Question
If $\text{y}=\mid\text{x}-\text{x}^2\mid,$ then find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$

Answer

We have
$\text{y}=\mid\text{x}-\text{x}^2\mid$
$\Rightarrow\text{y}=\mid\text{x}(1-\text{x})\mid$
$\Rightarrow\text{y}=\begin{cases}\text{x} -\text{x}^2 &\text{if}&0\leq\times\leq1\\\text{x}^2-\text{x} & \text{if}&\text{x}<0\ \text{or}\ \text{x}>1\end{cases}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\begin{cases}1-2\text{x}&\text{if}&0\leq\times\leq1\\2\text{x}-1&\text{if}&\text{x}\times0\ \text{or}\ \text{x}>1 \end{cases}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\begin{cases}-2&\text{if}&0\leq\times\leq1\\2&\text{if}&\text{x}<\ \text{or}\ \text{x}\geq\end{cases}$

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