Let A = [–1, 1]. Then, discuss whether the following functions defin

1.	Let A = [–1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective:(i) f(x) = x/2 (ii) g(x) = \|x\|(iii) h(x) = x\|x\| (iv) k(x) = x2
Answer» Given, A = [–1, 1] (i) f: [-1, 1] → [-1, 1], f (x) = x/2 Let f (x₁) = f(x₂) x₁/ 2 = x₂ So, f (x) is one-one. Also x ∈ [-1, 1] x/2 = f (x) = [-1/2, 1/2] Hence, the range is a subset of co-domain ‘A’ So, f (x) is not onto. Therefore, f (x) is not bijective. (ii) g (x) = \|x\| Let g (x₁) = g (x₂) \|x₁\| = \|x₂\| x₁ = ± x₂ So, g (x) is not one-one Also g (x) = \|x\| ≥ 0, for all real x Hence, the range is [0, 1], which is subset of co-domain ‘A’ So, f (x) is not onto. Therefore, f (x) is not bijective. (iii) h (x) = x\|x\| Let h (x₁) = h (x₂) x₁\|x₁\| = x₂\|x₂\| If x₁, x₂ > 0 x₁² = x₂² x₁² – x₂² = 0 (x₁ – x₂)(x₁ + x₂) = 0 x₁ = x₂ (as x₁ + x₂ ≠ 0) Similarly for x₁, x₂ < 0, we have x₁ = x₂ It’s clearly seen that for x₁ and x₂ of opposite sign, x₁ ≠ x₂. Hence, f (x) is one-one. For x ∈ [0, 1], f (x) = x² ∈ [0, 1] For x < 0, f (x) = -x² ∈ [-1, 0) Hence, the range is [-1, 1]. So, h (x) is onto. Therefore, h (x) is bijective. (iv) k (x) = x² Let k (x₁) = k (x₂) x₁² = x₂² x₁ = ± x₂ Therefore, k (x) is not one-one.

Let A = [–1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective:(i) f(x) = x/2 (ii) g(x) = |x|(iii) h(x) = x|x| (iv) k(x) = x2

Answer»

Given, A = [–1, 1]

(i) f: [-1, 1] → [-1, 1], f (x) = x/2

Let f (x₁) = f(x₂)

x₁/ 2 = x₂

So, f (x) is one-one.

Also x ∈ [-1, 1]

x/2 = f (x) = [-1/2, 1/2]

Hence, the range is a subset of co-domain ‘A’

So, f (x) is not onto.

Therefore, f (x) is not bijective.

(ii) g (x) = |x|

Let g (x₁) = g (x₂)

|x₁| = |x₂|

x₁ = ± x₂

So, g (x) is not one-one

Also g (x) = |x| ≥ 0, for all real x

Hence, the range is [0, 1], which is subset of co-domain ‘A’

So, f (x) is not onto.

Therefore, f (x) is not bijective.

(iii) h (x) = x|x|

Let h (x₁) = h (x₂)

x₁|x₁| = x₂|x₂|

If x₁, x₂ > 0

x₁² = x₂²

x₁² – x₂² = 0

(x₁ – x₂)(x₁ + x₂) = 0

x₁ = x₂ (as x₁ + x₂ ≠ 0)

Similarly for x₁, x₂ < 0, we have x₁ = x₂

It’s clearly seen that for x₁ and x₂ of opposite sign, x₁ ≠ x₂.

Hence, f (x) is one-one.

For x ∈ [0, 1], f (x) = x² ∈ [0, 1]

For x < 0, f (x) = -x² ∈ [-1, 0)

Hence, the range is [-1, 1].

So, h (x) is onto.

Therefore, h (x) is bijective.

(iv) k (x) = x²

Let k (x₁) = k (x₂)

x₁² = x₂²

x₁ = ± x₂

Therefore, k (x) is not one-one.