Question

For real x, let \[f(x)={{x}^{3}}+5x+1\], 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 will use the definition of one-one function and onto function to solve this question. A function f is one-to-one if every element of the range of f corresponds to exactly one element of the domain of f. If for every element of B there is at least one or more than one element matching with A, then the function is said to be onto.

Complete step-by-step answer:

Before proceeding with the question we should understand the concept of one-one function and onto function.

One to one function basically denotes the mapping of two sets. A function f is one-to-one if every element of the range of f corresponds to exactly one element of the domain of f. One-to-one is also written as 1-1.

If for every element of B there is at least one or more than one element matching with A, then the function is said to be onto
.
The given function \[f(x)={{x}^{3}}+5x+1\] maps R to R.

Now let \[y\in R\] then we get,

\[\begin{align}

  & f(x)=y={{x}^{3}}+5x+1 \\

 & \Rightarrow {{x}^{3}}+5x+1-y=0.....(1) \\

\end{align}\]

So we can see that the equation (1) is a cubic equation. And we know that any polynomial of odd degree always has at least one real root corresponding to any y belonging to the co-domain, then some x belongs to a domain such that \[f(x)=y\]. Hence, f is onto.

Also we can see that f is continuous on R because it is a polynomial function.

Now differentiating \[f(x)={{x}^{3}}+5x+1\] we get,

\[\Rightarrow 3{{x}^{2}}+5......(2)\]

From equation (2) we can see that the polynomial is strictly increasing as \[{{x}^{2}}\] is always positive. Hence f is one-one also.

Hence \[f(x)={{x}^{3}}+5x+1\] is both onto and one-one. Thus the correct answer is option (b).

Note: We have to be very clear with the definitions of one-one function and onto function. Also we have to remember the properties of cubic polynomials and square polynomials which is the key here.