Math Olympiad Test: Number System- 2 - Class 9 MCQ

α - (-3) is a factor of f(x)
∴ f(-3) = 0
⇒ α (-3)² + 18 = 0
⇒ α = -2

If x = 1 is a factor of f(x), then f(1) = 0
⇒ = f(1) = 4 (1)³ + 3(1)² - 4 (1) + k = 0
= 4 + 3 - 4 + k = 0
⇒ k = -3

Here x² + 6√2x + 10

Given x + 1/x = 3

If α = b = c, then
(α + b + c) = (α + α + α) = 3α
∴ (a + b + c)² - xα² = (3α)² - xα² = 0
⇒ xα² = 9α²
⇒ x = 9

9α² +  4b² + 16c² + 12αb - 24ac -16bc
= (3α)² + (2b)² + (- 4c)² + 2 (3α)(2b) + 2(3α)(-4c) + 2 (3a)(-4c)
= (3a + 2b - 4c)²
Putting α = 2, b = 1, c = -2, we have required expression as
[3(2) + 2 (1) - 4 (-2)]²
= [ 6 + 2 + 8]² = (16)² = 256

Given 3x + 2y = 13, xy = 6
∵ 3x + 2y = 13
∴ (3x + 2y)³ = (13)³
⇒ 27x³ + 8y³ + 3 × 3x × 2y (3x + 2y)
= (169) × 13
⇒ 27x³ + 8y³ + 18xy (3x + 2y) = (169) × 13
⇒ 27x³ + 8y³ + 18 × 6 × (13) = (169) × 13
⇒ 27x³ + 8y³ = 61 × 13 = 793

We have (x - 1) (x² + x + 1) = (x³ - 1)
Now,
Let x³ = p, then x⁶ = p²
∵ (x³ - 1) (x⁶ + x³ + 1)
= (p - 1)(p² + p + 1) = (p³ - 1)
= (x³)³ - 1
= x⁹ - 1

(x⁴ +  x² + 1) has two factors,
i.e, (x² +  x + 1) and (x² -  x + 1)

Given x + y = 4, xy = 4
⇒ (x + y)² = (x - y)² + 4xy
⇒ (4)² = (x - y)² + 4 × 4
⇒ (x - y)² = 16 - 16 = 0
⇒ x = y
∴ if x + y = 4
⇒ y = x = 2
Hence, 2x³ + y³ = 3x³ = 3(2)³ = 24

f(3) will be remained in both cases.
∴ α(3)³ + 4(3)³ + 3(3) - 4 = (3)³ - 4(3) + α
⇒ 27α + 36 + 9 - 4 = 27 - 12 + a
⇒ 26α = 15 - 5 - 36
⇒ 26α = -26
⇒ α = -1

α³ + b³ + c³ - 3αbc = (α + b + c) (α² + b² + c² - αb - bc - cα) = (15) (83 - 71)
= 15 (12)
= 180

Here (α + b + c)² = (15)²
⇒ (α² + b² + c²) = (15)² - 2(αb + bc + cα)
⇒ (83) - 225/2 = - (αb + bc + cα)
⇒ - 71 = - (αb + bc + cα) ...(i)
⇒ αb + bc + cα = 71 ...(i)

Let α - b = A, b - c = B,  c - α = C
Then, A + B + C = (α - b) + (b - c) + (c - α) = 0
∴ A³ + B³ + C³ = 3ABC
⇒ 3(α - b) (b - c) (c - α)

Given α + b + c = 6 , and, α³ + b³ + c³ = 18 + 3αbc
⇒ a³ + b³ + c³ - 3αbc = 18