Question
Prove that the function $f : R → R$ defined by $f (x) = 2x + 5$ is one-one.
✓
Answer
Note that a function $f$ is one-one if
$f(x_1 ) = f (x_2 )\Rightarrow x_1 = x_2 ($definition of one-one function$)$
Now, given that $f (x_1 ) = f (x_2 ), i.e., 2x_1 + 5 = 2x_2 + 5$
$\Rightarrow 2x_1 + 5 – 5 = 2x_2 + 5 – 5 ($adding the same quantity on both sides$)$
$2x_1 + 0 = 2x_2 + 0$
$2x_1 = 2x_2( $using additive identity of real number$)$
$\frac{2}{2} x_{1}=\frac{2}{2} x_{2}$ dividing by the same non zero quantity
$x_1 = x_2$
Hence, the given function is one one
$f(x_1 ) = f (x_2 )\Rightarrow x_1 = x_2 ($definition of one-one function$)$
Now, given that $f (x_1 ) = f (x_2 ), i.e., 2x_1 + 5 = 2x_2 + 5$
$\Rightarrow 2x_1 + 5 – 5 = 2x_2 + 5 – 5 ($adding the same quantity on both sides$)$
$2x_1 + 0 = 2x_2 + 0$
$2x_1 = 2x_2( $using additive identity of real number$)$
$\frac{2}{2} x_{1}=\frac{2}{2} x_{2}$ dividing by the same non zero quantity
$x_1 = x_2$
Hence, the given function is one one
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