If f(x) = xsqrt{1 - [x]^2},then (where [.] de | vedclass

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

Question

(A) $1$. For $x \in (0, 1)$,$[x] = 0 \Rightarrow f(x) = x\sqrt{1 - 0} = x$. Since $f'(x) = 1 > 0$,$f(x)$ is increasing in $(0, 1)$.$2$. At $x = 1$,$f(

Vedclass · Accepted Answer

$f(x)$ is increasing in $x \in (0, 1)$

Vedclass · Answer

(A) $1$. For $x \in (0, 1)$,$[x] = 0 \Rightarrow f(x) = x\sqrt{1 - 0} = x$. Since $f'(x) = 1 > 0$,$f(x)$ is increasing in $(0, 1)$.$2$. At $x = 1$,$f(1) = 1\sqrt{1 - 1^2} = 0$. For $x \in (0, 1)$,$f(x) = x$. As $x 	o 1^-$,$f(x) 	o 1$. Since $f(1) = 0$ and $f(x) > 0$ for $x$ near $1$ from the left,$x = 1$ is not a point of local maxima.$3$. The domain is determined by $1 - [x]^2 \ge 0 \Rightarrow [x]^2 \le 1 \Rightarrow -1 \le [x] \le 1$. This implies $x \in [-1, 2)$.$4$. The function is defined as:$f(x) = \begin{cases} 0, & -1 \le x Since $f(x) \ge 0$ for all $x$ in the domain,it is not a negative function.$5$. Since $f(x)$ is discontinuous at $x = 1$,Rolle's theorem is not applicable on $[0, 1]$.

If $f(x) = x\sqrt{1 - [x]^2}$,then (where $[.]$ denotes the greatest integer function):

Similar Questions

Let the function $f(x)$ be defined as: $f(x) = \begin{cases} [\tan(\frac{\pi}{4} + x)]^{\frac{1}{x}}, & x \neq 0 \\ k, & x = 0 \end{cases}$. If $f(x)$ is continuous at $x = 0$, then the value of $k$ is...

If function $f$ is continuous at point $x = \pi$ and $f(x) = \begin{cases} kx+1; & x \leq \pi \\ \cos x; & x > \pi \end{cases}$ then the value of $k$ is $\dots \dots \dots$

Discuss the continuity of the function $f$ given by $f(x) = |x|$ at $x = 0$.

If $f(x) = \begin{cases} \frac{x}{e^{1/x} + 1}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module