If the function
f( x ) = { k₁( x - π )² - 1,, x ≤ π; k₂cos x,, x > π is
twice differentiable, then the ordered pair (k₁, k₂) is equal to :

Solution

Maths sir · 4 steps

Match the two pieces at x = π three times: value, first derivative, then second derivative.

1. Continuity. The left piece gives f(π) = k₁(0)² - 1 = -1. The right piece tends to k₂ cos π = -k₂. Equating them gives -k₂ = -1, so k₂ = 1.
2. First derivative. On the left, f'(x) = 2k₁(x - π), which is 0 at x = π. On the right, f'(x) = -k₂ sin x, which tends to -k₂ sin π = 0. These match for every k₁, k₂, so this condition gives nothing new.
3. Second derivative. On the left, f''(x) = 2k₁. On the right, f''(x) = -k₂ cos x, which tends to -k₂ cos π = k₂. For f''(π) to exist, the two must be equal.

   2k₁ = k₂ = 1 ⇒ k₁ = 1/2

4. So. (k₁, k₂) = (1/2, 1), which is option (d).

The idea to keep. Twice differentiable at a joining point means value, first derivative and second derivative all agree from both sides. Each condition that is not automatically true gives one equation.

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