Let A = {x in R : -1 leq x leq 1} and f: A ri | vedclass

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

Question

(D) The function is defined as $f(x) = x|x|$.We can write this as a piecewise function:$f(x) = \begin{cases} -x^2, & -1 \leq x < 0 \ x^2, & 0 \leq x

Vedclass · Accepted Answer

bijective

Vedclass · Answer

(D) The function is defined as $f(x) = x|x|$.We can write this as a piecewise function:$f(x) = \begin{cases} -x^2, & -1 \leq x For injectivity: Since $f(x)$ is a strictly increasing function on the interval $[-1, 1]$ (as its derivative $f'(x) = 2|x| \geq 0$),it is injective.For surjectivity: The range of $f(x)$ for $x \in [-1, 0)$ is $(-1, 0]$ and for $x \in [0, 1]$ is $[0, 1]$. Combining these,the range is $[-1, 1]$,which is equal to the codomain $A$. Thus,the function is surjective.Since the function is both injective and surjective,it is bijective.

Let $A = \{x \in R : -1 \leq x \leq 1\}$ and $f: A \rightarrow A$ be a mapping defined by $f(x) = x|x|$. Then $f$ is

Similar Questions

Let $f: R \rightarrow R$ be defined by $f(x) = \begin{cases} x + 2, & x \leq -1 \\ x^2, & -1 < x < 1 \\ 2 - x, & x \geq 1 \end{cases}$. Then the value of $f(-1.75) + f(0.5) + f(1.5)$ is

Let $A = R - \{3\}$ and $B = R - \{1\}$. Consider the function $f: A \rightarrow B$ defined by $f(x) = \left(\frac{x-2}{x-3}\right)$. Is $f$ one-one and onto? Justify your answer.

If $f: R \rightarrow R$,such that $f(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$,then $f$ is

For the mapping $f: R-\{1\} \rightarrow R-\{2\}$,given by $f(x)=\frac{2x}{x-1}$,which of the following is correct?

Show that the function $f: N \rightarrow N$,given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2$,is onto but not one-one.

Vedclass Test Series

Exam Paper Generator

Online Exam Module