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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Find the values of $x$ for which $\text{y}=[\text{x}(\text{x}-2)]^2$ is an increasing function.
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Answer

Given: $\text{f}\text{(x)} = \text{y} =\text{[x}(\text{x}-2)]^{2}\ \Rightarrow \ \frac{\text{dy}}{\text{dx}}$ $=2\text{x}(\text{x}-2)\frac{\text{d}}{\text{dx}}\bigg[ \text{x}(\text{x}-2)\bigg]$
$\Rightarrow \ \frac{\text{dy}}{\text{dx}}=2\text{x}(\text{x}-2)$ $\bigg[x\frac{\text{d}}{\text{dx}}(\text{x}-2)+(\text{x}-2)\frac{\text{d}}{\text{dx}}\text{x}\bigg]\ [\text{Applying Product Rule}]$
$ \Rightarrow \ \frac{\text{dy}}{\text{dx}}=2\text{x}(\text{x}-2)[\text{x}+\text{x}-2]$ $=2\text{x}(\text{x}-2) (2\text{x}-2) =4\text{x}(\text{x}-2)(\text{x}-1)\ \dots\dots\text{(i)}$
Therefore, we have $(-\infty,\ 0),(0,\ 1),(1,\ 2),(2,\ \infty)$
$\text{For } (-\infty,\ 0)$ taking $x = -1 ($say$), \frac{\text{dy}}{\text{dx}}=(-)(-)(-)=(-)\leq 0 $
$\therefore f(x)$ is decreasing.
$\text{For }(0,\ 1)$ taking $\text{x} = \frac{1}{2} $ (say), $\frac{\text{dy}}{\text{dx}}=(+)(-)(-)=(+)\geq0$
$\therefore f(x)$ is increasing.
$\text{For }(1,2)$ taking $x = 1.5 ($say$) , \frac{\text{dy}}{\text{dx}}=(+)(-)(+)=(-)\leq0$
$\therefore f(x)$ is increasing.
$\text{For } (2,\ \infty)$ taking $x = 3 ($say$) \frac{\text{dy}}{\text{dx}}=(+)(+)(+)=(+)\geq0$
$\therefore f(x)$ is increasing.

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