Question

7. In each of the following cases state whether the function is bijective or not. Justify
your answer.
(1) f:R - R defined by f(x) = 2x +1 (ii) f:R - R defined by f(x)=3-4x^2​

Answer 1

Answer:

$f(x) = 2x + 1$

is bijective from f: R --> R whereas

$f(x) = 3 - 4 {x}^{2}$

is not bijective from f: R --> R .

Step-by-step explanation:

To prove that a function f(x) is one-one ( injective ) we have to show that

$f(a) = f(b)⇒a = b$

Let

$f(a) = f(b)⇒2a + 1 = 2b + 1⇒2a = 2b⇒a = b$

So, f is one-one. Now to prove that f is onto, we have to show for all y=f(x) there exists a value of x in the domain. So,

$y = 2x + 1⇒x = \frac{y - 1}{2}$

Note that domain and codomain both are R. You can see that for all real values of y, x is always real due to closure property of real numbers. Thus, we have proved that f(x) = 2x + 1 is one-one onto or bijection. In the same way, you can prove that the second function is neither one-one nor onto so it is not a bijection. It is many-one and into function.