Consider the following three statements for t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The function is $f(x) = |\log_{e} x| - |x - 1|$.For $x \in (0, 1)$, $\log_{e} x < 0$ and $x - 1 < 0$, so $f(x) = -\log_{e} x - (-(x - 1)) = -\log_

Vedclass · Accepted Answer

Only $(I)$ and $(III)$ are $TRUE$.

Vedclass · Answer

(D) The function is $f(x) = |\log_{e} x| - |x - 1|$.For $x \in (0, 1)$, $\log_{e} x Then $f'(x) = -\frac{1}{x} + 1 = \frac{x - 1}{x}$. Since $x \in (0, 1)$, $f'(x) For $x \in (1, \infty)$, $\log_{e} x > 0$ and $x - 1 > 0$, so $f(x) = \log_{e} x - (x - 1) = \log_{e} x - x + 1$.Then $f'(x) = \frac{1}{x} - 1 = \frac{1 - x}{x}$. Since $x > 1$, $f'(x) At $x = 1$, $f(x) = |\log_{e} x| - |x - 1|$. The left-hand derivative at $x = 1$ is $\lim_{h 	o 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h 	o 0^-} \frac{-\log_{e}(1+h) + h}{h} = -1 + 1 = 0$. The right-hand derivative at $x = 1$ is $\lim_{h 	o 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h 	o 0^+} \frac{\log_{e}(1+h) - h}{h} = 1 - 1 = 0$. Since $f'(1) = 0$, the function is differentiable at $x = 1$. However, at $x = 1$, the absolute value functions $|\log_{e} x|$ and $|x - 1|$ are differentiable. Checking $x=1$ specifically, $f$ is differentiable. But for $f$ to be differentiable at all $x > 0$, we check $x=1$. Since $f'(1)=0$, it is differentiable. However, the statement $(I)$ is $TRUE$. Thus, only $(I)$ and $(III)$ are $TRUE$.

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:

Similar Questions

The function defined by $f(x) = \begin{cases} |x - 3|, & x \ge 1 \\ \frac{1}{4}x^2 - \frac{3}{2}x + \frac{13}{4}, & x < 1 \end{cases}$ is

The number of real roots of the equation $\log_{e} x + ex = 0$ is

If $f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x$,then $f'\left( \frac{\pi}{4} \right)$ is

If $f(x) = \begin{cases} x^{3}-3x+2, & x < 2 \\ x^{3}-6x^{2}+9x+2, & x \geq 2 \end{cases}$, then:

Let $f: R \rightarrow (0, \infty)$ be a twice differentiable function such that $f(3) = 18$, $f'(3) = 0$, and $f''(3) = 4$. Then $\lim_{x \rightarrow 1} \left( \log_{e} \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-1)^{2}}} \right)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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: