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 $\beta = 6$.
Similar Questions
- Let $[\cdot]$ denote the greatest integer function, and let $f(x) = \min \{\sqrt{2}x, x^2\}$.
Let $S = \{x \in (-2, 2) : \text{the function } g(x) = |x|[x^2] \text{ is discontinuous at } x\}$.
Then $\sum_{x \in S} f(x)$ equals - If the function $f(x) = \frac{e^x\left(e^{\tan x - x} - 1\right) + \log_e(\sec x + \tan x) - x}{\tan x - x}$ is continuous at $x = 0$, then the value of $f(0)$ is equal to
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- Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_e x| - |x - 1|$:
(I) $f$ is differentiable at all $x > 0$.
(II) $f$ is increasing in $(0, 1)$.
(III) $f$ is decreasing in $(1, \infty)$.
Then. - If the domain of the function $f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x - 2} \right)$ is $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, then $\alpha + \beta + \gamma + \delta$ is equal to
Let $[t]$ denote the greatest integer less than or equal to $t$. If the function
$f(x) = \begin{cases} b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x \right] \right), & x < 0 \\ \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases}$
is continuous at $x = 0$, then $a^2 + b^2$ is equal to
- Let $f: \mathbb{R} \rightarrow (0, \infty)$ be a twice differentiable function such that $f(3) = 18, f'(3) = 0$ and $f''(3) = 4$. Then $\lim_{x \to 1} \left( \log_e \left( \frac{f(2+x)}{f(3)} \right)^{\frac{18}{(x-1)^2}} \right)$ is equal to :
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- Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ in m, has two equal roots for every $x \in \mathbb{R}$. If $f(0) = 1, f'(0) = 2$, and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to ____________.
- Let $f: \mathbf{R} \to \mathbf{R}$ be a twice differentiable function such that $f''(x) > 0$ for all $x \in \mathbf{R}$ and $f'(a-1) = 0$, where $a$ is a real number. Let $g(x) = f(\tan^2 x - 2\tan x + a)$, $0 < x < \frac{\pi}{2}$.
Consider the following two statements :
(I) $g$ is increasing in $\left(0, \frac{\pi}{4}\right)$
(II) $g$ is decreasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
Then,
Important Questions from Differential Calculus
- Let $[\cdot]$ denote the greatest integer function, and let $f(x) = \min \{\sqrt{2}x, x^2\}$.
Let $S = \{x \in (-2, 2) : \text{the function } g(x) = |x|[x^2] \text{ is discontinuous at } x\}$.
Then $\sum_{x \in S} f(x)$ equals - If the function $f(x) = \frac{e^x\left(e^{\tan x - x} - 1\right) + \log_e(\sec x + \tan x) - x}{\tan x - x}$ is continuous at $x = 0$, then the value of $f(0)$ is equal to
- Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x - 2) + \alpha x^2 + 4x - \alpha$, $x > 2$, is ______
- Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_e x| - |x - 1|$:
(I) $f$ is differentiable at all $x > 0$.
(II) $f$ is increasing in $(0, 1)$.
(III) $f$ is decreasing in $(1, \infty)$.
Then. - If the domain of the function $f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x - 2} \right)$ is $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, then $\alpha + \beta + \gamma + \delta$ is equal to
