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.
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
- 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
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 :
- If the domain of the function $f(x) = \cos^{-1}\left( \frac{2x - 5}{11 - 3x} \right) + \sin^{-1}(2x^2 - 3x + 1)$ is the interval $[\alpha, \beta]$, then $\alpha + 2\beta$ is equal to :
- 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 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 :
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
