Question

2x^4+3x^2*y^2+y^2+y^4если x^2+y^2=1Варианты ответов!1)32)53)24)1

Answer 1

2x^4+3x^2*y^2+y^2+y^4
x2+y2=1
2x4+3x2y2+y2+y4=x4+2x2y2+y4+x4+x2y2+y2=(x2+y2)2+x4+x2y2+y2 = 1+y2+x2(x2+y2)=1+y2+x2=1+1=2

Answer 2

To find the value of the expression 2x^4 + 3x^2*y^2 + y^2 + y^4, we can substitute x^2 + y^2 = 1 into the expression and simplify:

2x^4 + 3x^2y^2 + y^2 + y^4 = 2x^4 + 3x^2(1-x^2) + (1-x^2)^2 + (1-x^2)*x^2 = 2x^4 + 3x^2 - 3x^4 + 1 - 2x^2 + x^4 = -2x^4 + x^2 + 1

Now we need to find the maximum and minimum values of this expression as x varies over the range -1 to 1 (because x^2 + y^2 = 1). To do this, we can take the derivative of the expression with respect to x and set it equal to zero:

d/dx (-2x^4 + x^2 + 1) = -8x^3 + 2x = 0

This equation has three roots: x = 0, x = sqrt(2)/2, and x = -sqrt(2)/2. We can plug these values of x back into the expression and find the corresponding values of y (since x^2 + y^2 = 1) to get the maximum and minimum values:

• When x = 0, y = ±1, and the expression takes the value 1.
• When x = sqrt(2)/2, y = ±sqrt(2)/2, and the expression takes the value 5/2.
• When x = -sqrt(2)/2, y = ±sqrt(2)/2, and the expression takes the value 5/2.

Therefore, the possible answers are 1), 2), 3), and 4). However, we can see that the expression takes the value 1 at x = 0 regardless of the value of y, so the correct answer is 4) 1.