Ôn tập cuối năm môn Đại số 11

câu 1 khó nhìn đấy

1a.

\(\lim\limits_{x\rightarrow-\infty}\left(x^3+2x^2-1\right)=\lim\limits_{x\rightarrow-\infty}x^3\left(1+\dfrac{2}{x}-\dfrac{1}{x^3}\right)=-\infty.1=-\infty\)

b. \(\lim\limits_{x\rightarrow-\infty}\left(4x^6+5x^2+1\right)=\lim\limits_{x\rightarrow-\infty}x^6\left(4+\dfrac{5}{x^4}+\dfrac{1}{x^6}\right)=+\infty.4=+\infty\)

c. \(\lim\limits_{x\rightarrow1}\dfrac{2x^2-5x+3}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(2x-3\right)}{x-1}=\lim\limits_{x\rightarrow1}\left(2x-3\right)=-1\)

d. \(\lim\limits_{x\rightarrow-5}\dfrac{3-\sqrt{4-x}}{x+5}=\lim\limits_{x\rightarrow-5}\dfrac{x+5}{\left(x+5\right)\left(3+\sqrt{4-x}\right)}=\lim\limits_{x\rightarrow-5}\dfrac{1}{3+\sqrt{4-x}}=\dfrac{1}{6}\)

2.

a. \(y'=\dfrac{\left(2x+2\right)\left(x+2\right)-\left(x^2+2x-3\right)}{\left(x+2\right)^2}=\dfrac{x^2+4x+7}{\left(x+2\right)^2}\)

b. \(y'=-4x\sqrt{1+2x^2}+\dfrac{2x\left(1-2x^2\right)}{\sqrt{1+2x^2}}=-\dfrac{12x^3+2x}{\sqrt{1+2x^2}}\)

c. \(y'=-14x^6+\dfrac{1}{2\sqrt{x}}\)

3.

Ta có: \(\left\{{}\begin{matrix}SB\perp\left(ABC\right)\Rightarrow SB\perp AC\\AC\perp AB\end{matrix}\right.\) \(\Rightarrow AC\perp\left(SAB\right)\)

Do \(\left\{{}\begin{matrix}BH\in\left(SAB\right)\\AC\perp\left(SAB\right)\end{matrix}\right.\) \(\Rightarrow BH\perp AC\)

Mà \(BH\perp SA\Rightarrow BH\perp\left(SAC\right)\)

Đề không cho sẵn dãy tăng à? Vậy phải chứng minh nó tăng trước

\(u_{n+1}=\dfrac{u_n^2+2018u_n+1}{2020}\)

\(u_{n+1}-u_n=\dfrac{u_n^2+2018u_n+1}{2020}-u_n=\dfrac{\left(u_n-1\right)^2}{2020}\ge0\) \(\Rightarrow\) dãy tăng và không bị chặn trên \(\Rightarrow lim\left(u_n\right)=+\infty\)

\(\Rightarrow2020u_{n+1}=u_n^2+2018u_n+1\)

\(\Leftrightarrow2020u_{n+1}-2020=u_n^2+2018u_n-2019\)

\(\Leftrightarrow2020\left(u_{n+1}-1\right)=\left(u_n+2019\right)\left(u_n-1\right)\)

\(\Rightarrow\dfrac{1}{2020\left(u_{n+1}-1\right)}=\dfrac{1}{\left(u_n+2019\right)\left(u_n-1\right)}=\dfrac{1}{2020}\left(\dfrac{1}{u_n-1}-\dfrac{1}{u_n+2019}\right)\)

\(\Rightarrow\dfrac{1}{u_n+2019}=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}\)

Thế n=1;2;...;n ta được:

\(\dfrac{1}{u_1+2019}=\dfrac{1}{u_1-1}-\dfrac{1}{u_2-1}\)

\(\dfrac{1}{u_2+2019}=\dfrac{1}{u_2-1}-\dfrac{1}{u_3-1}\)

...

\(\dfrac{1}{u_n+2019}=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}\)

Cộng vế: \(S_n=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}=\dfrac{1}{2018}-\dfrac{1}{u_{n+1}-1}\)

\(\Rightarrow\lim\left(S_n\right)=\dfrac{1}{2018}-\dfrac{1}{\infty}=\dfrac{1}{2018}\)

Đặt \(u_n=v_n+1\Rightarrow v_{n+1}+1=\dfrac{2017+v_n+1}{2019-\left(v_n+1\right)}=\dfrac{2018+v_n}{2018-v_n}\)

\(\Rightarrow v_{n+1}=\dfrac{2018+v_n}{2018-v_n}-1=\dfrac{2v_n}{2018-v_n}\Rightarrow\dfrac{1}{v_{n+1}}=1009\dfrac{1}{v_n}-\dfrac{1}{2}\)

Đặt \(\dfrac{1}{v_n}=x_n\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{1}{v_1}=\dfrac{1}{u_1-1}=1\\x_{n+1}=1009x_n-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x_{n+1}-\dfrac{1}{2016}=1009\left(x_n-\dfrac{1}{2016}\right)\)

\(\Rightarrow x_n-\dfrac{1}{2016}\) là CSN với công bội 1009 \(\Rightarrow x_n-\dfrac{1}{2016}=\dfrac{2015}{2016}.1009^{n-1}\)

\(\Rightarrow x_n=\dfrac{2015}{2016}1009^{n-1}+\dfrac{1}{2016}\) 

\(\Rightarrow u_n=v_n+1=\dfrac{1}{x_n}+1=\dfrac{2016}{2015.1009^{n-1}+1}+1\)

\(\Rightarrow\lim\left(u_n\right)=1\)

Do 2 + 1 chia hết cho 3 nên theo bổ đề LTE ta có \(v_3\left(2^{3^n}+1\right)=v_3\left(2+1\right)+v_3\left(3^n\right)=n+1\).

