Question

Fungsi y=1/2x^2-x+a memenuhi persamaan y' . y' - y = 0. agar persamaan ini mempunyai tepat satu akar real, maka konstanta a adalah?

Answer 1

Y = (1/2) x² - x + a
y' = x - 1

y' . y' - y = 0
=> (x - 1)(x - 1) - ((1/2) x² - x + a) = 0
=> x² - 2x + 1 - (1/2) x² + x - a = 0
=> (1/2) x² - x + 1 - a = 0
a = 1/2
b = -1
c = 1 - a

agar mempunyai tepat satu akar real maka

D = 0
=> b² - 4ac = 0
=> (-1)² - 4(1/2)(1 - a) = 0
=> 1 - 2 + 2a = 0
=> -1 + 2a = 0
=> 2a = 1
=> a = 1/2