The mapping f: R to R defined as f(x) = cos x | vedclass

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

Question

(A) Let $x_1, x_2 \in R$. Then $f(x_1) = \cos x_1$ and $f(x_2) = \cos x_2$. For the function to be one-one,$f(x_1) = f(x_2)$ must imply $x_1 = x_2$. H

Vedclass · Accepted Answer

Neither one-one nor onto

Vedclass · Answer

(A) Let $x_1, x_2 \in R$. Then $f(x_1) = \cos x_1$ and $f(x_2) = \cos x_2$. For the function to be one-one,$f(x_1) = f(x_2)$ must imply $x_1 = x_2$. However,$\cos x_1 = \cos x_2$ implies $x_1 = 2n\pi \pm x_2$ for some integer $n$. Since $x_1$ does not necessarily equal $x_2$ (e.g.,$\cos(0) = \cos(2\pi) = 1$),the function is not one-one. For the function to be onto,the range must equal the codomain $R$. The range of $f(x) = \cos x$ is $[-1, 1]$,which is a proper subset of $R$. Since the range $
eq$ codomain,the function is not onto. Therefore,the mapping is neither one-one nor onto.

The mapping $f: R \to R$ defined as $f(x) = \cos x, x \in R$ is:

Similar Questions

The number of functions $f: \{1, 2, \ldots, 100\} \rightarrow \{0, 1\}$ that assign $1$ to exactly one of the positive integers less than or equal to $98$ is equal to $\qquad$.

$A$ function $f$ from the set of natural numbers $\mathbb{N}$ to the set of integers $\mathbb{Z}$ is defined by $f(n) = \begin{cases} \frac{n-1}{2}, & \text{if } n \text{ is odd} \\ -\frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$. The function $f$ is:

If the set $x$ contains $7$ elements and set $y$ contains $8$ elements,then the number of bijections from $x$ to $y$ is

The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2 + x$ is:

$f(x)=\frac{x}{e^x-1}+\frac{x}{2}+2 \cos ^3 \frac{x}{2}$ on $R-\{0\}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module