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RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRELATIONS AND FUNCTIONS5 Marks
Question
Show that the function $\text{f} : \text{R}\rightarrow\text{R}$ defined by $\text{f(x})=\frac{\text{x}}{\text{x}^2+1},\forall\text{ x}\in\text{R}$ is neither one-one nor onto. Also, if $\text{g} : \text{R}\rightarrow\text{R}$ is defined as g(x) = 2x – 1, find fog(x).
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Answer

$\text{f(x})=\frac{\text{x}}{\text{x}^2+1}$
For one-one f(x) = f(y)
$\frac{\text{x}}{\text{x}^2+1}=\frac{\text{y}}{\text{y}^2+1}$
$\text{xy}^2+\text{x}=\text{yx}^2+\text{y}$
$\text{xy}(\text{y}-\text{x})=\text{y}-\text{x}$
$\text{xy}=1$
$\text{x}=\frac{1}{\text{y}}$
$\text{x}\neq\text{y}$
So, not one-one
For onto f(x) = y
$\frac{\text{x}}{\text{x}^2+1}=\text{y}$
$\text{x}=\text{yx}^2+\text{y}$
$\text{x}^2\text{y}+\text{y}-\text{x}=0$
x cannot be express in y so not onto
As g(x) = 2x – 1
$\text{fog(x})=\text{f[g(x})]=\text{f}(2\text{x}-1)=\frac{2\text{x}-1}{(2\text{x}-1)^2+1}$
$=\frac{2\text{x}-1}{4\text{x}^2-4\text{x}+2}$

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