Mathematics - Application of Derivative Question with Solution | TestHub

The equation of the given curve is $\begin{array}{l}y=4x^3-2x^5\\ \frac{\mathrm{dy}}{\mathrm{dx}}=12x^2-10x^4\end{array}$ Therefore, the slope of the tangent at point $(x,y)$ is $12x^2-10x^4$. The equation of the tangent at $(x,y)$ is given by $\mathrm{Y}-y=\left(12x^2-10x^4\right)(\mathrm{X}-x)\dots \text{ (i) }$ When, the tangent passes through the origin $(0,0)$, then $\mathrm{X}=\mathrm{Y}=0$ Therefore, eq. (i) reduce to $\begin{array}{l}-y=\left(12x^2-10x^4\right)(-x)\\ \Rightarrow y=12x^3-10x^5\end{array}$ Also, we have $y=4x^3-2x^5$ $\begin{array}{l}\therefore 12x^3-10x^5\\ =4x^3-2x^5\\ \Rightarrow 8x^5-8x^3=0\\ \Rightarrow x^5-x^3=0\\ \Rightarrow x^3\left(x^2-1\right)=0\\ \Rightarrow x=0,\pm 1\end{array}$ When, $x=0$ $y=4(0{)}^3-2(0{)}^5=0$ When, $x=1$, $y=4(1{)}^3-2(1{)}^5=2$ When, $x=-1$, $y=4(-1{)}^3-2(-1{)}^5=-2$ Hence, the require points are $(0,0),(1,2)$ and $(-1,-2)$.