Matematika

Pertanyaan

tolong bantuannya~
sekaligus cara kerjanya
tolong bantuannya~ sekaligus cara kerjanya

2 Jawaban

  • a = (x + 1)i + xj
    |a| = √((x + 1)^2 + x^2) = √(x^2 + 2x + 1 + x^2) = √(2x^2 + 2x + 1)

    b = 2xi + (3x + 1)j

    b.a = (x + 1)(2x) + x(3x + 1) = 2x^2 + 2x + 3x^2 + x = 5x^2 + 3x

    p = proyeksi b pada a
    |p| = (b.a)/|a| = (5x^2 + 3x) / √(2x^2 + 2x + 1)

    |p| ≤ 2|a|
    (5x^2 + 3x)/√(2x^2 + 2x + 1) ≤ 2√(2x^2 + 2x + 1)
    Karena (2x^2 + 2x + 1) definit positif maka kita kali silang saja
    (5x^2 + 3x) ≤ 2(2x^2 + 2x + 1)
    5x^2 + 3x ≤ 4x^2 + 4x + 2
    x^2 - x - 2 ≤ 0
    (x - 2)(x + 1) ≤ 0
    x = 2 atau x = -1
    Garis bilangan
    +++ (-1) --- (2) +++
    -1 ≤ x ≤ 2
  • |a| = √(x + 1)^2 + x^2) =
    √((x + 1) (x + 1) + x^2) =
    √(x^2 + 2x + 1 + x^2) =
    √(2x^2 + 2x + 1)

    Proyeksi b ke a =
    b.a / |a| =
    2x+(3x + 1) . (x + 1) / √(2x^2 + 2x + 1) =
    2x^2 + 2x + 3x^2 + x / √(2x^2 + 2x + 1)=
    5x^2 + 3x / √(2x^2 + 2x + 1)

    |p| ≤ 2|a|
    5x^2 +3x/√(2x^2 + 2x + 1) ≤ 2√(2x^2 + 2x + 1)
    5x^2 + 3x ≤ 2(2x^2 + 2x + 1)
    5x^2 + 3x ≤ 4x^2 + 4x + 2
    5x^2 + 3x - ( 4x^2 + 4x + 2) ≤ 0
    5x^2 + 3x - 4x^2 - 4x - 2 ≤ 0
    x^2 - x - 2 ≤ 0
    (x + 1)(x - 2) ≤ 0

    x = -1 ≤ 0 atau
    0 ≤ x = 2

    -1 ≤ 0 ≤ 2 ( C )

    #semoga membantu.....