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Question
Let A = [–1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective:
h(x) = x|x|
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Solution
Given, A = [–1, 1]
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.
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