Giải SBT Toán 11 CTST Bài tập cuối chương VII có đáp án

Đáp án đúng là: B

Ta có ${y}^{'}=3{x}^{2}+6x$.

Tiếp tuyến với đồ thị của hàm số tại điểm M(‒1; ‒6) có hệ số góc bằng ${y}^{'}(-1)=-3$.

Đáp án đúng là: A

Ta có ${y}^{'}=3{x}^{2}-3$.

${y}^{'}(-1)=0$.

Ta có:

• ${f}^{'}(x)=9{x}^{2}-6x+6$

• ${f}^{''}(x)=18x-6$

• ${g}^{'}(x)=3{x}^{2}+2x$

Từ đó ${f}^{''}(x)-{f}^{'}(x)+{g}^{'}(x)-8\ge 0$

$\begin{matrix}\Leftrightarrow 18x-6-(9{x}^{2}-6x+6)+3{x}^{2}+2x-8\ge 0 \\ \Leftrightarrow -6{x}^{2}+26x-20\ge 0 \\ \Leftrightarrow 1\le x\le \frac{10}{3}\end{matrix}$

Vậy bất phương trình ${f}^{''}(x)-{f}^{'}(x)+{g}^{'}(x)-8\ge 0$ có tập nghiệm là $[1;\frac{10}{3}]$.

Ta có

${y}^{'}=\frac{{(x-1)}^{'}(x+1)-(x-1){(x+1)}^{'}}{{(x+1)}^{2}}=\frac{2}{{(x+1)}^{2}}$

${y}^{''}=-\frac{2{[{(x+1)}^{2}]}^{'}}{{(x+1)}^{2}}=-\frac{4}{{(x+1)}^{3}}$.

Vậy ${y}^{''}(1)=-\frac{4}{{(1+1)}^{3}}=-\frac{1}{2}$

Ta có ${f}^{'}(x)={e}^{\sqrt{{x}^{2}+4}}.\frac{2x}{2\sqrt{{x}^{2}+4}}=\frac{x{e}^{\sqrt{{x}^{2}+4}}}{\sqrt{{x}^{2}+4}}$.

Vậy ${f}^{'}(1)=\frac{{e}^{\sqrt{{1}^{2}+4}}}{\sqrt{{1}^{2}+4}}=\frac{{e}^{\sqrt{5}}}{\sqrt{5}}$.

a) Với ${x}_{0}\ge -\frac{1}{4}$, ta có:

${y}^{'}({x}_{0})=\underset{x\rightarrow {x}_{0}}{\lim }\frac{f(x)-f({x}_{0})}{x-{x}_{0}}=\underset{x\rightarrow {x}_{0}}{\lim }\frac{\sqrt{4x+1}-\sqrt{4{x}_{0}+1}}{x-{x}_{0}}$

$=\underset{x\rightarrow {x}_{0}}{\lim }\frac{4(\sqrt{4x+1}-\sqrt{4{x}_{0}+1})}{(4x+1)-(4{x}_{0}+1)}$

$=\underset{x\rightarrow {x}_{0}}{\lim }\frac{4(\sqrt{4x+1}-\sqrt{4{x}_{0}+1})}{(\sqrt{4x+1}-\sqrt{4{x}_{0}+1})(\sqrt{4x+1}+\sqrt{4{x}_{0}+1})}$

$=\underset{x\rightarrow {x}_{0}}{\lim }\frac{4}{(\sqrt{4x+1}+\sqrt{4{x}_{0}+1})}$

$=\frac{4}{2\sqrt{4{x}_{0}+1}}$.

Vậy ${y}^{'}(2)=\frac{4}{2\sqrt{4.2+1}}=\frac{2}{3}$.

b) Với ${x}_{0}\in$ ℝ, ta có:

${y}^{'}({x}_{0})=\underset{x\rightarrow {x}_{0}}{\lim }\frac{f(x)-f({x}_{0})}{x-{x}_{0}}=\underset{x\rightarrow {x}_{0}}{\lim }\frac{{x}^{4}-{{x}_{0}}^{4}}{x-{x}_{0}}$

