Question ID: #898
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
- (1) All (I), (II) and (III) are TRUE.
- (2) Only (I) is TRUE.
- (3) Only (II) and (III) are TRUE.
- (4) Only (I) and (III) are TRUE.
Solution:
$$f(x) = |\ln x| – |x-1|$$
Case 1: $x \ge 1$
$\ln x \ge 0$ and $x-1 \ge 0$.
$f(x) = \ln x – (x-1) = \ln x – x + 1$.
$f'(x) = \frac{1}{x} – 1$.
For $x > 1$, $\frac{1}{x} < 1 \Rightarrow f'(x) < 0$. So, f is decreasing in $(1, \infty)$. (Statement III is True).
Case 2: $0 < x < 1$ $\ln x < 0$ and $x-1 < 0$. $f(x) = -\ln x - (-(x-1)) = -\ln x + x - 1$. $f'(x) = -\frac{1}{x} + 1$. For $0 < x < 1$, $\frac{1}{x} > 1 \Rightarrow -\frac{1}{x} < -1 \Rightarrow f'(x) < 0$. So, f is decreasing in $(0, 1)$. (Statement II is False).
Check Differentiability at $x=1$:
$f(1) = 0$.
RHD ($x \to 1^+$): $\lim_{h \to 0} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0} \frac{(\ln(1+h)-(1+h)+1)-0}{h} = \lim_{h \to 0} (\frac{\ln(1+h)}{h} – 1) = 1 – 1 = 0$.
LHD ($x \to 1^-$): $\lim_{h \to 0} \frac{f(1-h)-f(1)}{-h} = \lim_{h \to 0} \frac{(-\ln(1-h)+(1-h)-1)-0}{-h} = \lim_{h \to 0} (\frac{\ln(1-h)}{h} + 1) = -1 + 1 = 0$.
LHD = RHD = 0. So f is differentiable at $x=1$.
Since f is differentiable in $(0, 1)$ and $(1, \infty)$ and at 1, Statement I is True.
Statements I and III are True.
Ans. (4)
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