Question

F(x)=asin4x - bcos2x f' (7П/12)=4 f' (3П/4)=2 a=? b=?

Answer 1

Ответ:

Объяснение:

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Answer 2

To find the values of 'a' and 'b', we can start by finding the first derivative of the function 'F(x)' and then substitute the given values of 'x' into the derivatives.

Let's find the first derivative of 'F(x)' and evaluate it at the given values of 'x':

F(x) = asin(4x) - bcos(2x)

To find F'(x), we can apply the chain rule:

F'(x) = a * d/dx[sin(4x)] - b * d/dx[cos(2x)]

The derivative of sin(4x) with respect to x is cos(4x) * 4, and the derivative of cos(2x) with respect to x is -sin(2x) * 2. Therefore, we have:

F'(x) = a * cos(4x) * 4 - b * (-sin(2x)) * 2 = 4acos(4x) + 2bsin(2x)

Now, we can substitute the given values of 'x' into the derivatives and solve for 'a' and 'b'.

1. For x = 7π/12:

F'(7π/12) = 4acos(4(7π/12)) + 2bsin(2(7π/12)) = 4acos(7π/3) + 2bsin(7π/6)

Since cos(7π/3) = -1/2 and sin(7π/6) = 1/2, we have:

4acos(7π/3) + 2bsin(7π/6) = 4a(-1/2) + 2b(1/2) = -2a + b = 4

2. For x = 3π/4:

F'(3π/4) = 4acos(4(3π/4)) + 2bsin(2(3π/4)) = 4acos(3π/2) + 2bsin(3π/2)

Since cos(3π/2) = 0 and sin(3π/2) = -1, we have:

4acos(3π/2) + 2bsin(3π/2) = 4a(0) + 2b(-1) = -2b = 2

From the equations above, we can solve for 'a' and 'b':

-2a + b = 4 -2b = 2

From the second equation, we find b = -1. Substituting this value into the first equation:

-2a + (-1) = 4 -2a - 1 = 4 -2a = 5 a = -5/2

Therefore, we have a = -5/2 and b = -1.