☰ Reja · Matematika

2-takrorlash: tengsizlik, funksiya, trigonometriya, daraja va logarifm, hosila

24-hafta · blok 1.11–1.13 · 3 dars
2

Aralash mashqlar: trigonometriya, daraja va logarifm (1.7, 1.8, 1.9)

maktab darsligi: matematika 10-sinf; 17–21-hafta mavzularini qayta ko‘ring
Maqsad: Trigonometrik, ko‘rsatkichli va logarifmik topshiriqlarni bir necha bosqichda yechish.
Qoida va g‘oya

Eslatma: sin^2 α + cos^2 α = 1; sin 2α = 2 sin α cos α; cos 2α = 1 - 2 sin^2 α = 2 cos^2 α - 1; tg(α + β) = (tg α + tg β)/(1 - tg α tg β). Chorak ishorasini hisobga olib ildizni tanlang. Trigonometrik tenglamada bir funksiyaga keltiring, so‘ng t = sin x (|t| ≤ 1) qo‘ying; kesmadagi ildizlarni n ga qiymat berib tanlang. Ko‘rsatkichli: a^f = a^g dan f = g (a > 1 da tengsizlik ishorasi saqlanadi, 0 < a < 1 da o‘zgaradi); a^(2x) va a^x qatnashsa t = a^x > 0. Logarifm: log_a x + log_a y = log_a (xy); tenglamani yechgach, logarifm ostidagi barcha ifodalar > 0 ekanini tekshiring.

Namunaviy masalalar
sin α = -4/5 va π < α < 3π/2. sin 2α va cos 2α ni toping.
Yechim: III chorakda cos α = -3/5. sin 2α = 2 · (-4/5) · (-3/5) = 24/25. cos 2α = 1 - 2 · 16/25 = -7/25.
Javob: sin 2α = 24/25, cos 2α = -7/25
2 cos^2 x - cos x - 1 = 0 tenglama [0; 2π] kesmada nechta ildizga ega?
Yechim: t = cos x: 2t^2 - t - 1 = 0, t = 1 yoki t = -1/2. cos x = 1: x = 0 va x = 2π; cos x = -1/2: x = 2π/3 va 4π/3. Jami 4 ta.
Javob: 4
log_2 (x + 1) + log_2 (x - 1) = 3 ni yeching.
Yechim: log_2 (x^2 - 1) = 3, x^2 - 1 = 8, x = ±3. x = -3 da x + 1 < 0, tashlanadi.
Javob: 3
Tipik xatolar
  • cos 2α ni cos 2 · cos α deb yozish.
  • t = sin x almashtirishda |t| > 1 ildizni tashlamaslik; ko‘rsatkichli almashtirishda t ≤ 0 ni tashlamaslik.
  • Logarifmik tenglamada kvadratdan chiqqan ikkala ildizni tekshirmay javobga yozish.
Mashq
1
(sin 70° cos 10° - cos 70° sin 10°)/(cos^2 15° - sin^2 15°) ni hisoblang.
2
2^(x + 1) + 2^(x - 1) = 20 ni yeching.
3
log_2 x · log_x 8 ni hisoblang (x > 0, x ≠ 1).
4
(1/2)^(x^2 - 3x) ≥ 1/16 ni yeching.