Let `f(x)=int_1^xsqrt(2-t^2)dtdot` Then the real roots of the equation , `x^2-f^(prime)(x)=0` are: `+-1` b. `+-1/(sqrt(2))` c.`+-1/2` d. `0&1`

Options

A. `pm 1`
B. `pm(1)/(sqrt2)`
C. `pm(1)/(2)`
D. `pmsqrt2`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Define the Function
We start with the function defined as:
\[ f(x) = \int_1^x \sqrt{2 - t^2} \, dt \]

### Step 2: Differentiate the Function
Using the Leibniz rule for differentiation under the integral sign, we find the derivative \( f'(x) \):
\[ f'(x) = \sqrt{2 - x^2} \]

### Step 3: Set Up the Equation
We need to solve the equation:
\[ x^2 - f'(x) = 0 \]
Substituting \( f'(x) \) into the equation gives:
\[ x^2 - \sqrt{2 - x^2} = 0 \]

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

### 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: Rearranging to Form a Polynomial
Rearranging gives us a polynomial equation:
\[ x^4 + x^2 - 2 = 0 \]

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

### Step 8: Factor the Quadratic
Factoring the quadratic, we have:
\[ (y - 1)(y + 2) = 0 \]

### Step 9: Solve for \( y \)
Setting each factor to zero gives:
1. \( y - 1 = 0 \) → \( y = 1 \)
2. \( y + 2 = 0 \) → \( y = -2 \) (not valid since \( y = x^2 \) cannot be negative)

### Step 10: Find \( x \)
Since \( y = x^2 \), we have:
\[ x^2 = 1 \implies x = \pm 1 \]

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

Thus, the answer is option **A: \( \pm 1 \)**.

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