 \ge 0 \implies \frac{3x+1}{x+2} \ge 0$$

The solution for this part is $x \in (-\infty, -2) \cup \left[ -\frac{1}{3}, \infty ight)$.

Next, we solve $\frac{2x-1}{x+2} \le 1$:

$$\frac{2x-1}{x+2} - 1 \le 0 \implies \frac{x-3}{x+2} \le 0$$

The solution for this part is $x \in (-2, 3]$. Taking the intersection of these two results, the domain for the inverse cosine component is:

$$x \in \left[ -\frac{1}{3}, 3 ight]$$

### The Grand Intersection

To find the domain of $f(x)$, we intersect the requirements from the logarithm and the inverse cosine:

$$	ext{Domain} = \left[ \left( -\frac{3}{2}, -1 ight) \cup \left( \frac{3}{4}, \infty ight) ight] \cap \left[ -\frac{1}{3}, 3 ight]$$

The interval $\left( -\frac{3}{2}, -1 ight)$ has no overlap with $\left[ -\frac{1}{3}, 3 ight]$. However, the interval $\left( \frac{3}{4}, \infty ight)$ overlaps with $\left[ -\frac{1}{3}, 3 ight]$ on the interval $\left( \frac{3}{4}, 3 ight]$.

Thus, the final domain is $\left( \frac{3}{4}, 3 ight]$.

### Final Calculation

Given the domain is $(\alpha, \beta]$, we identify $\alpha = \frac{3}{4}$ and $\beta = 3$. We are tasked to calculate $5\beta - 4\alpha$:

$$5(3) - 4\left( \frac{3}{4} ight) = 15 - 3 = 12$$

The final result is **12**.

Animated Solution for Mathematics - Functions: If the domain of the function $f (x) = lo g_{e} (\frac{2 x + 3}{4 x ^{2} + x - 3}) + cos^{- 1} (\frac{2 x - 1}{x + 2})$ is $(α, β]$ , then the value of $5 β - 4 α$ is equal to

Select Answer:

Visualized Solution

Understanding the Constraints

The Logarithm Condition

Factorizing the Quadratic

Solving the Log Inequality

Domain of Logarithm

The Inverse Cosine Condition

Solving the Left Inequality

Solving the Right Inequality

Domain of Inverse Cosine

Finding the Final Domain

Calculating $5 β - 4 α$

The Sigma Insight: Domain and Range of a Function

Analyzing the Setup

The Logarithm's Strict Demand

The Inverse Cosine's Boundary

The Grand Intersection

Final Calculation

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