For real x, let `f(x)=x^(3)+5x+1,` then

Options

A. f is one-one but not onto
B. f is onto but not one-one
C. f is one-one and onto R
D. is niether one-one nor onto R

Answer

C

Explanation

Let x, y `in R` be such that
f(x)=f(y)
`Rightarrow x^(3)=5x+1=y^(3)+5y+1`
`Rightarrow (x^(3)-y^(3))+5(x-y)=0`
`Rightarrow (x-y)(x^(2)+xy+y^(2)+5)=0`
`Rightarrow (x-y){(x+(y)/(2))^(2)+(3y^(2))/(4)+5}=0`
`Rightarrow x=y " "[therefore(x+(y)/(2))^(2)+(3y)^(2)/(4)+5 ne0]`
`therefore f:R to R` is one-one
Let y be an arbitirary element in R (Co-domain). Then f(x)=y i.e. `x^(3)+5x+1=y` has at least one real root, say `alpha, "in" R`.
`therefore alpha^(3)+5alpha+1=y Rightarrow f(alpha)=y`
Thus, for each `y in R` such that `f(alpha)=y`
So, `f:R to R` is onto. Hence, `f:R to R` is one-one onto.