Ôn tập cuối năm môn Đại số 11

\(a,f'\left(x\right)=3x^2-6x\\ f'\left(x\right)\le0\Leftrightarrow3x^2-6x\le0\\ \Leftrightarrow3x\left(x-2\right)\le0\Leftrightarrow0\le x\le2\)

Lời giải:

a. $f'(x)\leq 0$

$\Leftrightarrow 3x^2-6x\leq 0$

$\Leftrightarrow x(x-2)\leq 0$

$\Leftrightarrow 0\leq x\leq 2$

b.

$f'(x)=x^2-3x+2=0$

$\Leftrightarrow 3x^2-6x=x^2-3x+2=0$

$\Leftrightarrow 3x(x-2)=(x-1)(x-2)=0$

$\Leftrightarrow x-2=0$

$\Leftrightarrow x=2$

c.

$g(x)=f(1-2x)+x^2-x+2022$

$g'(x)=(1-2x)'f(1-2x)'_{1-2x}+2x-1$

$=-2[3(1-2x)^2-6(1-2x)]+2x-1$
$=-24x^2+2x+5$

$g'(x)\geq 0$

$\Leftrightarrow -24x^2+2x+5\geq 0$

$\Leftrightarrow (5-12x)(2x-1)\geq 0$

$\Leftrightarrow \frac{-5}{12}\leq x\leq \frac{1}{2}$

\(4^x+2^{x+1}-3=0\)

=>\(\left(2^x\right)^2+2\cdot2^x-3=0\)

=>\(\left(2^x+3\right)\left(2^x-1\right)=0\)

mà \(2^x+3>0\forall x\)

nên \(2^x-1=0\)

=>\(2^x=1\)

=>x=0

\(3^x+9\cdot3^{-x}=10\)

=>\(3^x+\dfrac{9}{3^x}=10\)

=>\(\dfrac{\left(3^x\right)^2+9}{3^x}=10\)

=>\(\left(3^x\right)^2+9=10\cdot3^x\)

=>\(\left(3^x\right)^2-10\cdot3^x+9=0\)

=>\(\left(3^x-1\right)\left(3^x-9\right)=0\)

=>\(\left[{}\begin{matrix}3^x-1=0\\3^x-9=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3^x=1\\3^x=9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

1:C 3:B 5:A 6:B 8:C 9:A 10:D 11:y=x

y' = 3x² +2x +1

a. y' <= 6

<=> 3x² + 2x +1 <=6

<=.> \(\dfrac{-5}{3}\) <= x <= 1

B. K=6

<=> 3x² + 2x +1 = 6

<=> \(\left[{}\begin{matrix}x=1\\x=\dfrac{-5}{3}\end{matrix}\right.\)

với x= 1 => y= -2

vập pttt: y = 6 (x-1) -2

<=> y = 6x - 8

với x=\(\dfrac{-5}{3}\) => y = \(\dfrac{-230}{27}\)

vập pttt : y= 6( x + \(\dfrac{5}{3}\) ) - \(\dfrac{230}{27}\)

<=> y = 6x + \(\dfrac{40}{27}\)

\(P=\left[\left(7+4\sqrt{3}\right)\left(4\sqrt{3}-7\right)\right]^{2016}\cdot\left(7+4\sqrt{3}\right)=\left(-1\right)^{2016}\cdot\left(7+4\sqrt{3}\right)=7+4\sqrt{3}\)

Câu 1:

\(S=\sqrt{2}+\frac{\sqrt{2}}{2}+....+\frac{\sqrt{2}}{2^n}\)

\(2S=2\sqrt{2}+\sqrt{2}+...+\frac{\sqrt{2}}{2^{n-1}}\)

\(S=2S-S=2\sqrt{2}-\frac{\sqrt{2}}{2^n}\)

\(\lim S=\lim (2\sqrt{2}-\frac{\sqrt{2}}{2^n})=2\sqrt{2}\)

Đáp án C.

Câu 2:

\(\lim\limits_{x\to +\infty}\frac{3-2x-5x^3}{x^3-1}=\lim\limits_{x\to +\infty}\frac{\frac{3}{x^3}-\frac{2}{x^2}-5}{1-\frac{1}{x^3}}=\frac{-5}{1}=-5\)

Đáp án B.

Câu 3:

** Tứ diện $ABCD$ có $SA\perp (BCD)$???? Bạn coi lại đề!

f'(x)=(m-1)*x^2+2

Đặt f'(x)=0

=>\(x^2=\dfrac{-2}{m-1}\)

=>\(x=\pm\sqrt{-\dfrac{2}{m-1}}\)

Để phương trình có nghiệm dương thì m-1<0

=>m<1