Question

At what points of curve $y=\frac{2}{3}{x}^3+\frac{1}{2}{x}^2$, the tangent makes equal angle with the axis

• A. $\left(\frac{1}{2},\frac{5}{24}\right)$ and $(-1,-\frac{1}{6})$
• B. $\left(\frac{1}{2},\frac{4}{9}\right)$ and $(-1,0)$
• C. $\left(\frac{1}{3},\frac{1}{7}\right)$and $(-3,\frac{1}{2})$
• D. $\left(\frac{1}{3},\frac{4}{47}\right)$ and $(-1,-\frac{1}{3})$

Answer 1

The correct option is A $\left(\frac{1}{2},\frac{5}{24}\right)$ and $(-1,-\frac{1}{6})$

Given equation of curve is

$y=\frac{2}{3}{x}^3+\frac{1}{2}{x}^2$

$\frac{dy}{dx}=2x^2+x$

Since ,the tangent makes equal angle with x-axis,

$\frac{dy}{dx}=1$

$\Rightarrow 2x^2+x-1=0$

$\Rightarrow x=-1,\frac{1}{2}$

At $x=-1\Rightarrow y=-\frac{1}{6}$

At $x=\frac{1}{2}\Rightarrow y=\frac{5}{24}$