Do đó \(2^{3^n}+1⋮3^{n+1}\) nhưng không chia hết cho \(3^{n+2}\).

a) đồ thị (c) cắt hàm số tại điểm M có hoành độ bằng 2

→ x_o = 2 → y_o = 14

→M_(2;14)

ta có : y^' = 2x + 3x² → y^'(2) = 16 = f^'(x_o)

phương trình tiếp tuyến có dạng :

y = f^'(xo) (x - x_o) +y_o = 16(x-2) +14 = 16x-18

b) vì pttt song song với đường thẳng y=-\(\dfrac{1}{8}\)x +5

→ K =- ​​​\(\dfrac{1}{8}\) ​

y^' = \(\dfrac{-2}{\left(x-1\right)^2}\) = \(\dfrac{-1}{8}\) → \(x\left[{}\begin{matrix}xo=5\\xo=-3\end{matrix}\right.\) → \(\left[{}\begin{matrix}y_o=\dfrac{3}{2}\\y_o=\dfrac{1}{2}\end{matrix}\right.\) 

- phương trình tiếp tuyến có dạng :

+ y = \(\dfrac{-1}{8}\)(x-5) +\(\dfrac{3}{2}\) = \(\dfrac{-1}{8}x\) + \(\dfrac{17}{8}\)

+ y= \(\dfrac{-1}{8}\)(x+3) +\(\dfrac{1}{2}\) = \(\dfrac{-1}{8}x\) + \(\dfrac{1}{8}\)

Đặt \(g\left(x\right)=f\left(x+\dfrac{1}{3}\right)-f\left(x\right)\)

Hiển nhiên \(g\left(x\right)\) cũng liên tục trên R

Ta có: \(g\left(0\right)=f\left(\dfrac{1}{3}\right)-f\left(0\right)\)

\(g\left(\dfrac{2}{3}\right)=f\left(1\right)-f\left(\dfrac{2}{3}\right)\)

\(g\left(\dfrac{1}{3}\right)=f\left(\dfrac{2}{3}\right)-f\left(\dfrac{1}{3}\right)\)

Cộng vế với vế:

\(g\left(0\right)+g\left(\dfrac{1}{3}\right)+g\left(\dfrac{2}{3}\right)=f\left(1\right)-f\left(0\right)=0\)

- Nếu tồn tại 1 trong 3 giá trị \(g\left(0\right);g\left(\dfrac{1}{3}\right);g\left(\dfrac{2}{3}\right)\) bằng 0 thì hiển nhiên pt có nghiệm

- Nếu cả 3 giá trị đều khác 0 \(\Rightarrow\) tồn tại ít nhất 2 trong 3 giá trị \(g\left(0\right)\) ; \(g\left(\dfrac{1}{3}\right)\) ; \(g\left(\dfrac{2}{3}\right)\) trái dấu

\(\Rightarrow\) Luôn tồn tại ít nhất 1 trong 3 tích số: \(g\left(0\right).g\left(\dfrac{1}{3}\right)\) ; \(g\left(0\right).g\left(\dfrac{2}{3}\right)\) ; \(g\left(\dfrac{1}{3}\right).g\left(\dfrac{2}{3}\right)\) âm

\(\Rightarrow\) Pt \(g\left(x\right)=0\) luôn có ít nhất 1 nghiệm thuộc \(\left[0;1\right]\)

\(\left\{{}\begin{matrix}SA\perp\left(ABC\right)\Rightarrow SA\perp BC\\AB\perp BC\left(gt\right)\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\)

Mà \(BC\in\left(SBC\right)\Rightarrow\left(SBC\right)\perp\left(SAB\right)\)

b.

Đường thẳng AM cắt (SBC) tại S, mà \(AS=2MS\)

\(\Rightarrow d\left(M;\left(SBC\right)\right)=\dfrac{1}{2}d\left(A;\left(SBC\right)\right)\)

Kẻ \(AH\perp SB\Rightarrow AH\perp\left(SBC\right)\)

\(\Rightarrow AH=d\left(A;\left(SBC\right)\right)\)

\(\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AB^2}\Rightarrow AH=\dfrac{SA.AB}{\sqrt{SA^2+AB^2}}=\dfrac{2a\sqrt{21}}{7}\)

\(\Rightarrow d\left(M;\left(SBC\right)\right)=\dfrac{a\sqrt{21}}{7}\)

\(M\left(2;4\right)\)

\(y'=\dfrac{-3}{\left(x-1\right)^2}\Rightarrow y'\left(2\right)=-3\)

Phương trình tiếp tuyến \(y=-3\left(x-2\right)+4\Leftrightarrow y=-3x+10\)

\(\Rightarrow b-a^2=1\)

\(y'=2\sqrt{x+1}+\dfrac{2x-1}{2\sqrt{x+1}}=\dfrac{6x+3}{2\sqrt{x+1}}\)