If y = Σₖ₌₁⁶ kcos ⁻¹{ 3/5 cos kx - 4/5 sin kx } ,
then dy/dx at x = 0 is _______.

Solution

Maths sir · 4 steps

Read 3/5, 4/5 as cos α, sin α: each bracket becomes cos(kx+α).

1. Bring in an angle. Take α with cos α = 3/5 and sin α = 4/5, so α is in (0, π/2). Then 3/5 cos kx - 4/5 sin kx = cos kx cos α - sin kx sin α = cos(kx + α).
2. Remove the inverse. cos⁻¹(cos θ) = θ only when θ is in [0, π]. Near x = 0, kx + α is close to α, which is inside (0, π). So each term is k(kx + α).

   y = Σₖ₌₁⁶ k(kx + α) ⇒ dy/dx = Σₖ₌₁⁶ k²

3. Add the squares. Σₖ₌₁⁶ k² = (6 × 7 × 13)/6 = 91.
4. So. dy/dx at x = 0 is 91.

The common slip: 21. Differentiating kx to 1 and forgetting the outer factor k gives Σ k = 21, not 91. Each term has a k outside and a k inside.

The idea to keep. When you see a cos θ - b sin θ with a² + b² = 1, write it as cos(θ + α). Then check the range before cancelling cos⁻¹ with cos.

Answer from the NTA final answer key. Solution by PrepWiser.