If the domain of the function f(x) = sin^-1 ((5-x)/(3+2x)) + 1/loge(10-x) is (-∞, α] ∪ [β, γ) - {δ},

Q: Why do we square both sides of the inequality $\left| \frac{5-x}{3+2x} \right| \leq 1$? Is it always safe?

That is a great question! Squaring both sides of an inequality is safe only when both sides are non-negative. In this case, the absolute value is always non-negative, and $1$ is positive. Since both sides are non-negative, squaring them preserves the inequality sign, allowing us to avoid splitting the problem into multiple cases based on the sign of the denominator.

Q: Why is the domain of $\log_e(10-x)$ not just $x < 10$? Why do we exclude $9$?

You are absolutely right that the argument of the logarithm must be positive, which gives $x < 10$. However, look at the structure of the function again. The term $\frac{1}{\log_e(10-x)}$ is a fraction. If the denominator, $\log_e(10-x)$, equals zero, the function becomes undefined. Since $\log_e(10-x) = 0$ when $10-x = 1$, we must exclude the value where $x = 9$.

Q: What happens to the constraint $x \neq -\frac{3}{2}$ from the denominator of the sine inverse argument?

Excellent observation! We must always check our constraints. The condition $3+2x \neq 0$ implies $x \neq -1.5$. When we solved the inequality for the sine inverse, we found the valid interval to be $(-\infty, -8] \cup [\frac{2}{3}, \infty)$. Since $-1.5$ is not in this interval, it is automatically excluded by the domain of the sine inverse, so we don't need to worry about it separately.

Question

If the domain of the function $f(x) = \sin^{-1} \left(\frac{5-x}{3+2x}\right) + \frac{1}{\log_e(10-x)}$ is $(-\infty, \alpha] \cup [\beta, \gamma) - \{\delta\}$, then $6(\alpha + \beta + \gamma + \delta)$ is equal to

Options:

(A) 67

(B) 66

(C) 70

(D) 68

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the domain of the function $f(x) = \sin^{-1} \left( \frac{5-x}{3+2x} ight) + \frac{1}{\log_e(10-x)}$, we must satisfy the conditions imposed by both the inverse sine component and the logarithmic denominator. The domain is the intersection of all valid regions for $x$.

### The Inverse Sine Constraint

The first gatekeeper is the inverse sine function, $\sin^{-1}(u)$, which requires the input $u$ to be trapped within the closed interval $[-1, 1]$. We set up the inequality:

$$-1 \leq \frac{5-x}{3+2x} \leq 1$$

This is equivalent to the condition $\left| \frac{5-x}{3+2x} ight| \leq 1$. Since the denominator must not be zero, we note $x 
eq -\frac{3}{2}$. Squaring both sides of the inequality yields:

$$\frac{(5-x)^2}{(3+2x)^2} \leq 1$$

Expanding the squares, we obtain $25 - 10x + x^2 \leq 9 + 12x + 4x^2$. Rearranging all terms to one side results in the quadratic inequality:

$$3x^2 + 22x - 16 \geq 0$$

Factoring this quadratic, we find $(3x-2)(x+8) \geq 0$. Using the wavy curve method, we identify the valid intervals for this constraint as $x \in (-\infty, -8] \cup [\frac{2}{3}, \infty)$, while excluding the vertical asymptote at $x = -\frac{3}{2}$.

### The Logarithmic Constraint

Next, we examine the term $\frac{1}{\log_e(10-x)}$. Logarithms require a strictly positive argument, so we must have $10-x > 0$, which simplifies to $x < 10$.

Furthermore, because this term is in the denominator, the expression $\log_e(10-x)$ cannot equal zero. Since $\log_e(10-x) = 0$ when $10-x = 1$, we must exclude the value $x = 9$. Thus, the second gatekeeper demands $x < 10$ and $x 
eq 9$.

### The Intersection and Final Victory

We now combine our findings. We require $x$ to be in $(-\infty, -8] \cup [\frac{2}{3}, \infty)$ while simultaneously satisfying $x < 10$ and $x 
eq 9$. Note that the value $x = -\frac{3}{2}$ is already excluded by the first interval.

Visualizing this on a number line, the overlap is $(-\infty, -8] \cup [\frac{2}{3}, 10) \setminus \{9\}$. Comparing this to the form $(-\infty, \alpha] \cup [\beta, \gamma) \setminus \{\delta\}$, we identify the parameters:

$$\alpha = -8, \quad \beta = \frac{2}{3}, \quad \gamma = 10, \quad \delta = 9$$

The final step is to calculate the requested value:

$$6(\alpha + \beta + \gamma + \delta) = 6\left(-8 + \frac{2}{3} + 10 + 9ight)$$

$$= 6\left(11 + \frac{2}{3}ight) = 6\left(\frac{35}{3}ight) = 70$$

Through careful analysis and logical deduction, we have determined the final result is **70**.

Animated Solution for Mathematics - Functions: If the domain of the function $f (x) = sin^{- 1} (\frac{5 - x}{3 + 2 x}) + \frac{1}{l o g _{e} ( 10 - x )}$ is $(- \infty, α] \cup [β, γ) - {δ}$ , then $6 (α + β + γ + δ)$ is equal to

Select Answer:

Visualized Solution

Analyzing the Function Components

Condition 1: Domain of $sin^{- 1} (u)$

Solving the Rational Inequality

Expanding the Inequality

Factoring the Quadratic

Interval for Part 1

Condition 2: Domain of $lo g_{e} (v)$

Condition 3: Denominator Constraint

Finding the Intersection

Identifying $α, β, γ, δ$

Final Calculation

The Sigma Insight: Domain and Range of a Function

Analyzing the Setup

The Inverse Sine Constraint

The Logarithmic Constraint

The Intersection and Final Victory

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