(I) $f$ is differentiable at all $x > 0$.
(II) $f$ is increasing in $(0, 1)$.
(III) $f$ is decreasing in $(1, \infty)$.
Then.
Differentiability Analysis of Function $f(x)$
The function is given by $f(x) = |\log_e x| - |x - 1|$ for $x \in (0, \infty)$. We analyze its differentiability, focusing on the point $x=1$ where the absolute value arguments become zero.
First, we express $f(x)$ piecewise:
- For $0 < x < 1$: $\log_e x < 0$ and $x - 1 < 0$. So, $f(x) = -(\log_e x) - (-(x - 1)) = x - \log_e x - 1$.
- For $x = 1$: $f(1) = |\log_e 1| - |1 - 1| = |0| - |0| = 0$.
- For $x > 1$: $\log_e x > 0$ and $x - 1 > 0$. So, $f(x) = (\log_e x) - (x - 1) = \log_e x - x + 1$.
Statement (I): $f$ is differentiable at all $x > 0$.
We check differentiability at $x=1$ by comparing the left-hand derivative (LHD) and the right-hand derivative (RHD).
LHD at $x=1$:
Using the derivative for $0 < x < 1$, $f'(x) = 1 - \frac{1}{x}$.
LHD = $\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} \left(1 - \frac{1}{x}\right) = 1 - 1 = 0$.
RHD at $x=1$:
Using the derivative for $x > 1$, $f'(x) = \frac{1}{x} - 1$.
RHD = $\lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} \left(\frac{1}{x} - 1\right) = 1 - 1 = 0$.
Since LHD = RHD = 0, the function $f$ is differentiable at $x=1$. For $x \neq 1$, the function is composed of standard functions whose absolute values are differentiable away from zero. Therefore, $f$ is differentiable for all $x > 0$.
Conclusion for Statement (I): Statement (I) is TRUE.
Monotonicity Analysis in $(0, 1)$
Statement (II): $f$ is increasing in $(0, 1)$.
For $0 < x < 1$, $f(x) = x - \log_e x - 1$.
The derivative is $f'(x) = 1 - \frac{1}{x}$.
In the interval $(0, 1)$, we have $0 < x < 1$, which means $\frac{1}{x} > 1$.
Therefore, $f'(x) = 1 - \frac{1}{x} < 0$ for $x \in (0, 1)$.
A negative derivative indicates the function is decreasing.
Conclusion for Statement (II): Statement (II) is FALSE.
Monotonicity Analysis in $(1, \infty)$
Statement (III): $f$ is decreasing in $(1, \infty)$.
For $x > 1$, $f(x) = \log_e x - x + 1$.
The derivative is $f'(x) = \frac{1}{x} - 1$.
In the interval $(1, \infty)$, we have $x > 1$, which means $0 < \frac{1}{x} < 1$.
Therefore, $f'(x) = \frac{1}{x} - 1 < 0$ for $x \in (1, \infty)$.
A negative derivative indicates the function is decreasing.
Conclusion for Statement (III): Statement (III) is TRUE.
Overall Conclusion
Summarizing the findings:
- Statement (I) is TRUE.
- Statement (II) is FALSE.
- Statement (III) is TRUE.
Thus, only statements (I) and (III) are TRUE.
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 __________.
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- 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 ________
