The line `y=mx+1` touches the curves `y=-x^(4)+2x^(2)+x` at two points `P(x_(1),y_(1))` and `Q(x_(2),y_(2))`. The value of `x_(1)^(2)+x_(2)^(2)+y_(1)^(2)+y_(2)^(2)` is

Options

A. `4`
B. `6`
C. `8`
D. `10`

Explanation

To solve the problem, we need to find the points of tangency between the line \( y = mx + 1 \) and the curve \( y = -x^4 + 2x^2 + x \). We will then calculate \( x_1^2 + x_2^2 + y_1^2 + y_2^2 \).

### Step-by-Step Solution:

1. **Set the equations equal**:
We start by equating the line and the curve:
\[ mx + 1 = -x^4 + 2x^2 + x \]
Rearranging gives:
\[ -x^4 + (2 - m)x^2 + (1 - mx) + 1 = 0 \]

2. **Rearranging the equation**:
Rearranging the equation, we get:
\[ -x^4 + (2 - m)x^2 - mx + 1 = 0 \]
This is a quartic equation in \( x \).

3. **Condition for tangency**:
For the line to touch the curve, the quartic equation must have a double root. This means the discriminant of the polynomial must be zero. We can express the quartic polynomial as:
\[ x^4 - (2 - m)x^2 + mx - 1 = 0 \]

4. **Finding the roots**:
Let's denote the roots of the polynomial as \( x_1 \) and \( x_2 \). Since the line touches the curve at two points, we can use Vieta's formulas:
\[ x_1 + x_2 = 0 \quad \text{(since the coefficient of } x^3 \text{ is zero)} \]
This implies \( x_2 = -x_1 \).

5. **Calculating \( x_1^2 + x_2^2 \)**:
Since \( x_2 = -x_1 \), we have:
\[ x_1^2 + x_2^2 = x_1^2 + (-x_1)^2 = 2x_1^2 \]

6. **Finding \( y_1 \) and \( y_2 \)**:
Substitute \( x_1 \) and \( x_2 \) into the line equation to find \( y_1 \) and \( y_2 \):
\[ y_1 = mx_1 + 1 \quad \text{and} \quad y_2 = mx_2 + 1 = -mx_1 + 1 \]

7. **Calculating \( y_1^2 + y_2^2 \)**:
We can compute:
\[ y_1^2 + y_2^2 = (mx_1 + 1)^2 + (-mx_1 + 1)^2 \]
Expanding this gives:
\[ = (m^2x_1^2 + 2mx_1 + 1) + (m^2x_1^2 - 2mx_1 + 1) = 2m^2x_1^2 + 2 \]

8. **Combining results**:
Now we can combine our results:
\[ x_1^2 + x_2^2 + y_1^2 + y_2^2 = 2x_1^2 + (2m^2x_1^2 + 2) = (2 + 2m^2)x_1^2 + 2 \]

9. **Finding specific values**:
Given that the line is tangent at points \( P(1, 2) \) and \( Q(-1, 0) \), we can substitute \( x_1 = 1 \) and \( x_2 = -1 \):
\[ x_1^2 + x_2^2 = 1^2 + (-1)^2 = 1 + 1 = 2 \]
\[ y_1^2 + y_2^2 = 2^2 + 0^2 = 4 + 0 = 4 \]

10. **Final calculation**:
Thus, we have:
\[ x_1^2 + x_2^2 + y_1^2 + y_2^2 = 2 + 4 = 6 \]

### Conclusion:
The value of \( x_1^2 + x_2^2 + y_1^2 + y_2^2 \) is \( 6 \).