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 $(2\alpha, \alpha)$ 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 ______
- 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 ____________.
- The number of points in the interval $[2, 4]$, at which the function $f(x) = \left[ x^2 - x - \frac{1}{2} \right]$, where $[\cdot]$ denotes the greatest integer function, is discontinuous, is __________.
- The number of points of discontinuity of the function $f (x) = \left[\frac{x^2}{2}\right] - [\sqrt{x}], x \in [0,4]$, where $[.]$denotes the greatest integer function, is ________
- The number of functions $f: \{1, 2, 3, 4\} \rightarrow \{a, b, c\}$, which are not onto, is:
- If $y = \tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) + 2\tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right)$, then $\frac{dy}{dx}$ at $x = \frac{\sqrt{3}}{2}$ is equal to:
- Let $f(x) = \lim_{\theta \rightarrow 0} \left(\frac{\cos\pi x - x^{\left(\frac{2}{\theta}\right)}\sin(x - 1)}{1 + x^{\left(\frac{2}{\theta}\right)}(x - 1)}\right)$, $x \in \mathbf{R}$. Consider the following two statements :
(I) $f(x)$ is discontinous at $x = 1$.
(II) $f(x)$ is continous at $x = -1$.
Then, - The sum of all the elements in the range of $f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x)$,
$x \neq \frac{n\pi}{2}, n \in \mathbf{Z}$, where $\text{Sgn}(t) = \begin{cases} 1, & \text{if } t > 0 \\ -1, & \text{if } t < 0 \end{cases}$, is : - 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, - 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 :
Important Questions from Differential Calculus
- 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 ______
- 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 ____________.
For a real number $y$, consider $[y]$ denotes the greatest integer less than or equal to $y$.
If $f(x) = \frac{\tan(\pi[x-\pi])}{1+[x]^2}$, then- The number of points in the interval $[2, 4]$, at which the function $f(x) = \left[ x^2 - x - \frac{1}{2} \right]$, where $[\cdot]$ denotes the greatest integer function, is discontinuous, is __________.
- The number of points of discontinuity of the function $f (x) = \left[\frac{x^2}{2}\right] - [\sqrt{x}], x \in [0,4]$, where $[.]$denotes the greatest integer function, is ________
