Question

Tolong bantuannya~
sekaligus cara kerjanya

Answer 1

Verified answer

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

Answer 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 )

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