If f : R → R be a function defined as f (x) = x4, then(a) f is one-o

1.	If f : R → R be a function defined as f (x) = x4, then(a) f is one-one onto (b) f is many-one onto (c) f is one-one but not onto (d) f is neither one-one nor onto
Answer» Answer : (d) = f is neither one -one nor onto For all x, y ∈ R, f (x) = f (y) ⇒ x⁴ = y⁴ ⇒ x⁴ – y⁴ = 0 ⇒ (x² – y²) (x²+ y²) = 0 ⇒ (x – y) (x + y) (x² + y²) = 0 ∴ (x – y) = 0 or (x + y) = 0 or (x² + y²) = 0 ⇒ x = y or x = – y or x = ± y ∴ x = – y ⇒ f is a many-one function Also let z = f (x) = x⁴ ⇒ x = (z)^1/4 Now z = – 1 ∈ R has no pre-image in R as (–1)^1/4 ∉ R. ∴ f : R → R is not an onto function.

If f : R → R be a function defined as f (x) = x4, then(a) f is one-one onto (b) f is many-one onto (c) f is one-one but not onto (d) f is neither one-one nor onto

Answer»

Answer : (d) = f is neither one -one nor onto

For all x, y ∈ R,

f (x) = f (y)

⇒ x⁴ = y⁴ ⇒ x⁴ – y⁴ = 0

⇒ (x² – y²) (x²+ y²) = 0 ⇒ (x – y) (x + y) (x² + y²) = 0

∴ (x – y) = 0 or (x + y) = 0 or (x² + y²) = 0

⇒ x = y or x = – y or x = ± y

∴ x = – y

⇒ f is a many-one function

Also let z = f (x) = x⁴ ⇒ x = (z)^1/4

Now z = – 1 ∈ R has no pre-image in R as (–1)^1/4 ∉ R.

∴ f : R → R is not an onto function.