Show that the function f : R → R : f(x) = 1 + x2 is many-one into.

1.	Show that the function f : R → R : f(x) = 1 + x2 is many-one into.
Answer» To prove: function is many-one into Given: f : R → R : f(x) = 1 + x² We have, f(x) = 1 + x² For, f(x₁) = f(x₂) ⇒ 1 + x₁² = 1 + x₂² ⇒ x₁² = x₂² ⇒ x₁² - x₂²= 0 ⇒ (x₁ – x₂) (x₁ + x₂) = 0 ⇒ x₁ = x₂ or, x₁ = –x₂ Clearly x₁has more than one image ∴ f(x) is many-one f(x) = 1 + x² Let f(x) = y such that \(y\in R\) ⇒ y = 1 + x² ⇒ x² = y – 1 \(\Rightarrow x=\sqrt{y-1}\) If y = 3, as \(y \in R\) Then x will be undefined as we can’t place the negative value under the square root Hence f(x) is into Hence Proved

Answer»

To prove: function is many-one into

Given: f : R → R : f(x) = 1 + x²

We have,

f(x) = 1 + x²

For, f(x₁) = f(x₂)

⇒ 1 + x₁² = 1 + x₂²

⇒ x₁² = x₂²

⇒ x₁² - x₂²= 0

⇒ (x₁ – x₂) (x₁ + x₂) = 0

⇒ x₁ = x₂ or, x₁ = –x₂

Clearly x₁has more than one image

∴ f(x) is many-one

f(x) = 1 + x²

Let f(x) = y such that \(y\in R\)

⇒ y = 1 + x²

⇒ x² = y – 1

\(\Rightarrow x=\sqrt{y-1}\)

If y = 3, as \(y \in R\)

Then x will be undefined as we can’t place the negative value under the square root

Hence f(x) is into

Hence Proved

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