Baholash usuli: chap va o'ng tomon bir songa teng: matematika darsi

$x^2-4x+7=3\cos\pi x$ tenglamada algebraik va trigonometrik ifodalar aralashgan, uni odatdagi almashtirishlar bilan yechib bo'lmaydi. Lekin chap tomon $(x-2)^2+3$ doim kamida $3$ , o'ng tomon esa ko'pi bilan $3$ . Ular faqat ikkalasi ham $3$ ga teng bo'lganda uchrashadi: $x=2$ . Bu darsda ifodalarni baholab, shunday tenglama va tengsizliklarni yechamiz.

Asosiy qoida

Agar barcha $x$ uchun $f(x)\ge A$ va $g(x)\le A$ bo'lsa, $f(x)=g(x)$ tenglama $f(x)=A$ , $g(x)=A$ sistemasiga teng kuchli. Ikkala shart bitta $x$ da bajarilishi kerak.

Sababi: $f(x)=g(x)$ bo'lsa, $A\le f(x)=g(x)\le A$ , ya'ni ikkalasi ham $A$ . Amalda ishlash tartibi: bir tomonning chegarasini va unga erishiladigan nuqtani topamiz (odatda algebraik tomon), keyin shu nuqtani ikkinchi tomonga qo'yib tekshiramiz.

Usul har doim ham natija bermaydi. Chegaralarga qarab uch holat bor:

Holat	Misol	Xulosa
HolatChegaralar ustma-ust tushadi	Misol $(x-2)^2+3=3\cos\pi x$ : $f\ge 3$ , $g\le 3$	XulosaSistemaga keladi, $x=2$
HolatChegaralar orasida bo'shliq bor	Misol $6x-x^2-7=2^{x^2-6x+11}$ : chap $\le 2$ , o'ng $\ge 4$	XulosaYechim yo'q
HolatChegaralar kesishadi	Misol $x^2+1=2\cos x$ : chap $\ge 1$ , o'ng $\le 2$	XulosaBaholash yetmaydi, tenglamaning ikki ildizi bor, lekin ular bu usulda topilmaydi

Chegaralar ustma-ust tushgani yechim borligini ham bildirmaydi. $(x-1)^2+2=2\cos x$ da $f\ge 2$ , $g\le 2$ , lekin $f=2$ faqat $x=1$ da, u yerda esa $2\cos 1\ne 2$ . Yechim yo'q. Shuning uchun nuqta topilgach, ikkinchi tomon albatta tekshiriladi.

Baholash uchun tayyor chegaralar

Ifoda	Chegara	Tenglik sharti
Ifoda $\sin x$ , $\cos x$	Chegara $[-1;\,1]$	Tenglik shartiHar bir chetki qiymat davriy takrorlanadi
Ifoda $a\sin x+b\cos x$	Chegara $[-\sqrt{a^2+b^2};\,\sqrt{a^2+b^2}]$	Tenglik shartiIkkala chetki qiymatga erishiladi
Ifoda $(x-a)^2+b$	Chegara $\ge b$	Tenglik sharti $x=a$
Ifoda $b-(x-a)^2$	Chegara $\le b$	Tenglik sharti $x=a$
Ifoda $\sqrt{(x-a)^2+b}$ , $b\ge 0$	Chegara $\ge\sqrt{b}$	Tenglik sharti $x=a$
Ifoda $\sqrt{c-x^2}$ , $c\ge 0$	Chegara $\le\sqrt{c}$	Tenglik sharti $x=0$
Ifoda $t+\dfrac{1}{t}$ , $t>0$	Chegara $\ge 2$	Tenglik sharti $t=1$
Ifoda $t+\dfrac{1}{t}$ , $t<0$	Chegara $\le -2$	Tenglik sharti $t=-1$
Ifoda $a^x+a^{-x}$ , $a>0$ , $a\ne 1$	Chegara $\ge 2$	Tenglik sharti $x=0$
Ifoda $\operatorname{tg}^2x+\operatorname{ctg}^2x$	Chegara $\ge 2$	Tenglik sharti $\operatorname{tg}^2x=1$
Ifoda $a^{(x-p)^2+q}$ , $a>1$	Chegara $\ge a^q$	Tenglik sharti $x=p$
Ifoda $\lg\left(10-(x-p)^2\right)$	Chegara $\le 1$	Tenglik sharti $x=p$
Ifoda $\arccos x$	Chegara $[0;\,\pi]$	Tenglik sharti $\arccos 1=0$ , $\arccos(-1)=\pi$
Ifoda $\arcsin x$	Chegara $\left[-\dfrac{\pi}{2};\,\dfrac{\pi}{2}\right]$	Tenglik sharti $\arcsin(-1)=-\dfrac{\pi}{2}$ , $\arcsin 1=\dfrac{\pi}{2}$

$t>0$ da $t+\dfrac{1}{t}\ge 2$ ning isboti:

t+\frac{1}{t}-2=\frac{(t-1)^2}{t}\ge 0

t+\frac{1}{t}+2=\frac{(t+1)^2}{t}\le 0

t<0

da kasrning maxraji manfiy

$a\sin x+b\cos x$ uchun yordamchi burchak ishlatiladi:

a\sin x+b\cos x

=\sqrt{a^2+b^2}\sin(x+\varphi)

Masalan,

3\sin x-4\cos x

ning qiymatlari

[-5;\,5]

oraliqda

Kvadrat va trigonometrik funksiya

Algebraik tomonda to'liq kvadrat ajratamiz, trigonometrik tomonni $[-1;\,1]$ bilan baholaymiz.

x^2-4x+7=(x-2)^2+3\ge 3

Chap tomon

x=2

da eng kichik qiymat

3

ni oladi

3\cos\pi x\le 3

3\cos 2\pi=3

O'ng tomon

x=2

da ham

3

. Demak

x=2

yagona ildiz

$1+\operatorname{tg}^2x=\cos^23x$ da chap tomon $\ge 1$ , o'ng tomon $\le 1$ . Chap tomon $\operatorname{tg}x=0$ , ya'ni $x=\pi k$ da $1$ ga teng. Bu nuqtalarda $\cos^23\pi k=1$ , o'ng tomon ham $1$ . Javob: $x=\pi k$ , $k\in\mathbb{Z}$ . Ildizlarni sanashda shu formuladan foydalanamiz: $[-\pi;\,\pi]$ da $k=-1,\ 0,\ 1$ , uchta ildiz.

Ba'zan tenglamaning bir tomoni avval songa keltiriladi. Buning uchun quyidagi formula kerak:

\cos a\cos 2a\cos 4a=\frac{\sin 8a}{8\sin a}

\sin a\ne 0

8\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}=-1

Chunki

\sin\dfrac{8\pi}{7}=-\sin\dfrac{\pi}{7}

Shunda quyidagi tenglamaning o'ng tomoni $-1$ ga teng:

\operatorname{tg}^2x-\cos^24x

=8\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}

Tenglama

1+\operatorname{tg}^2x=\cos^24x

ga aylanadi va

x=\pi k

beradi

Ildizli va ko'rsatkichli ifodalar

Ildiz ostida ham kvadrat ajratiladi: $(x-3)^2+4$ va $(x-3)^2+9$ .

\sqrt{x^2-6x+13}\ge 2

\sqrt{x^2-6x+18}\ge 3

Ikki ildizning yig'indisi kamida

5

. Tenglik

x=3

6x-x^2-4=5-(x-3)^2\le 5

Ildizlar yig'indisini shu ifodaga tenglashtirsak, yagona ildiz

x=3

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