Question

For real \[x,left(x) = {x^3} + 5x + 1,\] then
A. $f$ is onto $R$ but not one-one.
B. $f$ is one-one and onto $R$.
C. $f$ is neither one-one nor onto $R$.
D. $f$ is one-one but not onto $R$.

Answer 1

Hint: We have given a function to check the function that it is one-one or onto or both. So firstly we check if the function is continuous or not then we check the function is differentiable. If yes then we differentiate it if differentiation is greater than zero then it will be one-one function. Next we check if it is onto or not.

Complete step by step answer:
We have given that
$f(x) = {x^3} + 5x + 1,Vx \in R$
We have to check whether the function is one-one and onto. One-one but not onto neither one-one nor onto. Onto but not one-one on R.
We have $f(x) = {x^3} + 5x + 1$
Since $f(x)$is the polynomial so it is continuous on $R$. $f(x)$ is also differentiable.
Differentiating both sides w.r.t. $'x'$
$f'(x) = 3{x^3} + 5$
$Vx \in R$ $f'(x) = 3{x^3} + 5 > 0$
Therefore $f(x)$ is strictly increasing function.
Now $f(x)$ is continuous and strictly increasing function.
So $f(x)$ is one – one function.
Now $f(x) = {x^3} + 5x + 1$
Let $y \in R$ then $f(x) = {x^3} + 5x + 1$
🡪 ${x^3} + 5x + 1 - f(x) = 0$
${x^3} + 5x + 1 - f(x) = 0$
As we know that polynomial with ad degree has at least one real root corresponding to any $y \in R$
So for some $K \in $ domain there exist a $y \in R$such that
Hence function is onto
From the above we have the result function $f(x)$is one-one and onto over $R$

Therefore option (B) is correct.

Note: One-one function: A function is said to be a one-on function if every element in the domain has a unique image in range.
Onto function: Function is said to be onto if every element is range there is an element in domain.