Question Detail
Question
What is the value of K for which the function f(x) = Ksinx + sin3x/3 has an extremum at x = π/3?
Correct Answer
Option C is the correct answer.
Detailed Explanation
For a function to have an extremum (local maximum or minimum) at a point, its first derivative at that point must be zero.Given f(x) = Ksinx + (1/3)sin3x.First, find the derivative f'(x):f'(x) = d/dx(Ksinx) + d/dx((1/3)sin3x)f'(x) = Kcosx + (1/3) * (cos3x * 3)f'(x) = Kcosx + cos3xNow, set f'(x) = 0 at x = π/3:Kcos(π/3) + cos(3 * π/3) = 0K(1/2) + cos(π) = 0K(1/2) - 1 = 0K/2 = 1K = 2
Hint
An extremum occurs where the first derivative of the function is zero. Differentiate f(x) with respect to x, set the derivative to zero at x = π/3, and solve for K. Remember cos(π/3) = 1/2 and cos(π) = -1.