If `y^2+ln(cos^2x)=y`, then

Options

A. `|f''(0)|=0`
B. `|f''(0)+f'(0)|=3`
C. `|f''(0)+f'(0)|=2`
D. `|f''(0)+f'(0)|=1`

Answer

To solve the equation \( y^2 + \ln(\cos^2 x) = y \), we will perform the following steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ y^2 + \ln(\cos^2 x) = y \]
Rearranging gives:
\[ y^2 - y + \ln(\cos^2 x) = 0 \]

### Step 2: Applying the Quadratic Formula
This is a quadratic equation in \( y \). We can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = -1 \), and \( c = \ln(\cos^2 x) \).

Substituting these values:
\[ y = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot \ln(\cos^2 x)}}{2 \cdot 1} \]
\[ y = \frac{1 \pm \sqrt{1 - 4\ln(\cos^2 x)}}{2} \]

### Step 3: Finding the First Derivative
To find the first derivative \( y' \), we differentiate both sides of the original equation with respect to \( x \):
\[ \frac{d}{dx}(y^2) + \frac{d}{dx}(\ln(\cos^2 x)) = \frac{dy}{dx} \]
Using the chain rule:
\[ 2y \frac{dy}{dx} + \frac{1}{\cos^2 x} \cdot (-2\cos x \sin x) = \frac{dy}{dx} \]
This simplifies to:
\[ 2y \frac{dy}{dx} - 2\tan x = \frac{dy}{dx} \]

Rearranging gives:
\[ (2y - 1) \frac{dy}{dx} = 2\tan x \]
Thus,
\[ \frac{dy}{dx} = \frac{2\tan x}{2y - 1} \]

### Step 4: Finding the Second Derivative
Now we differentiate \( \frac{dy}{dx} \) again to find the second derivative \( \frac{d^2y}{dx^2} \):
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(2y - 1)(2\sec^2 x) - (2\tan x)(2\frac{dy}{dx})}{(2y - 1)^2} \]

### Step 5: Evaluating at Specific Points
We need to evaluate \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) at specific points.

1. **At \( x = 0 \)**:
- \( \tan(0) = 0 \)
- From the original equation, when \( x = 0 \), \( \cos(0) = 1 \) and \( \ln(1) = 0 \), so \( y^2 - y = 0 \) gives \( y = 0 \) or \( y = 1 \).

2. **For \( y = 0 \)**:
- \( \frac{dy}{dx} = \frac{2 \cdot 0}{2 \cdot 0 - 1} = 0 \)
- \( \frac{d^2y}{dx^2} = \frac{(2 \cdot 0 - 1)(2 \cdot 1) - (2 \cdot 0)(2 \cdot 0)}{(2 \cdot 0 - 1)^2} = \frac{-2}{1} = -2 \)

3. **For \( y = 1 \)**:
- \( \frac{dy}{dx} = \frac{2 \cdot 0}{2 \cdot 1 - 1} = 0 \)
- \( \frac{d^2y}{dx^2} = \frac{(2 \cdot 1 - 1)(2 \cdot 1) - (2 \cdot 0)(2 \cdot 0)}{(2 \cdot 1 - 1)^2} = \frac{2}{1} = 2 \)

### Conclusion
Thus, we have:
- \( \frac{dy}{dx} = 0 \) at both points.
- \( \frac{d^2y}{dx^2} = -2 \) when \( y = 0 \) and \( \frac{d^2y}{dx^2} = 2 \) when \( y = 1 \).