Which one of the following is a bijective function on the set of real

1.	Which one of the following is a bijective function on the set of real numbers?(a) 2x – 5 (b) \| x \| (c) x2 + 1 (d) x4 – x2 + 1
Answer» Answer : (a) = 2x - 5 Let f (x) = 2x – 5 f : R → R Then, for all x, y ∈ R, f (x) = f (y) ⇒ 2x – 5 = 2y – 5 ⇒ x = y ⇒ f is one-one Also, let z = f (x) = 2x – 5 ⇒ \(x= \frac{z+5}{2}\) ∴ For all z ∈ R, there exists an x ∈ R such that z = f (x) = 2x + 5 ⇒ f is onto. ∴ f is one-one onto ⇒ f is bijective. Now, check for other functions: Let g(x) = \| x \|, g : R → R Then g(– 1) and g (1) = 1 ⇒ Both 1, and – 1 have the same image in R ⇒ g is not one-one Similarly, for the functions x² + 1 and x² – x² + 1; – 1 and 1 have the same image in R, so both the functions are not one-one, hence bijective.

Which one of the following is a bijective function on the set of real numbers?(a) 2x – 5 (b) | x | (c) x2 + 1 (d) x4 – x2 + 1

Answer»

Answer : (a) = 2x - 5

Let f (x) = 2x – 5 f : R → R

Then, for all x, y ∈ R, f (x) = f (y)

⇒ 2x – 5 = 2y – 5 ⇒ x = y

⇒ f is one-one

Also, let z = f (x) = 2x – 5

⇒ \(x= \frac{z+5}{2}\)

∴ For all z ∈ R, there exists an x ∈ R such that z = f (x) = 2x + 5

⇒ f is onto.

∴ f is one-one onto

⇒ f is bijective.

Now, check for other functions: Let g(x) = | x |, g : R → R

Then g(– 1) and g (1) = 1

⇒ Both 1, and – 1 have the same image in R

⇒ g is not one-one

Similarly, for the functions x² + 1 and x² – x² + 1;

– 1 and 1 have the same image in R,

so both the functions are not one-one,

hence bijective.