Question

The domain of the function $f(z) = \log_e \left\{ \log_{|\sin x|} (x^2 - 8x + 23) - \frac{3}{\log_2 |\sin x|} \right\}$ contains which of the following interval/ intervals.

• A. $(3, \pi)$
• B. $(\pi, \frac{3\pi}{2})$
• C. $(\frac{3\pi}{2}, 5)$
• D. None

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The function given is $f(z) = \log_e \left\{ \log_{|\sin x|} (x^2 - 8x + 23) - \frac{3}{\log_2 |\sin x|} \right\}$.

For $f(z)$ to be defined, we must satisfy the following conditions:

1. Argument of the outermost logarithm must be positive:
   Let $Y = \log_{|\sin x|} (x^2 - 8x + 23) - \frac{3}{\log_2 |\sin x|}$. We need $Y > 0$.

2. Base of the inner logarithms must be positive and not equal to 1:
   $|\sin x| > 0 \implies \sin x \neq 0 \implies x \neq n\pi$ for any integer $n$.
   $|\sin x| \neq 1 \implies \sin x \neq \pm 1 \implies x \neq (2n+1)\frac{\pi}{2}$ for any integer $n$.
   Combining these, $x \neq \frac{m\pi}{2}$ for any integer $m$.

3. Argument of the inner logarithm must be positive:
   $x^2 - 8x + 23 > 0$.
   To check this quadratic, consider its discriminant $D = (-8)^2 - 4(1)(23) = 64 - 92 = -28$.
   Since $D < 0$ and the leading coefficient (1) is positive, the quadratic $x^2 - 8x + 23$ is always positive for all real $x$. This condition is satisfied for all $x \in \mathbb{R}$.

Now, let's simplify the expression $Y$:
$Y = \log_{|\sin x|} (x^2 - 8x + 23) - \frac{3}{\log_2 |\sin x|}$
Using the change of base formula for logarithms, $\frac{1}{\log_a b} = \log_b a$.
So, $\frac{1}{\log_2 |\sin x|} = \log_{|\sin x|} 2$.
Substitute this into the expression for $Y$:
$Y = \log_{|\sin x|} (x^2 - 8x + 23) - 3 \log_{|\sin x|} 2$
Using the logarithm property $k \log_b a = \log_b a^k$:
$Y = \log_{|\sin x|} (x^2 - 8x + 23) - \log_{|\sin x|} 2^3$
$Y = \log_{|\sin x|} (x^2 - 8x + 23) - \log_{|\sin x|} 8$
Using the logarithm property $\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)$:
$Y = \log_{|\sin x|} \left( \frac{x^2 - 8x + 23}{8} \right)$

We need $Y > 0$, so $\log_{|\sin x|} \left( \frac{x^2 - 8x + 23}{8} \right) > 0$.
From condition 2, we know that $0 < |\sin x| < 1$ (since $|\sin x|$ cannot be greater than 1).
When the base $b$ of a logarithm $\log_b A$ is between 0 and 1 ($0 < b < 1$), the logarithm function is decreasing. Therefore, $\log_b A > 0$ implies $A < b^0$, which means $A < 1$.
Applying this to our inequality:
$\frac{x^2 - 8x + 23}{8} < 1$
$x^2 - 8x + 23 < 8$
$x^2 - 8x + 15 < 0$
Factor the quadratic: $(x-3)(x-5) < 0$.
This inequality holds when $x$ is between the roots, i.e., $3 < x < 5$.

Finally, we must combine this interval $(3, 5)$ with the restrictions from condition 2 ($x \neq \frac{m\pi}{2}$).
In the interval $(3, 5)$:

• $\pi \approx 3.14159$. Since $3 < \pi < 5$, we must exclude $x = \pi$.
• $\frac{3\pi}{2} \approx 4.71239$. Since $3 < \frac{3\pi}{2} < 5$, we must exclude $x = \frac{3\pi}{2}$.

No other values of $\frac{m\pi}{2}$ fall within the interval $(3, 5)$. (e.g., $\frac{\pi}{2} \approx 1.57$, $2\pi \approx 6.28$)

Therefore, the domain of the function $f(z)$ is $(3, 5)$ excluding $\pi$ and $\frac{3\pi}{2}$.
This can be written as the union of three intervals: $(3, \pi) \cup (\pi, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, 5)$.

Now, let's check the given options:
A $(3, \pi)$: This interval is a part of the calculated domain.
B $(\pi, \frac{3\pi}{2})$: This interval is a part of the calculated domain.
C $(\frac{3\pi}{2}, 5)$: This interval is a part of the calculated domain.
D None: This is incorrect as A, B, and C are all valid parts of the domain.

The question asks "contains which of the following interval/ intervals", implying that multiple options can be correct. All three intervals listed in options A, B, and C are contained within the domain of the function.