If $f:\Bbb R\to \Bbb R$ defined by $f(x)=x^2+11, x\in \Bbb R$ then which of the following arguments is not true? If $f:\Bbb R\to\Bbb R$ defined by $f(x)=x^2+11, x\in R$ then which of the following arguments is not true? State with justification.


*

*It is one to one.

*It is many to one

*It is onto.

*It is not bijective.
My Effort:
I guess the ans is $1$. But I neither know calculation nor the justification. Please help.
 A: Hint
Note that for $f:\Bbb R\to\Bbb R :x \mapsto x^2+11$, (hoover for extra hint):


*

*$f(x) = f(-x)$, so...



 the function can't be one to one because...



*

*$f(x)=x^2+11 \ge 11$, so...



 the function can't be onto because...

I can elaborate if this doesn't help; let me know via comments.

Addition after comment, referring to the hints above in the same order:


*

*if different $x$-values are mapped to the same value, a function is not one to one;

*a function is onto if all elements of the codomain are the image of some $x$-value(s).

A: *

*$f(1)=f(-1)=12$ so 2 different $x$ map to the same value, so it is not 1-to-1

*Many $x$ map to the same value (see 1.) so it is many-one

*There is no real $x$ that maps to the value say $0$ ( in fact no $x$ maps to anything $\lt 11)$, thus it is not onto. onto meaning every element of the codomain ($\Bbb R$) is the image of some element in the domain (also $\Bbb R$)

*It is not 1-1 or onto, so it is not bijective.

A: In general if $f$ is a polynomial with distinct real roots then it cannot be one-one.
So, it is many-to-one.
So, it is not bijective.
And any polynomial of even degree having real coefficients cannot be onto, since it has to be bounded on one side (positive if leading coefficient is negative, and negative if leading coefficient is positive) .
Try to apply this on your polynomials and get your result.
