Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIncreasing and Decreasing Functions5 Marks
Question
Show that the function $x^2 - x + 1$ is neither increasing nor decreasing on $(0, 1).$
✓
Answer
$f(x) = x^2 - x + 1 $
$f'(x) = 2x - 1$
For, $f(x)$ to be increasing, we must have
$f'(x) > 0$
$\Rightarrow 2x - 1 > 0$
$\Rightarrow 2x > 1$
$\Rightarrow\text{x}>\frac{1}{2}$
$\Rightarrow\text{x}\in\Big(\frac{1}{2},1\Big)$
So, $f(x)$ is increasing on $\text{x}\in\Big(\frac{1}{2},1\Big).$
For, $f(x)$ to be decreasing, we must have
$f'(x) < 0$
$\Rightarrow 2x - 1 < 0$
$\Rightarrow 2x < 1$
$\Rightarrow\text{x}<\frac{1}{2}$
$\Rightarrow\text{x}\in\Big(0,\frac{1}{2}\Big)$
So, $f(x)$ is increasing on $\Big(0,\frac{1}{2}\Big)$
Since, $f(x)$ is increasing on $\Big(\frac{1}{2},1\Big)$ and decreasing on $\Big(0,\frac{1}{2}\Big),$
$\text{f}(\text{x})$ is neither increasing nor decreasing on $(0, 1).$
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