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[matt43]

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Find the acute angle that the curve y = 1-3x² cut the x-axis.
A. 77°
B. 79°
C. 75°
D. 120°

 

[MAZ3]

To find the acute angle that the curve y = 1-3x² cuts the x-axis, we need to find the slope of the tangent line at the point where the curve intersects the x-axis.

When the curve intersects the x-axis, y = 0. Therefore, we need to solve the equation 0 = 1 - 3x² for x:

3x² = 1 x² = 1/3 x = ±sqrt(1/3)

We can see that there are two points where the curve intersects the x-axis: (-sqrt(1/3), 0) and (sqrt(1/3), 0).

To find the slope of the tangent line at each of these points, we need to take the derivative of y with respect to x:

dy/dx = -6x

At (-sqrt(1/3), 0), the slope of the tangent line is dy/dx = -6(-sqrt(1/3)) = 2sqrt(3).

At (sqrt(1/3), 0), the slope of the tangent line is dy/dx = -6(sqrt(1/3)) = -2sqrt(3).

The acute angle between the x-axis and a line with slope m is given by arctan(m). Therefore, the acute angle that the tangent lines at these points make with the x-axis are:

arctan(2sqrt(3)) ≈ 75° and arctan(-2sqrt(3)) ≈ -75°

Since we're looking for the acute angle, the answer is 75°, which is option C.

 

[Aizy02893]

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