$=\underset{x\rightarrow {x}_{0}}{\lim }\frac{{x}^{4}-{{x}_{0}}^{4}}{x-{x}_{0}}=\underset{x\rightarrow {x}_{0}}{\lim }\frac{({x}^{2}-{{x}_{0}}^{2})({x}^{2}+{{x}_{0}}^{2})}{x-{x}_{0}}$

$=\underset{x\rightarrow {x}_{0}}{\lim }\frac{(x-{x}_{0})(x+{x}_{0})({x}^{2}+{{x}_{0}}^{2})}{x-{x}_{0}}$

$=\underset{x\rightarrow {x}_{0}}{\lim }(x+{x}_{0})({x}^{2}+{{x}_{0}}^{2})=2{x}_{0}.2{{x}_{0}}^{2}=4{{x}_{0}}^{3}$

Vậy ${y}^{'}(-1)=4.{(-1)}^{3}=-4$.

c) Với ${x}_{0}\ne -1$, ta có:

${y}^{'}({x}_{0})=\underset{x\rightarrow {x}_{0}}{\lim }\frac{f(x)-f({x}_{0})}{x-{x}_{0}}=\underset{x\rightarrow {x}_{0}}{\lim }\frac{\frac{1}{x+1}-\frac{1}{{x}_{0}+1}}{x-{x}_{0}}$

$=\underset{x\rightarrow {x}_{0}}{\lim }\frac{\frac{1}{x+1}-\frac{1}{{x}_{0}+1}}{x-{x}_{0}}=\underset{x\rightarrow {x}_{0}}{\lim }\frac{({x}_{0}+1)-(x+1)}{(x-{x}_{0})(x+1)({x}_{0}+1)}$

$=\underset{x\rightarrow {x}_{0}}{\lim }\frac{{x}_{0}-x}{(x-{x}_{0})(x+1)({x}_{0}+1)}==\underset{x\rightarrow {x}_{0}}{\lim }-\frac{1}{(x+1)({x}_{0}+1)}$

$=-\frac{1}{{({x}_{0}+1)}^{2}}$.

Vậy ${y}^{'}(x)=-\frac{1}{{(x+1)}^{2}}\ (x\ne -1).$

d) Với ${x}_{0}\in$ ℝ, ta có:

${y}^{'}({x}_{0})=\underset{x\rightarrow {x}_{0}}{\lim }\frac{f(x)-f({x}_{0})}{x-{x}_{0}}=\underset{x\rightarrow {x}_{0}}{\lim }\frac{\sqrt[3]{{x}^{2}+1}-\sqrt[3]{{x}_{0}^{2}+1}}{x-{x}_{0}}$

$=\underset{x\rightarrow {x}_{0}}{\lim }\frac{(x+{x}_{0})(\sqrt[3]{{x}^{2}+1}-\sqrt[3]{{x}_{0}^{2}+1})}{(x-{x}_{0})(x+{x}_{0})}$

$=\underset{x\rightarrow {x}_{0}}{\lim }\frac{(x+{x}_{0})(\sqrt[3]{{x}^{2}+1}-\sqrt[3]{{x}_{0}^{2}+1})}{{x}^{2}-{x}_{0}^{2}}$ $=\underset{x\rightarrow {x}_{0}}{\lim }\frac{(x+{x}_{0})(\sqrt[3]{{x}^{2}+1}-\sqrt[3]{{x}_{0}^{2}+1})}{({x}^{2}+1)-({x}_{0}^{2}+1)}$

$=\underset{x\rightarrow {x}_{0}}{\lim }\frac{(x+{x}_{0})(\sqrt[3]{{x}^{2}+1}-\sqrt[3]{{x}_{0}^{2}+1})}{(\sqrt[3]{{x}^{2}+1}-\sqrt[3]{{x}_{0}^{2}+1})[\sqrt[3]{{({x}^{2}+1)}^{2}}+\sqrt[3]{{x}^{2}+1}\sqrt[3]{{x}_{0}^{2}+1}+\sqrt[3]{{({x}_{0}^{2}+1)}^{2}}]}$

$=\underset{x\rightarrow {x}_{0}}{\lim }\frac{x+{x}_{0}}{\sqrt[3]{{({x}^{2}+1)}^{2}}+\sqrt[3]{{x}^{2}+1}\sqrt[3]{{x}_{0}^{2}+1}+\sqrt[3]{{({x}_{0}^{2}+1)}^{2}}}=\underset{x\rightarrow {x}_{0}}{\lim }\frac{2{x}_{0}}{3\sqrt[3]{{({x}_{0}^{2}+1)}^{2}}}$.

