In the following [x] denotes the greatest int | vedclass

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

Question

(B) $f(x) = x|x|$. This is continuous and differentiable everywhere,including $x=0$. Since $f'(x) = 2|x| \ge 0$,it is strictly increasing. Thus,$(A) \

Vedclass · Accepted Answer

$A-(p, q, r), B-(p, s), C-(r, s), D-(p, q)$

Vedclass · Answer

(B) $f(x) = x|x|$. This is continuous and differentiable everywhere,including $x=0$. Since $f'(x) = 2|x| \ge 0$,it is strictly increasing. Thus,$(A) 	o (p, q, r)$.$(B)$ $f(x) = \sqrt{|x|}$. This is continuous everywhere. At $x=0$,it has a cusp,so it is not differentiable at $x=0$. Thus,$(B) 	o (p, s)$.$(C)$ $f(x) = x + [x]$. This function is discontinuous at every integer point. In $(-1, 1)$,it is discontinuous at $x=0$. It is strictly increasing on $(-1, 0)$ and $[0, 1)$. It is not differentiable at $x=0$. Thus,$(C) 	o (r, s)$.$(D)$ $f(x) = |x - 1| + |x + 1|$. In $(-1, 1)$,$f(x) = (1 - x) + (x + 1) = 2$. This is a constant function,which is continuous and differentiable everywhere. Thus,$(D) 	o (p, q)$.

Column $I$	Column $II$
$(A)$ $f(x) = x\|x\|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{\|x\|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = \|x - 1\| + \|x + 1\|$	$(s)$ not differentiable at least at one point in $(-1, 1)$

Similar Questions

Let $f: R \rightarrow R$ be defined as $f(x)=\begin{cases} \frac{a-b \cos 2 x}{x^2} & ; x<0 \\ x^2+c x+2 & ; 0 \leq x \leq 1 \\ 2 x+1 & ; x>1 \end{cases}$. If $f$ is continuous everywhere in $R$ and $m$ is the number of points where $f$ is $NOT$ differentiable,then $m+a+b+c$ equals:

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:

$\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\frac{{{e^{{e^{{x^2}}}}} - e}}{x}} \right)$ is

Let $y = f(x) = \begin{cases} e^{-\frac{1}{x^2}}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$. Then which of the following can best represent the graph of $y = f(x)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module