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
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.
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