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.

1. It is one to one.

2. It is many to one

3. It is onto.

4. It is not bijective.

My Effort:

I guess the ans is $1$. But I neither know calculation nor the justification. Please help.

• your guess is wrong !! – adityaguharoy May 12 '17 at 10:47

1. $f(1)=f(-1)=12$ so 2 different $x$ map to the same value, so it is not 1-to-1
2. Many $x$ map to the same value (see 1.) so it is many-one
3. 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 $\Bbb R$)
4. It is not 1-1 or onto, so it is not bijective.
• @ÉvaristeGalois see update – PM. Mar 31 '17 at 14:01

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).
In general if $f$ is a polynomial with distinct real roots then it cannot be one-one.