Let λ * be the largest value of λ for which the function f_λ (x) = 4λ x³ - 36λ x² + 36x + 48 is increasing for all x ∈ R. Then f_(λ *)(1) + f_(λ *)( - 1) is equal to :

Solution

Maths sir · 5 steps

Increasing for all x means the quadratic f'(x) never goes negative, so its discriminant is at most 0.

1. Differentiate. f'_λ(x) = 12λ x² - 72λ x + 36 = 12(λ x² - 6λ x + 3). For f_λ to be increasing on all of ℝ, we need λ x² - 6λ x + 3 ≥ 0 for every x.
2. Find the range. A quadratic stays non-negative everywhere when its leading coefficient is positive and its discriminant is at most 0. At λ = 0 the expression is just 3 > 0, which also works, so only the upper limit matters.

   36λ² - 12λ ≤ 0 ⇒ 12λ(3λ - 1) ≤ 0 ⇒ 0 ≤ λ ≤ 1/3

3. Largest value. So λ^* = 1/3. Here f'(x) = 4x² - 24x + 36 = 4(x-3)², which is zero only at x = 3, so the function is still increasing.
4. Add the values. In f(1) + f(-1) the odd-power terms cancel, leaving 2(-36λ + 48). With λ = 1/3 this is 2(-12 + 48) = 72.
5. So. f_(λ^*)(1) + f_(λ^*)(-1) = 72, which is option (d).

The idea to keep. Increasing everywhere means f'(x) ≥ 0 for all x. When f' is a quadratic, ask for a positive leading coefficient and D ≤ 0. Equality is allowed, because f' can touch zero at a single point.

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