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 $f(x)$ be a differentiable function, $f^{\prime}(x) > f(x)$ and $f(0) = 0$. Then

If $f(x) = \begin{cases} ax+b, & \text{if } x \leq 1 \\ ax^2+c, & \text{if } 1 < x \leq 2 \\ \frac{dx^2+1}{x}, & \text{if } x > 2 \end{cases}$ is differentiable on $\mathbb{R}$, then $ad-bc = $

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$ equals

Let $g(x)$ be a linear function and $f(x) = \begin{cases} g(x) & , x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & , x > 0 \end{cases}$ is continuous at $x = 0$. If $f^{\prime}(1) = f(-1)$,then the value of $g(3)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module