Question

Find the values of x for which $y=\left[x\right(x-2){]}^2$ is an increasing function.

OR

Find the equation of tangent and normal to the curve $\frac{x^2}{a}-\frac{y^2}{b^2}=1$ at the point $(\sqrt{2}a,b)$.

Answer 1

We have $y=\left[x\right(x-2){]}^2\Rightarrow y=[x^2-2x{]}^2\therefore \frac{dy}{dx}=2[x^2-2x][2x-2]$

$\Rightarrow \frac{dy}{dx}=4x[x-2][x-1]\text{For}\frac{dy}{dx}=0\Rightarrow 4x[x-2][x-1]=0$

$\therefore x=0,1,2$

$\begin{array}{ccc}\text{Interval}& \text{sign of}\frac{dy}{dx}& \text{y is}\\ (-\infty ,0)& \text{Negative}& \text{Decreasing}\\ (0,1)& \text{Positive}& \text{Increasing}\\ (1,2)& \text{Negative}& \text{Decreasing}\\ (2,\infty )& \text{Positive}& \text{Increasing}\end{array}$

Since $\frac{dy}{dx}>0$ in $(0,1)\cup (2,\infty )$ so, y is increasing in $(0,1)\cup (2,\infty )$.

OR

Given $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\Rightarrow \frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$

$\therefore$ Slope of tangent at $(\sqrt{2}a,b)=\frac{b^2\left(\sqrt{2}a\right)}{a^{2}b}=\frac{\sqrt{2}b}{a}$ and, Slope of normal at $(\sqrt{2}a,b)=-\frac{a}{\sqrt{2}b}$

So, eq. of tangent is :$y-b=\frac{\sqrt{2}b}{a}(x-\sqrt{2}a)\Rightarrow \sqrt{2}bx-ay=ab$

and, eq. of normal is : $y-b\frac{a}{\sqrt{2}b}(x-\sqrt{2}a)\Rightarrow ax+\sqrt{2}by=\sqrt{2}(a^2+b^2)$