Let f : R -> R be defined as f(x) = 2x -1 and g : R - {1} -> R be defined as g(x) = $\frac{x - \frac{1}{2}}{x - 1} .$ Then the composition function f(g(x)) is:

Option 1 -

Neither one-one nor onto

Option 2 -

Onto but not one-one

Option 3 -

Both one-one and onto

Option 4 -

One-one but not onto

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

$f (x) = 2 x - 1 g : R - {1} \to R$

$g (x) = \frac{x - 1 / 2}{x - 1}$

$f {g (x)} = 2 (\frac{x - 1 / 2}{x - 1}) - 1 = \frac{2 x - 1 - x + 1}{x - 1} = \frac{x}{x - 1}, x \neq 1$

$y = \frac{x}{x - 1} \Rightarrow x = \frac{y}{y - 1} ∴ y \neq 1$

One – one but not onto

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Let f : R -> R be defined as f(x) = 2x -1 and g : R - {1} -> R be defined as g(x) = $\frac{x - \frac{1}{2}}{x - 1} .$ Then the composition function f(g(x)) is:

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Let f : R -> R be defined as f(x) = 2x -1 and g : R - {1} -> R be defined as g(x) = x − 1 2 x − 1 . Then the composition function f(g(x)) is:

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Let f : R -> R be defined as f(x) = 2x -1 and g : R - {1} -> R be defined as g(x) = $\frac{x - \frac{1}{2}}{x - 1} .$ Then the composition function f(g(x)) is: