#### Explain solution RD Sharma class 12 chapter 11 Higher Order Derivatives exercise very short answer type question 8 maths

$\frac{d^{2} y}{d x^{2}}=\left\{\begin{array}{l} -2,01, x<0 \end{array}\right.$

Hint:

$\text { if } y=|p| \: {\text {then }} y=\left\{\begin{array}{c} p, \text { if } 0 \leq p \leq 1 \\ -p, \text { if } p<0 \end{array}\right.$

Given:

$y=\left | x-x^{2} \right |$

Explanation:

It is given that

$y=\left | x-x^{2} \right |$

This can be written as

$y=\left\{\begin{array}{c} x-x^{2} \text { if } 0 \leq x \leq 1 \\ -\left(x-x^{2}\right) \text { if } x<0 \text { or } x>1 \end{array}\right.$

Diff $y$ w.r to $x$

$\frac{d y}{d x}=\left\{\begin{array}{c} 1-2 x \text { if } 0<\mathrm{x} \leq 1 \\ -(1-2 x) \text { if } x>0 \text { or } x>1 \end{array}\right.$

$Di\! f\! f\; w.r.t\; to\; x$

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\left\{\begin{array}{c} -2 \text { if } 0 \leq x \leq 1 \\ -(-2) \text { if } x<0 \text { or } x>1 \end{array}\right. \end{aligned}

Thus,

$\frac{d^{2} y}{d x^{2}}=\left\{\begin{array}{l} -2,01, x<0 \end{array}\right.$