Let [tex]\( m \)[/tex] and [tex]\( M \)[/tex] be the minimum and maximum values of

[tex]\[
f(x) = \sin^{-1}(2x) + \sin(2x) + \cos^{-1}(2x) + \cos(2x), \quad x \in \left[9 \frac{\pi}{8}\right]
\][/tex]

What is [tex]\( m + M \)[/tex]?



Answer :

To analyze the function [tex]\( f(x) = \sin^{-1}(2x) + \sin(2x) + \cos^{-1}(2x) + \cos(2x) \)[/tex] over the interval [tex]\( x \in \left[9 \frac{\pi}{8}\right] \)[/tex], we need to first understand the domains and behavior of each component separately.

### Step-by-Step Solution

1. Understanding the domains of the functions:

- [tex]\( \sin^{-1}(u) \)[/tex] (or [tex]\( \arcsin(u) \)[/tex]) is defined for [tex]\( -1 \leq u \leq 1 \)[/tex].
- [tex]\( \cos^{-1}(u) \)[/tex] (or [tex]\( \arccos(u) \)[/tex]) is defined for [tex]\( -1 \leq u \leq 1 \)[/tex].
- [tex]\( \sin(2x) \)[/tex] and [tex]\( \cos(2x) \)[/tex] are defined and periodic for all [tex]\( x \)[/tex].

For [tex]\( \sin^{-1}(2x) \)[/tex] and [tex]\( \cos^{-1}(2x) \)[/tex] to be defined, [tex]\( 2x \)[/tex] must lie within [tex]\( [-1, 1] \)[/tex]. Hence:
[tex]\[ -1 \leq 2x \leq 1 \implies -\frac{1}{2} \leq x \leq \frac{1}{2} \][/tex]

2. Checking the given interval [tex]\( x \in \left[9 \frac{\pi}{8}\right] \)[/tex]:

- Clearly, [tex]\( 9 \frac{\pi}{8} \approx 3.534 \)[/tex] (which is out of the domain of [tex]\( \arcsin(2x) \)[/tex] and [tex]\( \arccos(2x) \)[/tex] since it demands the domain from [tex]\([-0.5 \leq x \leq 0.5]\)[/tex]).
- Therefore, we observe that neither [tex]\( \sin^{-1}(2x) \)[/tex] nor [tex]\( \cos^{-1}(2x) \)[/tex] can be evaluated at [tex]\( x = 9 \frac{\pi}{8} \)[/tex].

### Analysis:
Since the function [tex]\( f(x) = \sin^{-1}(2x) + \sin(2x) + \cos^{-1}(2x) + \cos(2x) \)[/tex] contains terms that are not defined for the given [tex]\( x \)[/tex] in [tex]\(\left[9 \frac{\pi}{8}\right] \)[/tex], the correct interpretation of the given interval may possibly have a typographical error.

Given this contradiction, we assume redefining proper interval for evaluating [tex]\( f(x) \)[/tex].

### Conclusion:
As the correct domain for [tex]\( x \)[/tex] must fit within [tex]\( -\frac{1}{2} \leq x \leq \frac{1}{2} \)[/tex], recheck the problem statement or another way both minimum (m) and maximum (M) cannot be determined at given [tex]\( x \)[/tex].

To summarise: [tex]\( f(x) \in \left[ -0.5 , 0.5 \right] \not \left[ 9 \frac{\pi}{8}\right] \)[/tex] for validity.

Without further provided corrections to the interval mentioned, solving for [tex]\( m + M \)[/tex] with existing data may not work as function components induce errors.