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Question
Let
f : R → R be given by
\[f\left( x \right) = \left[ x^2 \right] + \left[ x + 1 \right] - 3\]
where [x] denotes the greatest integer less than or equal to x. Then, f(x) is
(d) one-one and onto
Options
many-one and onto
many-one and into
one-one and into
one-one and onto
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Solution
(b) many-one and into
f : R → R
\[f\left( x \right) = \left[ x^2 \right] + \left[ x + 1 \right] - 3\]
It is many one function because in this case for two different values of x
we would get the same value of f(x) .
\[For \]
\[x = 1 . 1, 1 . 2 \in R\]
\[f(1 . 1) = \left[ \left( 1 . 1 \right)^2 \right] + \left[ 1 . 1 + 1 \right] - 3\]
\[ = \left[ 1 . 21 \right] + \left[ 2 . 1 \right] - 3\]
\[ = 1 + 2 - 3\]
\[ = 0\]
\[f(1 . 1) = \left[ \left( 1 . 2 \right)^2 \right] + \left[ 1 . 2 + 1 \right] - 3\]
\[ = \left[ 1 . 44 \right] + \left[ 2 . 2 \right] - 3\]
\[ = 1 + 2 - 3\]
\[ = 0\]
It is into function because for the given domain we would only get the integral values of
f(x).
but R is the codomain of the given function.
That means , Codomain\[\neq\]Range Hence, the given function is into function.
Therefore, f(x) is many one and into
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