The number of real solutions of `y+y^(2)=sinx and y+y^(3)=cos^(-1)(cos x)` is/are (a) 0 (b) 1 (c) 3 (d) infinite

Options

A. 0
B. 1
C. 3
D. Infinite

Answer

To find the number of real solutions for the equations \( y + y^2 = \sin x \) and \( y + y^3 = \cos^{-1}(\cos x) \), we will analyze each equation step by step.

### Step 1: Analyze the first equation
The first equation is:
\[ y + y^2 = \sin x \]
We can rearrange this to:
\[ y^2 + y - \sin x = 0 \]
This is a quadratic equation in \( y \). The solutions for \( y \) can be found using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = -\sin x \). Thus, we have:
\[ y = \frac{-1 \pm \sqrt{1 + 4\sin x}}{2} \]
For \( y \) to be real, the discriminant must be non-negative:
\[ 1 + 4\sin x \geq 0 \]
This implies:
\[ \sin x \geq -\frac{1}{4} \]

### Step 2: Analyze the second equation
The second equation is:
\[ y + y^3 = \cos^{-1}(\cos x) \]
The expression \( \cos^{-1}(\cos x) \) is equal to \( x \) (modulo \( 2\pi \)) in the range \( [0, \pi] \). Thus, we can write:
\[ y + y^3 = x \quad \text{for } x \in [0, \pi] \]
Rearranging gives:
\[ y^3 + y - x = 0 \]
This is a cubic equation in \( y \). A cubic equation can have either one or three real roots.

### Step 3: Find the intersection of solutions
Now we need to find the values of \( x \) for which both equations have solutions.

1. From the first equation, we established that \( \sin x \geq -\frac{1}{4} \). The values of \( x \) that satisfy this condition can be found in the range of \( x \) where \( \sin x \) is greater than or equal to \(-\frac{1}{4}\).

2. For the second equation, we need to analyze the cubic function \( y^3 + y - x = 0 \). The behavior of this function will determine how many real solutions \( y \) can take for given values of \( x \).

### Step 4: Check specific values of \( x \)
Let’s evaluate specific values of \( x \) that satisfy both equations.

- For \( x = 0 \):
- \( y + y^2 = 0 \) gives \( y(y + 1) = 0 \) → \( y = 0 \) or \( y = -1 \).
- \( y + y^3 = 0 \) gives \( y(y^2 + 1) = 0 \) → \( y = 0 \) is a solution.

- For \( x = \pi \):
- \( y + y^2 = 0 \) gives \( y(y + 1) = 0 \) → \( y = 0 \) or \( y = -1 \).
- \( y + y^3 = \pi \) → \( y^3 + y - \pi = 0 \) has one real solution.

- For \( x = 2\pi \):
- Similar analysis shows \( y = 0 \) is a solution.

### Conclusion
The analysis shows that for each \( x \) that satisfies \( \sin x \geq -\frac{1}{4} \), there are infinite pairs \( (x, y) \) that satisfy both equations. Thus, the number of real solutions is infinite.

### Final Answer
The number of real solutions is:
**(d) infinite**