Let f: R rightarrow R be defined by f(x)=5x^4 | vedclass

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

Question

(D) Given the function $f: R \rightarrow R$ defined by $f(x) = 5x^4 + 2$. To check for one-one: Consider $f(x_1) = f(x_2)$. $5x_1^4 + 2 = 5x_2^4 + 2$

Vedclass · Accepted Answer

$f$ is neither one-one nor onto

Vedclass · Answer

(D) Given the function $f: R ightarrow R$ defined by $f(x) = 5x^4 + 2$. To check for one-one: Consider $f(x_1) = f(x_2)$. $5x_1^4 + 2 = 5x_2^4 + 2$ $x_1^4 = x_2^4$ $x_1 = \pm x_2$. Since $f(1) = 5(1)^4 + 2 = 7$ and $f(-1) = 5(-1)^4 + 2 = 7$,we have $f(1) = f(-1)$ but $1 
eq -1$. Therefore,$f$ is not one-one. To check for onto: Since $x^4 \geq 0$ for all $x \in R$,it follows that $5x^4 \geq 0$. Thus,$f(x) = 5x^4 + 2 \geq 2$. The range of $f$ is $[2, \infty)$,which is not equal to the codomain $R$. Therefore,$f$ is not onto. Hence,$f$ is neither one-one nor onto.

Let $f: R \rightarrow R$ be defined by $f(x)=5x^4+2$. Then

Similar Questions

$f(x) = x + \sqrt{x^2}$ is a function from $R \to R$,then $f(x)$ is

If a set $A$ has $m$ elements and the set $B$ has $n$ elements,then the number of injections from $A$ to $B$ is

Let $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right) & \text{when } x \neq 0 \\ 1 & \text{when } x = 0 \end{cases}$ and $A = \{x \in \mathbb{R} : f(x) = 1\}$. Then,$A$ has

Show that the function $f : R \rightarrow \{ x \in R : -1 < x < 1 \}$ defined by $f(x) = \frac{x}{1+|x|}$ for all $x \in R$ is a one-one and onto function.

$f(x)=ax^2+bx+c$ is an even function and $g(x)=px^3+qx^2+rx$ is an odd function. If $h(x)=f(x)+g(x)$ and $h(-2)=0$, then $8p+4q+2r=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module