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Question
Which of the following functions is NOT one-one?
Options
f : R `rightarrow` R defined by f(x) = 6x – 1.
f : R `rightarrow` R defined by f(x) = x2 + 7.
f : R `rightarrow` R defined by f(x) = x3.
f : R – {7} `rightarrow` R defined by f(x) = `(2x + 1)/(x - 7)`
Solution
f : R `rightarrow` R defined by f(x) = x2 + 7.
Explanation:
(a) We have f(x) = 6x – 1, x ∈ R.
Let f(x1) = f(x2), x1, x2 ∈ R
`\implies` 6x1 – 1 = 6x2 – 1
`\implies` 6x1 = 6x2
`\implies` x1 = x2.
∴ ‘f’ is one-one.
(b) We have f(x) = x2 + 7, x ∈ R
f(–2) = (–2)2 + 7 = 11,
f(2) = (2)2 + 7 = 11
∴ The images of distinct elements –2 and 2 of R are equal.
∴ ‘f’ cannot be one-one.
(c) f(x) = x3, x ∈ R.
Let f(x1) = f(x2), x1, x2 ∈ R
`\implies x_1^3 = x_2^3`
`\implies x_1^3 - x_2^3` = 0
`\implies (x_1 - x_2) (x_1^2 + x_1x_2 + x_2^2)` = 0
`\implies` x1 – x2 = 0
`\implies` x1 = x2,
because the other factor cannot be zero
∴ ‘f’ is one-one
(d) We have f(x) = `(2x + 1)/(x - 7), x ∈ R - {7}`.
Let f(x1) = f(x2), x1, x2 ∈ R – {7}.
`\implies (2x_1 + 1)/(x_1 - 7) = (2x_2 + 1)/(x_2 - 7)`
`\implies` – 15x1 = – 15x2
`implies` x1 = x2
∴ f is one-one.
