At what point of curve `y=2/3x^3+1/2x^2,` the tangent makes equal angle with the axis? (a) `(1/2,5/(24))a n d(-1,-1/6)` (b) `(1/2,4/9)a n d(-1,0)` (c) `(1/3,1/7)a n d(-3,1/2)` (d) `(1/3,4/(47))a n d(-1,-1/3)`

Options

A. `((1)/(2),(4)/(24))and (-1,-(1)/(6))`
B. `((1)/(2),(4)/(9))and(-1,0)`
C. `((1)/(3),(1)/(7))and(-3,(1)/(2))`
D. `((1)/(3),(4)/(47))and (-1,-(1)/(3))`

Answer

To find the points on the curve \( y = \frac{2}{3}x^3 + \frac{1}{2}x^2 \) where the tangent makes equal angles with the x-axis, we need to follow these steps:

### Step 1: Find the derivative of the function
The first step is to calculate the derivative \( \frac{dy}{dx} \) of the given function.

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

Using the power rule for differentiation:

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{2}{3}x^3\right) + \frac{d}{dx}\left(\frac{1}{2}x^2\right) \]

Calculating the derivatives:

\[ \frac{dy}{dx} = 2x^2 + x \]

### Step 2: Set the derivative equal to 1
Since the tangent makes equal angles with the x-axis, we set the derivative equal to 1:

\[ 2x^2 + x = 1 \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:

\[ 2x^2 + x - 1 = 0 \]

### Step 4: Solve the quadratic equation
Now, we can solve this quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 2 \), \( b = 1 \), and \( c = -1 \):

\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]

Calculating the discriminant:

\[ x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]

This gives us two solutions:

\[ x = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad x = \frac{-4}{4} = -1 \]

### Step 5: Find corresponding y-values
Now we need to find the corresponding y-values for both x-values.

1. For \( x = \frac{1}{2} \):

\[ y = \frac{2}{3}\left(\frac{1}{2}\right)^3 + \frac{1}{2}\left(\frac{1}{2}\right)^2 \]

Calculating:

\[ y = \frac{2}{3} \cdot \frac{1}{8} + \frac{1}{2} \cdot \frac{1}{4} = \frac{2}{24} + \frac{1}{8} = \frac{2}{24} + \frac{3}{24} = \frac{5}{24} \]

So, one point is \( \left(\frac{1}{2}, \frac{5}{24}\right) \).

2. For \( x = -1 \):

\[ y = \frac{2}{3}(-1)^3 + \frac{1}{2}(-1)^2 \]

Calculating:

\[ y = \frac{2}{3}(-1) + \frac{1}{2}(1) = -\frac{2}{3} + \frac{1}{2} \]

Finding a common denominator (6):

\[ y = -\frac{4}{6} + \frac{3}{6} = -\frac{1}{6} \]

So, the other point is \( (-1, -\frac{1}{6}) \).

### Final Answer
The points on the curve where the tangent makes equal angles with the x-axis are:

\[ \left(\frac{1}{2}, \frac{5}{24}\right) \quad \text{and} \quad (-1, -\frac{1}{6}) \]

Thus, the correct option is **(a)** \( \left(\frac{1}{2}, \frac{5}{24}\right) \) and \( (-1, -\frac{1}{6}) \).

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