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

Đáp án đúng là: B

Ta có  $y^{\text{'}}=3x^2+6x$.

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

Đáp án đúng là: A

Ta có  $y^{\text{'}}=3x^2-3$.

 $y^{\text{'}}\left(-1\right)=0$.

Ta có:

•  $f^{\text{'}}\left(x\right)=9x^2-6x+6$

•  $f^{\text{'}\text{'}}\left(x\right)=18x-6$

•  $g^{\text{'}}\left(x\right)=3x^2+2x$

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

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

$y^{\text{'}}=\frac{{\left(2x-1\right)}^{\text{'}}\left(3x+2\right)-\left(2x-1\right){\left(3x+2\right)}^{\text{'}}}{{\left(3x+2\right)}^2}=\frac{7}{{\left(3x+2\right)}^2}$.

Ta có

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

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

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

Ta có  $y^{\text{'}}=3^{x^2+1}\ln 3.2x=2x{3}^{x^2+1}\ln 3$.

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

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

a) Với  $x_0\ge -\frac{1}{4}$, ta có:

 $y^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\sqrt{4x+1}-\sqrt{4x_0+1}}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\frac{4\left(\sqrt{4x+1}-\sqrt{4x_0+1}\right)}{\left(4x+1\right)-\left(4x_0+1\right)}$

 $=\underset{x\to x_0}{\lim }\frac{4\left(\sqrt{4x+1}-\sqrt{4x_0+1}\right)}{\left(\sqrt{4x+1}-\sqrt{4x_0+1}\right)\left(\sqrt{4x+1}+\sqrt{4x_0+1}\right)}$

 $=\underset{x\to x_0}{\lim }\frac{4}{\left(\sqrt{4x+1}+\sqrt{4x_0+1}\right)}$

 $=\frac{4}{2\sqrt{4x_0+1}}$.

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

b) Với  $x_0\in$ ℝ, ta có:

 $y^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}=\underset{x\to x_0}{\lim }\frac{x^4-{x_0}^4}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\frac{x^4-{x_0}^4}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\left(x^2-{x_0}^2\right)\left(x^2+{x_0}^2\right)}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\frac{\left(x-x_0\right)\left(x+x_0\right)\left(x^2+{x_0}^2\right)}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\left(x+x_0\right)\left(x^2+{x_0}^2\right)=2x_0.2{x_0}^2=4{x_0}^3$

Vậy  $y^{\text{'}}(-1)=4.{(-1)}^3=-4$.

c) Với  $x_0\ne -1$, ta có:

 $y^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\frac{1}{x+1}-\frac{1}{x_0+1}}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\frac{\frac{1}{x+1}-\frac{1}{x_0+1}}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\left(x_0+1\right)-\left(x+1\right)}{\left(x-x_0\right)\left(x+1\right)\left(x_0+1\right)}$

 $=\underset{x\to x_0}{\lim }\frac{x_0-x}{\left(x-x_0\right)\left(x+1\right)\left(x_0+1\right)}==\underset{x\to x_0}{\lim }-\frac{1}{\left(x+1\right)\left(x_0+1\right)}$

 $=-\frac{1}{{\left(x_0+1\right)}^2}$.

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

d) Với  $x_0\in$ ℝ, ta có:

 $y^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\frac{\left(x+x_0\right)\left(\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}\right)}{\left(x-x_0\right)\left(x+x_0\right)}$

 $=\underset{x\to x_0}{\lim }\frac{\left(x+x_0\right)\left(\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}\right)}{x^2-x_0^2}$ $=\underset{x\to x_0}{\lim }\frac{\left(x+x_0\right)\left(\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}\right)}{\left(x^2+1\right)-\left(x_0^2+1\right)}$

 $=\underset{x\to x_0}{\lim }\frac{\left(x+x_0\right)\left(\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}\right)}{\left(\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}\right)\left[\sqrt[3]{{\left(x^2+1\right)}^2}+\sqrt[3]{x^2+1}\sqrt[3]{x_0^2+1}+\sqrt[3]{{\left(x_0^2+1\right)}^2}\right]}$

 $=\underset{x\to x_0}{\lim }\frac{x+x_0}{\sqrt[3]{{\left(x^2+1\right)}^2}+\sqrt[3]{x^2+1}\sqrt[3]{x_0^2+1}+\sqrt[3]{{\left(x_0^2+1\right)}^2}}=\underset{x\to x_0}{\lim }\frac{2x_0}{3\sqrt[3]{{\left(x_0^2+1\right)}^2}}$.

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

Gọi tiếp tuyến là d và tiếp điểm M(x₀,f(x₀)).

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

Vậy hệ số góc của d nhỏ nhất bằng  $\frac{11}{6}$ khi  $x_0=\frac{1}{6}$.

Phương trình đường tiếp tuyến d:

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

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

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

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

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

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

 $s^{\text{'}\text{'}}\left(1\right)=18.1=18$ (m/s²)

Vậy khi t = 1, vận tốc và gia tốc của vật đó lần lượt bằng 14 m/s và 18 m/s².

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

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

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

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

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

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

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

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

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

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

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

 $=3\sin 6x$.

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

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

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

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

a)  $y^{\text{'}}=f^{\text{'}}\left(x^3\right).{\left(x^3\right)}^{\text{'}}=3x^2{f}^{\text{'}}\left(x^3\right)$.

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

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

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

Ta có  $f^{\text{'}}\left(x\right)=3x^2+4x-m$

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

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

b)  $f^{\text{'}}\left(x\right)\ge 0$ với mọi x khi  $\Delta =\frac{4+3m}{2}\ge 0$ hay  $m\le -\frac{4}{3}$.

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

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

$f^{\text{'}}\left(x\right)=-\frac{2}{3}\iff \frac{x-1}{\sqrt{x^2-2x+8}}=-\frac{2}{3}$

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

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

 $\iff 4\left(x^2-2x+8\right)=9\left(x^2-2x+1\right)$

 $\iff 5x^2-10x-23=0$

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

Vậy   $x=\frac{5-2\sqrt{35}}{5}$.

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

 $y^{\text{'}\text{'}}={\left(2x+1\right)}^{\text{'}}{e}^{2x}+\left(2x+1\right){\left(e^{2x}\right)}^{\text{'}}=2e^{2x}+\left(2x+1\right)e^{2x}.2$

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