Question

Let f(x)= ax² + bx + c be such that f(1) = 3, f(–2) = λ and f(3) = 4. If f(0) + f(1) + f(–2) + f(3) = 14, then λ is equal to ______.

Options

• –4

• `13/2`

• `23/2`

• 4

Answer 1

Let f(x)= ax² + bx + c be such that f(1) = 3, f(–2) = λ and f(3) = 4. If f(0) + f(1) + f(–2) + f(3) = 14, then λ is equal to 4.

Explanation:

For R to be reflexive ⇒ x R x

⇒ 3x + αx = 7x

⇒ (3 + α)x = 7K

⇒ 3 + α = 7λ

⇒ α = 7λ – 3 = 7N + 4, K, λ, N ∈ I

∴ when α divided by 7, remainder is 4.

R to be symmetric xRy ⇒ yRx

3x + αy = 7N₁, 3y + αx = 7N₂

⇒ (3 + α)(x + y) = 7(N₁ + N₂) = 7N₃

Which holds when 3 + α is multiple of 7

∴ α = 7N + 4 (as did earlier)

R to be transitive xRy and yRz ⇒ xRz.

3x + αy = 7N₁ and 3y + αz = 7N₂ and 3x + αz = 7N₃

∴ 3x + 7N₂ – 3y = 7N₃

∴ 7N₁ – αy + 7N₂ – 3y = 7N₃

∴ 7(N1₂ + N₂) – (3 + α)y = 7N₃

∴ (3 + α)y = 7N

Which is true again when 3 + α divisible by 7, i.e. when α divided by 7, remainder is 4.