Question Detail

For a function to have an extremum (local maximum or minimum) at a point, its first derivative must be zero at that point.Given f(x) = p sin 3x + K/(3x).First, find the derivative f'(x):f'(x) = d/dx(p sin 3x) + d/dx(K/(3x))f'(x) = p * (cos(3x) * 3) + K * d/dx((1/3)x-1)f'(x) = 3p cos(3x) + K * (1/3) * (-1)x-2f'(x) = 3p cos(3x) - K/(3x2)Now, set f'(x) = 0 at x = π/3:3p cos(3 * (π/3)) - K/(3 * (π/3)2) = 03p cos(π) - K/(3 * (π2/9)) = 03p (-1) - K/(π2/3) = 0-3p - 3K/π2 = 0This equation gives a relationship between p and K: 3p = -3K/π2, which simplifies to p = -K/π2.Since the question asks for a specific value of K without providing a value for p, it implies that the value of K is independent of p, or that p is such that K takes one of the options. Given option C (K=2) is marked, if K=2, then p = -2/π2, which is a valid value for p. Thus, K=2 is a possible value for which the condition holds.