Let for f(x) = a₀x² + a₁x + a₂, f'(0) = 1 and f'(1) = 0. If a₀, a₁, a₂ are in an arithmatico-geometric progression, whose corresponding A.P. has common difference 1 and corresponding G.P. has common ratio 2, then f(4) is equal to _____________.

Solution

Maths sir · 4 steps

Get a₀ and a₁ from the derivative conditions, then let the AGP pattern give you a₂.

1. Use the derivative. f'(x) = 2a₀x + a₁. From f'(0) = 1 you get a₁ = 1. From f'(1) = 0 you get 2a₀ + a₁ = 0, so a₀ = -1/2.
2. Write the AGP terms. Each AGP term is an A.P. term times the matching G.P. term. With A.P. first term a, difference 1, and G.P. first term b, ratio 2, the terms are below.

   a₀ = ab, a₁ = (a+1)(2b), a₂ = (a+2)(4b)

3. Find a and b. ab = -1/2, and a₁ = 2ab + 2b = -1 + 2b = 1, so b = 1 and a = -1/2. Then a₂ = (3/2)(4) = 6.
4. So. f(x) = -1/2x² + x + 6, so f(4) = -8 + 4 + 6 = 2.

The common slip: -4. Dropping the 2 in f'(x) = 2a₀x + a₁ gives a₀ = -1, then b = 3/2, a₂ = 8 and f(4) = -4.

The idea to keep. In an AGP, write each term as (A.P. term) × (G.P. term). Expanding the product often brings in a known term like ab, and that is the quickest way to the unknowns.

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