Let `f(x)= int_(1)^(x) sqrt(2-t^(2)) dt`. Then the real roots of the equation `x^(2)- f'(x)= 0` are

Options

A. `+-1`
B. `+- (1)/(sqrt2)`
C. `+- (1)/(2)`
D. 0 and 1

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript.

### Step 1: Define the function
We are given the function:
\[ f(x) = \int_{1}^{x} \sqrt{2 - t^2} \, dt \]

### Step 2: Differentiate \( f(x) \)
To find \( f'(x) \), we will use the Fundamental Theorem of Calculus, which states that if \( F(x) = \int_{a}^{x} g(t) \, dt \), then \( F'(x) = g(x) \). Here, \( g(t) = \sqrt{2 - t^2} \).

Thus, we have:
\[ f'(x) = \sqrt{2 - x^2} \]

### Step 3: Set up the equation
We need to find the real roots of the equation:
\[ x^2 - f'(x) = 0 \]
Substituting \( f'(x) \) into the equation gives:
\[ x^2 - \sqrt{2 - x^2} = 0 \]

### Step 4: Rearrange the equation
Rearranging the equation, we get:
\[ x^2 = \sqrt{2 - x^2} \]

### Step 5: Square both sides
To eliminate the square root, we square both sides:
\[ (x^2)^2 = (2 - x^2) \]
This simplifies to:
\[ x^4 = 2 - x^2 \]

### Step 6: Rearrange into standard polynomial form
Rearranging gives us:
\[ x^4 + x^2 - 2 = 0 \]

### Step 7: Substitute \( y = x^2 \)
Let \( y = x^2 \). Then the equation becomes:
\[ y^2 + y - 2 = 0 \]

### Step 8: Factor the quadratic equation
Factoring the quadratic, we have:
\[ (y + 2)(y - 1) = 0 \]
Setting each factor to zero gives:
\[ y + 2 = 0 \quad \text{or} \quad y - 1 = 0 \]
Thus:
\[ y = -2 \quad \text{or} \quad y = 1 \]

### Step 9: Solve for \( x \)
Since \( y = x^2 \), we have:
1. \( x^2 = -2 \) (no real solution)
2. \( x^2 = 1 \)

From \( x^2 = 1 \), we find:
\[ x = \pm 1 \]

### Conclusion
The real roots of the equation \( x^2 - f'(x) = 0 \) are:
\[ x = 1 \quad \text{and} \quad x = -1 \]