#### Explain solution rd sharma class 12 chapter 16 Increasing and decreasing function exercise multiple choice question, question 32

1<x<3, Option (a)

Hint: Take a derivative of given equation

Given: $y=x(x-3)^{2}$ decreases for the values

Solution: We have, $y=x(x-3)^{2}$

\begin{aligned} &y=x\left(x^{2}-6 x+9\right) \\ &y=x^{3}-6 x^{2}+9 x \end{aligned}

\begin{aligned} \frac{\mathrm{dy}}{\mathrm{dx}} &=3 \mathrm{x}^{2}-12 \mathrm{x}+9 \\ \frac{\mathrm{dy}}{\mathrm{dx}} &=3\left(\mathrm{x}^{2}-4 \mathrm{x}+3\right) \\ &=3(\mathrm{x}-3)(\mathrm{x}-1) \end{aligned}

Y = f(x) decreases when \begin{aligned} \frac{\mathrm{dy}}{\mathrm{dx}}<0 \end{aligned}

The sign scheme of \begin{aligned} \frac{\mathrm{dy}}{\mathrm{dx}} \end{aligned} is shown,

∴ f’(x) =

From the sign scheme \begin{aligned} \frac{\mathrm{dy}}{\mathrm{dx}}<0 \end{aligned}

For $x\epsilon (1,3)$

$y=x(x-3)^{2}$  decreases when $x\epsilon 1

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