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,
Function Behavior Analysis
We are given a twice differentiable function $f: \mathbf{R} \to \mathbf{R}$ with $f''(x) > 0$ for all $x$. This condition implies that $f$ is strictly convex, and its first derivative, $f'(x)$, is strictly increasing.
We are also given $f'(a-1) = 0$. Since $f'(x)$ is strictly increasing, this means $f'(x) < 0$ for $x < a-1$ and $f'(x) > 0$ for $x > a-1$. The point $x = a-1$ corresponds to the global minimum of $f(x)$.
A composite function $g(x)$ is defined as $g(x) = f(\tan^2 x - 2\tan x + a)$ for $0 < x < \frac{\pi}{2}$.
Derivative of the Inner Function
Let the inner function be $h(x) = \tan^2 x - 2\tan x + a$. We can rewrite $h(x)$ by completing the square:
$ h(x) = (\tan x - 1)^2 + a - 1 $To determine the behavior of $g(x)$, we first find the derivative of $g(x)$ using the chain rule:
$ g'(x) = \frac{d}{dx} f(h(x)) = f'(h(x)) \cdot h'(x) $Now, let's find the derivative of $h(x)$:
$ h'(x) = \frac{d}{dx} (\tan^2 x - 2\tan x + a) $ $ h'(x) = 2\tan x \sec^2 x - 2\sec^2 x $ $ h'(x) = \sec^2 x (2\tan x - 2) $ $ h'(x) = 2\sec^2 x (\tan x - 1) $Analysis of $h'(x)$ Sign
For $x \in \left(0, \frac{\pi}{2}\right)$, we know $\sec^2 x > 0$. Therefore, the sign of $h'(x)$ depends on the term $(\tan x - 1)$.
- In the interval $\left(0, \frac{\pi}{4}\right)$, $\tan x < 1$, so $(\tan x - 1) < 0$. Thus, $h'(x) < 0$.
- In the interval $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, $\tan x > 1$, so $(\tan x - 1) > 0$. Thus, $h'(x) > 0$.
Analysis of the Inner Function $h(x)$
The minimum value of $h(x) = (\tan x - 1)^2 + a - 1$ occurs when $(\tan x - 1)^2$ is minimal, which is $0$. This happens when $\tan x = 1$, corresponding to $x = \frac{\pi}{4}$.
The minimum value of $h(x)$ is $h(\frac{\pi}{4}) = (1-1)^2 + a - 1 = a - 1$.
Therefore, $h(x) \ge a - 1$ for all $x$ in the domain $\left(0, \frac{\pi}{2}\right)$.
Analysis of $f'(h(x))$ Sign
Since $h(x) \ge a - 1$ and $f'(y)$ is strictly increasing with $f'(a-1) = 0$, we have:
$ f'(h(x)) \ge f'(a-1) = 0 $Thus, $f'(h(x))$ is non-negative for all $x \in \left(0, \frac{\pi}{2}\right)$.
Determining the Behavior of $g(x)$
Now we analyze the sign of $g'(x) = f'(h(x)) \cdot h'(x)$:
- Interval $\left(0, \frac{\pi}{4}\right)$: Here, $f'(h(x)) \ge 0$ and $h'(x) < 0$. So, $g'(x) \le 0$. This means $g(x)$ is decreasing in $\left(0, \frac{\pi}{4}\right)$.
- Interval $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$: Here, $f'(h(x)) \ge 0$ and $h'(x) > 0$. So, $g'(x) \ge 0$. This means $g(x)$ is increasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$.
Evaluating the Statements
- Statement (I): $g$ is increasing in $\left(0, \frac{\pi}{4}\right)$. This is False, as $g(x)$ is decreasing in this interval.
- Statement (II): $g$ is decreasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. This is False, as $g(x)$ is increasing in this interval.
Since both statements (I) and (II) are false, neither (I) nor (II) is true.
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