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

Solution:

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.