Question

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

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

Answer 1

The correct option is A

$\left(\frac{1}{2},\frac{5}{24}\right)\mathrm{and}\left(-1,\frac{-1}{6}\right)$

Explanation for correct option:

Step-1: Simplify the given data.

Given, equation of curve is $y=\frac{2}{3}{x}^3+\frac{1}{2}{x}^2,$and tangent makes equal angle with axes.

It means tangent is $\pm 1$

$\Rightarrow$$\frac{dy}{dx}=\pm 1.......\left(i\right)$

Step-2: Differentiate with respect to $x$.

$\Rightarrow$$\frac{dy}{dx}=\frac{2}{3}\times 3x^2+\frac{1}{2}\times 2x$

$\Rightarrow$$\frac{dy}{dx}=2x^2+x.........\left(ii\right)$

Step-3: Finding value of $x$ with the help of equation $\left(i\right)$ and $\left(ii\right)$.

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

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

$\Rightarrow$$\left(2x-1\right)\left(x+1\right)$

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

Step-4: Finding value of $y$ at value of $x$.

At- $x=\frac{1}{2}$

$\Rightarrow$$y=\frac{2}{3}{\left(\frac{1}{2}\right)}^3+\frac{1}{2}{\left(\frac{1}{2}\right)}^2$

$\Rightarrow$$y=\frac{1}{12}+\frac{1}{8}$

$\Rightarrow$$y=\frac{5}{24}$

At- $x=-1$

$\Rightarrow$$y=\frac{2}{3}{\left(-1\right)}^3+\frac{1}{2}{\left(-1\right)}^2$

$\Rightarrow$$y=-\frac{2}{3}+\frac{1}{2}$

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

Therefore, required point is $\left(\frac{1}{2},\frac{5}{24}\right)\mathrm{and}\left(-1,\frac{-1}{6}\right)$

Hence, option $\left(A\right)$ is correct.