If f : R to R is defined by f(x) = 2x + cos x | vedclass

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

Question

(A) Given $f(x) = 2x + \cos x$. To check for one-one,we find the derivative: $f'(x) = 2 - \sin x$. Since $-1 \le \sin x \le 1$,it follows that $1 \le

Vedclass · Accepted Answer

One-one and onto

Vedclass · Answer

(A) Given $f(x) = 2x + \cos x$. To check for one-one,we find the derivative: $f'(x) = 2 - \sin x$. Since $-1 \le \sin x \le 1$,it follows that $1 \le 2 - \sin x \le 3$. Thus,$f'(x) > 0$ for all $x \in R$. Since the derivative is strictly positive,$f(x)$ is a strictly increasing function,which implies $f$ is one-one. To check for onto,we observe the limits: $\lim_{x 	o \infty} (2x + \cos x) = \infty$ and $\lim_{x 	o -\infty} (2x + \cos x) = -\infty$. Since $f(x)$ is a continuous function and its range is $(-\infty, \infty) = R$,the function $f$ is onto. Therefore,$f$ is one-one and onto.

If $f : R \to R$ is defined by $f(x) = 2x + \cos x$,then $f$ is

Similar Questions

Let $f :(0,1) \rightarrow R$ be a function defined by $f(x)=\frac{1}{1-e^{-x}}$,and $g(x)=(f(-x)-f(x))$. Consider two statements:
$(I)$ $g$ is an increasing function in $(0,1)$
$(II)$ $g$ is one-one in $(0,1)$
Then,

For $0 \leq x \leq 1$,what type of function is $f(x) = |x| + |x - 1|$?

The number of onto functions from the set $\{1, 2, \ldots, 11\}$ to the set $\{1, 2, \ldots, 10\}$ is

Let $A = \{1, 2, 3, 4\}$ and $R : A \to A$ be the relation defined by $R = \{ (1, 1), (2, 3), (3, 4), (4, 2) \}$. The correct statement is

The number of bijective functions $f: Z \rightarrow Z$ such that $f(x+y)=f(x)+f(y)$ for all $x, y \in Z$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module