Vậy ${y}^{'}(x)=\frac{2x}{3\sqrt[3]{{({x}^{2}+1)}^{2}}}$ ( $x\in$ ℝ).

Gọi tiếp tuyến là d và tiếp điểm M(x$_{0}$, f(x$_{0}$)).

Ta có ${f}^{'}(x)=6{x}^{2}-2x+2=6{(x-\frac{1}{6})}^{2}+\frac{11}{6}\ge \frac{11}{6}$.

Vậy hệ số góc của d nhỏ nhất bằng $\frac{11}{6}$ khi ${x}_{0}=\frac{1}{6}$.

Phương trình đường tiếp tuyến d:

$y-f(\frac{1}{6})={f}^{'}(\frac{1}{6})(x-\frac{1}{6})$

$\Leftrightarrow y-\frac{71}{54}=\frac{11}{6}(x-\frac{1}{6})$

$\Leftrightarrow y=\frac{11}{6}x+\frac{109}{108}$

Vậy tiếp tuyến với (C) có hệ số góc nhỏ nhất là d: $y=\frac{11}{6}x+\frac{109}{108}$.

Ta có ${s}^{'}(t)=9{t}^{2}+5$ nên ${{s}^{'}}^{'}(t)=18t$.

${s}^{'}(1)={9.1}^{2}+5=14$ (m/s)

${s}^{''}(1)=18.1=18$ (m/s$^{2}$)

Vậy khi t = 1, vận tốc và gia tốc của vật đó lần lượt bằng 14 m/s và 18 m/s$^{2}$.

a) ${y}^{'}={(\sqrt{x})}^{'}({x}^{2}-\sqrt{x}+1)+\sqrt{x}{({x}^{2}-\sqrt{x}+1)}^{'}$

$=\frac{({x}^{2}-\sqrt{x}+1)}{2\sqrt{x}}+\sqrt{x}(2x-\frac{1}{2\sqrt{x}})$

$=\frac{x\sqrt{x}}{2}-\frac{1}{2}+\frac{1}{2\sqrt{x}}+2x\sqrt{x}-\frac{1}{2}$

$=\frac{5}{2}x\sqrt{x}+\frac{1}{2\sqrt{x}}-1$.

b) ${y}^{'}=\frac{{({x}^{2}-3x+1)}^{'}}{{({x}^{2}-3x+1)}^{2}}$

$=\frac{2x-3}{{({x}^{2}-3x+1)}^{2}}$.

c, ${y}^{'}=\frac{{(2x+3)}^{'}(3x+2)-(2x+3){(3x+2)}^{'}}{{(3x+2)}^{2}}$

$=\frac{2(3x+2)+(2x+3)3}{{(3x+2)}^{2}}=-\frac{5}{{(3x+2)}^{2}}$

a, ${y}^{'}=\frac{(x\sin x)'(1-\tan x)+(x\sin x){(1-\tan x)}^{'}}{{(1-\tan x)}^{2}}$

$=\frac{(\sin x+x\cos x)(1-\tan x)+(x\sin x)(-\frac{1}{{\cos }^{2}x})}{{(1-\tan x)}^{2}}\ =\frac{\sin x+x\cos x-\sin x\tan x-x\sin x+\frac{x\sin x}{{\cos }^{2}x}}{{(1-\tan x)}^{2}}\ =\frac{\sin x+x\cos x-\sin x\tan x+x\sin x(-1+\frac{1}{{\cos }^{2}x})}{{(1-\tan x)}^{2}}$

