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.
$$\left( {\frac{1}{2},\,\frac{5}{{24}}} \right){\text{ and }}\left( { - 1,\, - \frac{1}{6}} \right)$$
B.
$$\left( {\frac{1}{2},\,\frac{4}{9}} \right){\text{ and }}\left( { - 1,\,0} \right)$$
C.
$$\left( {\frac{1}{3},\,\frac{1}{7}} \right){\text{ and }}\left( { - 3,\,\frac{1}{2}} \right)$$
D.
$$\left( {\frac{1}{3},\,\frac{4}{{47}}} \right){\text{ and }}\left( { - 1,\, - \frac{1}{3}} \right)$$
Answer :
$$\left( {\frac{1}{2},\,\frac{5}{{24}}} \right){\text{ and }}\left( { - 1,\, - \frac{1}{6}} \right)$$
Solution :
$$\eqalign{
& y = \frac{2}{3}{x^3} + \frac{1}{2}{x^2} \cr
& \therefore \,\frac{{dy}}{{dx}} = \frac{2}{3}3{x^2} + \frac{1}{2}2x \cr
& = 2{x^2} + x \cr} $$
Since the tangent makes equal angles with the axes.
$$\eqalign{
& \therefore \,\frac{{dy}}{{dx}} = \pm 1 \cr
& {\text{or 2}}{x^2} + x = \pm 1 \cr
& {\text{or 2}}{x^2} + x - 1 = 0\,\,\,\,\,\left( {{\text{2}}{x^2} + x + 1 = 0{\text{ has no real roots}}} \right) \cr
& {\text{or }}\left( {2x - 1} \right)\left( {x + 1} \right) = 0 \cr
& {\text{i}}{\text{.e}}{\text{., }}x = \frac{1}{2}{\text{ or }}x = - 1 \cr} $$