If the function f: R rightarrow R is defined | vedclass

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

Question

(C) Given,$f(x) = (x^{2} + 1)^{35}$ for all $x \in R$.For one-one: Check if $f(x_1) = f(x_2)$ implies $x_1 = x_2$.$f(1) = (1^{2} + 1)^{35} = 2^{35}$ a

Vedclass · Accepted Answer

neither one-one nor onto

Vedclass · Answer

(C) Given,$f(x) = (x^{2} + 1)^{35}$ for all $x \in R$.For one-one: Check if $f(x_1) = f(x_2)$ implies $x_1 = x_2$.$f(1) = (1^{2} + 1)^{35} = 2^{35}$ and $f(-1) = ((-1)^{2} + 1)^{35} = 2^{35}$.Since $f(1) = f(-1)$ but $1 
eq -1$,the function is not one-one.For onto: The range of $f(x)$ must be equal to the codomain $R$.Since $x^{2} \geq 0$,we have $x^{2} + 1 \geq 1$,which implies $(x^{2} + 1)^{35} \geq 1^{35} = 1$.Thus,the range is $[1, \infty)$,which is not equal to the codomain $R$.Therefore,the function is neither one-one nor onto.

If the function $f: R \rightarrow R$ is defined by $f(x) = (x^{2} + 1)^{35}, \forall x \in R,$ then $f$ is

Similar Questions

Show that an onto function $f: \{1, 2, 3\} \rightarrow \{1, 2, 3\}$ is always one-one.

If $f(x)=|x-1|+|x-2|+|x-3|$ for $2 < x < 3$,then $f$ is

If $f: S \rightarrow R$ where $S$ is the set of all non-singular matrices of order $2$ over $R$ and $f\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = ad - bc$,then:

$A$ function from $A = \{x : -1 \leq x \leq 1\}$ to itself which is not a bijection is

Prove that the function $f: R \rightarrow R$,defined by $f(x)=2x$,is one-one and onto.

Vedclass Test Series

Exam Paper Generator

Online Exam Module