Quiz about Балауса 5 нұсқа

Question 1

^{$$\lim_{x\rightarrow\infty}\ \frac{x^2+5x-12}{3x^2-18x+7}$$}

Accepted Answer

$$\frac{1}{3}$$

Answer

$$\frac{2}{3}$$

Answer

$$\frac{1}{2}$$

Answer

1

Question 2

$$f\left(x\right)=-\ \frac{1-x}{\sqrt[]{x^2-2}}\ кему\ аралығы$$

Accepted Answer

$$\left(-\infty;-\sqrt[]{2}\right)\cup\left(\sqrt[]{2;2}\right]$$

Answer

$$\left(\sqrt[]{2};2\right)\cup\left(2;+\infty\right)$$

Answer

$$\left(-\infty;-\sqrt[]{2}\right)\cup\left(4;+\infty\right)$$

Answer

$$\left(-\infty;-\sqrt[]{2}\right]\cup\left[\sqrt[]{2;2}\right]$$

Question 3

$$\int_{ }^{ }\sin5x\cos7xdx$$

Accepted Answer

$$\frac{1}{24}\cos12x+\frac{1}{4}\cos2x+c$$

Answer

$$-\frac{1}{2}\cos6x+\frac{1}{2}\cos x+c$$

Answer

$$\frac{1}{2}\cos6x+\frac{1}{2}\cos x+c$$

Answer

$$-\frac{1}{12}\cos6x+\frac{1}{2}\cos2x+c$$

Question 4

$$\frac{\cos2\beta+1-8\sin^2\beta\left(1-\frac{1-\cos2\beta}{2}\right)}{\cos^4\frac{\beta}{2}-\sin^4\frac{\beta}{2}}\ өрнегінің\ ең\ үлкен\ және\ ең\ кіші\ мәні\ \left(көбейтінді\right)$$

Accepted Answer

-4

Answer

-9

Answer

-1

Answer

0

Question 5

$$\frac{\cos2\beta+1-8\sin^2\beta\left(1-\frac{1-\cos2\beta}{2}\right)}{\cos^4\frac{\beta}{2}-\sin^4\frac{\beta}{2}}\ өрнегінің\ ең\ үлкен\ және\ ең\ кіші\ мәні\ \left(қосынды\right)$$

Accepted Answer

0

Answer

-9

Answer

-1

Answer

-4

Question 6

$$Теңсіздікті\ шешіңіз:3^{x^2-x^{ }}\ge\left(5^{x-1}\right)^x$$

Accepted Answer

[0;1]

Answer

$$\left[-1;1\right]$$

Answer

[1;2]

Answer

[-2;0]

Question 7

$$\log_2\left(9-2^x\right)=10\log_{10}\left(3-x\right)\ теңдеудің\ түбірі\ жатқан\ аралық$$

Accepted Answer

(0,2;2)

Answer

(0,2;5)

Answer

(0,1;1)

Answer

(-3;5)

Question 8

$$f\left(x\right)=\sqrt[]{x\sin2x\ }\ функциясының\ туындысы$$

Accepted Answer

$$\frac{\sin2x+2x\cos2x}{2\sqrt[]{x\sin x}}$$

Answer

$$\frac{\sin2x+2x\cos2x}{\sqrt[]{x\sin x}}$$

Answer

$$\frac{\sin x\cos x+x\left(1-2\sin^2x\right)}{2\sqrt[]{x+\sin2x}}$$

Answer

$$\frac{\sin2x+2x\cos2x}{2\sqrt[]{x\sin2x}}$$

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