$=\frac{\sin x+x\cos x-\sin x\tan x+x\sin x\frac{1-{\cos }^{2}x}{{\cos }^{2}x}}{{(1-\tan x)}^{2}}\ =\frac{\sin x+x\cos x-\sin x\tan x+x\sin x\frac{{\sin }^{2}x}{{\cos }^{2}x}}{{(1-\tan x)}^{2}}\ =\frac{\sin x+x\cos x-\sin x\tan x+x\sin x{\tan }^{2}x}{{(1-\tan x)}^{2}}$

b) ${y}^{'}=-\sin \sqrt{{x}^{2}-x+1}.{(\sqrt{{x}^{2}-x+1})}^{'}$

$=-\sin \sqrt{{x}^{2}-x+1}.\frac{1}{2\sqrt{{x}^{2}-x+1}}.{({x}^{2}-x+1)}^{'}$

$=-\frac{\sin \sqrt{{x}^{2}-x+1}}{2\sqrt{{x}^{2}-x+1}}.(2x-1)$

$=-\frac{(2x-1)\sin \sqrt{{x}^{2}-x+1}}{2\sqrt{{x}^{2}-x+1}}$.

c) ${y}^{'}=2sin3x.{(\sin 3x)}^{'}=2sin3x.\cos 3x.3$

$=3\sin 6x$.

d) $y=co{s}^{2}(cos3x)=2\cos (\cos 3x).{(\cos (\cos 3x))}^{'}$

$=2\cos (\cos 3x).(-\sin (\cos 3x)).{(\cos 3x)}^{'}$

$=-2\cos (\cos 3x).\sin (\cos 3x).(-\sin 3x).3$

$=3\sin 3x\sin (2\cos 3x)$.

b) ${y}^{'}=\frac{1}{2\sqrt{{f}^{2}(x)+{g}^{2}(x)}}.{({f}^{2}(x)+{g}^{2}(x))}^{'}$

$=\frac{(2f(x){f}^{'}(x)+2g(x){g}^{'}(x))}{2\sqrt{{f}^{2}(x)+{g}^{2}(x)}}$

$=\frac{f(x){f}^{'}(x)+g(x){g}^{'}(x)}{\sqrt{{f}^{2}(x)+{g}^{2}(x)}}$.

Ta có ${f}^{'}(x)=3{x}^{2}+4x-m$

$Δ=\frac{{4}^{2}-4.3.(-m)}{2.4}=\frac{16+12m}{8}=\frac{4+3m}{2}$.

a) ${f}^{'}(x)=0$ có nghiệm kép khi $Δ=\frac{4+3m}{2}=0$ hay $m=-\frac{4}{3}$.

Ta có ${f}^{'}(x)=\frac{1}{2\sqrt{{x}^{2}-2x+8}}.{({x}^{2}-2x+8)}^{'}$

$=\frac{2x-2}{2\sqrt{{x}^{2}-2x+8}}=\frac{x-1}{\sqrt{{x}^{2}-2x+8}}$.

${f}^{'}(x)=-\frac{2}{3}\Leftrightarrow \frac{x-1}{\sqrt{{x}^{2}-2x+8}}=-\frac{2}{3}$

$\Leftrightarrow 2\sqrt{{x}^{2}-2x+8}=-3(x-1)\ (x<1)$

$\Leftrightarrow 4({x}^{2}-2x+8)=9{(x-1)}^{2}$

$\Leftrightarrow 4({x}^{2}-2x+8)=9({x}^{2}-2x+1)$

$\Leftrightarrow 5{x}^{2}-10x-23=0$

$\Leftrightarrow x=\frac{5+2\sqrt{35}}{5}$ (loại); $x=\frac{5-2\sqrt{35}}{5}$ (nhận).

Vậy $x=\frac{5-2\sqrt{35}}{5}$.

b, $y=\sqrt{3x+2}=\frac{{(3x+2)}^{'}}{2\sqrt{3x+2}}=\frac{3}{2\sqrt{3x+2}}$

${y}^{''}=\frac{3.(-\frac{1}{2}).{(3x+2)}^{'}}{2\sqrt{{(3x+2)}^{3}}}=-\frac{3.3}{4\sqrt{{(3x+2)}^{3}}}$

$=-\frac{9}{4\sqrt{{(3x+2)}^{3}}}$

c) ${y}^{'}=x{e}^{2x}={e}^{2x}+x{({e}^{2x})}^{'}={e}^{2x}+x.{e}^{2x}.2=(2x+1){e}^{2x}$.

${y}^{''}={(2x+1)}^{'}{e}^{2x}+(2x+1){({e}^{2x})}^{'}=2{e}^{2x}+(2x+1){e}^{2x}.2$

$=2{e}^{2x}+(4x+2){e}^{2x}=4(x+1){e}^{2x}$.