Answer is: option4
\( (1, 1) \)Solution:
To find the point on the curve \( y = 2 - x^2 \) that is nearest to the point \( (3, 2) \), we can follow these steps:
The distance \( D \) between a point \( (x, y) \) on the curve and the point \( (3, 2) \) is given by:
\[ D = \sqrt{(x - 3)^2 + (y - 2)^2} \]
Since \( y = 2 - x^2 \), substitute \( y \) into the distance formula:
\[ D = \sqrt{(x - 3)^2 + ((2 - x^2) - 2)^2} = \sqrt{(x - 3)^2 + (-x^2)^2} = \sqrt{(x - 3)^2 + x^4} \]
To minimize \( D \), it's equivalent to minimize \( D^2 \):
\[ D^2 = (x - 3)^2 + x^4 \]
Take the derivative of \( D^2 \) with respect to \( x \) and set it to zero:
\[ \frac{d}{dx} \left((x - 3)^2 + x^4\right) = 2(x - 3) + 4x^3 \]
Set the derivative equal to zero:
\[ 2(x - 3) + 4x^3 = 0 \quad \Rightarrow \quad 2x - 6 + 4x^3 = 0 \quad \Rightarrow \quad 4x^3 + 2x - 6 = 0 \]
Simplify the equation:
\[ 2x^3 + x - 3 = 0 \]
Test possible rational roots (e.g., \( x = 1 \)):
\[ 2(1)^3 + 1 - 3 = 0 \Rightarrow 0 = 0 \]
So, \( x = 1 \) is a root. Factor the polynomial:
\[ 2x^3 + x - 3 = (x - 1)(2x^2 + 2x + 3) \]
The quadratic \( 2x^2 + 2x + 3 \) has no real roots (discriminant < 0), so the only real critical point is at \( x = 1 \).
Check the second derivative to ensure it's a minimum:
\[ \frac{d^2}{dx^2} (D^2) = \frac{d}{dx} \left(4x^3 + 2x - 6\right) = 12x^2 + 2 \]
Since \( 12x^2 + 2 > 0 \) for all \( x \), the critical point at \( x = 1 \) is indeed a minimum.
Substitute \( x = 1 \) into the curve equation: \[ y = 2 - (1)^2 = 1 \]
Thus, the point is \( (1, 1) \). The point \( (1, 1) \) corresponds to option (D).