Function Domain Analysis
The function is given by $f(x) = \log_3 \log_5 \left(7 - \log_2 \left(x^2 - 10x + 85\right)\right) + \sin^{-1} \left(\left|\frac{3x - 7}{17 - x}\right|\right)$. The domain is specified as $(\alpha, \beta]$. We need to determine the constraints on $x$ imposed by the function's components.
Logarithmic Constraints
The function involves nested logarithms ($\log_3$, $\log_5$, $\log_2$). The arguments of these logarithms must be positive.
- Argument of $\log_2$: $x^2 - 10x + 85$. The discriminant is $\Delta = (-10)^2 - 4(1)(85) = 100 - 340 = -240$. Since $\Delta < 0$ and the leading coefficient ($1$) is positive, the quadratic $x^2 - 10x + 85$ is always positive for all real $x$. This condition is satisfied for all $x \in \mathbb{R}$.
- Argument of $\log_5$: $7 - \log_2(x^2 - 10x + 85)$. This must be positive. $7 - \log_2(x^2 - 10x + 85) > 0$ $7 > \log_2(x^2 - 10x + 85)$ $2^7 > x^2 - 10x + 85$ $128 > x^2 - 10x + 85$ $x^2 - 10x - 43 < 0$ The roots of $x^2 - 10x - 43 = 0$ are $x = \frac{10 \pm \sqrt{100 - 4(1)(-43)}}{2} = 5 \pm 2\sqrt{17}$. Thus, the inequality holds for $x \in (5 - 2\sqrt{17}, 5 + 2\sqrt{17})$.
- Argument of $\log_3$: $\log_5(7 - \log_2(x^2 - 10x + 85))$. This must be positive. $\log_5(7 - \log_2(x^2 - 10x + 85)) > 0$ $7 - \log_2(x^2 - 10x + 85) > 5^0 = 1$ $6 > \log_2(x^2 - 10x + 85)$ $2^6 > x^2 - 10x + 85$ $64 > x^2 - 10x + 85$ $x^2 - 10x + 21 < 0$ $(x - 3)(x - 7) < 0$ Thus, the inequality holds for $x \in (3, 7)$.
Comparing the intervals derived from conditions 2 and 3, $x \in (3, 7)$ is the more restrictive condition because $3 > 5 - 2\sqrt{17}$ and $7 < 5 + 2\sqrt{17}$. Therefore, the logarithmic constraints collectively require $x \in (3, 7)$.
Inverse Sine Constraints
The function involves $\sin^{-1}(y)$, which is defined only for $y \in [-1, 1]$.
In this case, $y = \left|\frac{3x - 7}{17 - x}\right|$. The condition is $-1 \le \left|\frac{3x - 7}{17 - x}\right| \le 1$.
Since the absolute value $\left|\frac{3x - 7}{17 - x}\right|$ is always non-negative, the condition simplifies to $\left|\frac{3x - 7}{17 - x}\right| \le 1$. This is equivalent to two simultaneous inequalities:
$-1 \le \frac{3x - 7}{17 - x} \le 1$We solve these separately:
- First inequality: $\frac{3x - 7}{17 - x} \le 1$ $\frac{3x - 7}{17 - x} - 1 \le 0$ $\frac{3x - 7 - (17 - x)}{17 - x} \le 0$ $\frac{4x - 24}{17 - x} \le 0 \implies \frac{x - 6}{17 - x} \le 0$ The critical points are $x=6$ (numerator is 0) and $x=17$ (denominator is 0). Analyzing the sign intervals shows the solution is $x \in (-\infty, 6] \cup (17, \infty)$.
- Second inequality: $\frac{3x - 7}{17 - x} \ge -1$ $\frac{3x - 7}{17 - x} + 1 \ge 0$ $\frac{3x - 7 + (17 - x)}{17 - x} \ge 0$ $\frac{2x + 10}{17 - x} \ge 0 \implies \frac{x + 5}{17 - x} \ge 0$ The critical points are $x=-5$ (numerator is 0) and $x=17$ (denominator is 0). Analyzing the sign intervals shows the solution is $x \in [-5, 17)$.
To satisfy the condition $|\frac{3x - 7}{17 - x}| \le 1$, we need the intersection of the solutions from both inequalities:
$( (-\infty, 6] \cup (17, \infty) ) \cap [-5, 17)$The intersection yields the interval $x \in [-5, 6]$.
Combining Constraints for Domain
The domain of the function $f(x)$ is the intersection of the intervals derived from all constraints.
- Logarithmic constraints require $x \in (3, 7)$.
- Inverse sine constraint requires $x \in [-5, 6]$.
We find the intersection of these two intervals:
$(3, 7) \cap [-5, 6] = (3, 6]$The domain of the function $f(x)$ is $(3, 6]$. This matches the given format $(\alpha, \beta]$.
By comparison, $\alpha = 3$ and $\beta = 6$.
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Important Questions from Differential Calculus
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