255 Bài trắc nghiệm Hàm số mũ và Logarit cơ bản nâng cao cực hay có lời giải (P4)

Đáp án C

Đáp án D

$$\begin{gathered} y={\log }_2\left(\sqrt{x^2+1}+x\right) \\ y\text{'}=\frac{\left(\sqrt{x^2+1}+x\right)\text{'}}{\left(\sqrt{x^2+1}+x\right)\ln 2}=\frac{{\displaystyle \frac{x}{\sqrt{x^2+1}}}+1}{\left(\sqrt{x^2+1}+x\right)\ln 2}=\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}\left(\sqrt{x^2+1}+x\right)\ln 2} \\ =\frac{1}{\sqrt{x^2+1}.\ln 2} \end{gathered}$$

Đáp án B

$$\begin{gathered} ĐK: 2x-3x^2>0\iff 0<x<\frac{2}{3} \\ y\text{'}=\frac{2-6x}{\left(2x-3x^2\right).\ln 9} \\ y\text{'}=0\iff x=\frac{1}{3}\left(TM\right) \\ Ta có f\left(\frac{1}{3}\right)=\frac{-1}{2} \\ \underset{x\to 0}{\lim }{\log }_9\left(2x-3x^2\right)=\underset{x\to \frac{2}{3}}{\lim }{\log }_9\left(2x-3x^2\right)=-\infty \\ \Rightarrow TGT G=(-\infty ;\frac{1}{2}] \end{gathered}$$

Đáp án A

$$\begin{gathered} Đặt t=2^x (t>0) \\ BPT\hspace{0.17em}trở thành 3t^2-(9m+1)t+3m\le 0 \\ \iff 3t^2-9mt-t+3m\le 0 \\ \iff (3t-1)(t-3m)\le 0 \\ Ta thấy BPT luôn có nghiệm TMĐK khi t\ge \frac{1}{3} hoặc t\le \frac{1}{3} \\ \Rightarrow BPT\hspace{0.17em}luôn có nghệm với mọi m\in \mathrm{ℝ} \end{gathered}$$

Đáp án D

Đáp án C

$$\begin{gathered} HD:Ta chứng minh với f\left(x\right)=\frac{1}{1+c^x}khi a+b=0 thì f\left(a\right)+f\left(b\right)=1 \\ Ta có: f\left(a\right)=\frac{1}{1+c^a} \\ f\left(b\right)=\frac{1}{1+c^b}=\frac{1}{1+c^{-a}} \\ \Rightarrow f\left(a\right)+f\left(b\right)=\frac{1}{1+c^a}+\frac{1}{1+c^{-a}} \\ =\frac{1}{1+c^a}+\frac{1}{1+{\displaystyle \frac{1}{c^a}}}=\frac{1}{1+c^a}+\frac{c^a}{1+c^a}=\frac{1+c^a}{1+c^a}=1 \\ Theo ĐB: f\left(x\right)=\frac{1}{1+{2017}^x} \\ \Rightarrow S=f(-1999)+f(-1998)+...+f\left(0\right)+f\left(1\right)+f\left(2\right)+...+f\left(1999\right) \\ =1+1+...+1+f\left(0\right) \\ =1\times 1999+\frac{1}{2}=\frac{3999}{2} \end{gathered}$$

Đáp án C

Đáp án C

$$\begin{gathered} 3^{-x}=1-\sqrt{2} \\ \iff 3^x=\frac{1}{1-\sqrt{2}} \\ \iff x={\log }_3\left(\frac{1}{1-\sqrt{2}}\right)=-{\log }_3\left(1-\sqrt{2}\right) \end{gathered}$$

Đáp án B

$$\begin{gathered} f\left(x\right)=4^{1-x}.\ln 4 \\ f\text{'}\left(x\right)=-\ln 4.4^{1-x}=4^{1-x}.\ln \left(\frac{1}{4}\right) \end{gathered}$$

Đáp án D

ĐK: x > 1

$$\begin{gathered} {\log }_{2-\sqrt{3}}(x-1)\ge 0 \\ \iff x-1\le 1\iff x\le 2 \\ KHĐK: 1<x\le 2 \end{gathered}$$

Đáp án A

$$\begin{gathered} 9^x-5.3^x+6<0 \\ \iff 2<3^x<3 \\ \iff {\log }_{3}2<x<1 \end{gathered}$$

Đáp án C

$$\begin{gathered} 3^{x^2-x}=4^x \\ \iff x={\log }_4\left(3^{x^2-x}\right) \\ \iff x=(x^2-x){\log }_{4}3 \\ \iff x[1-(x-1\left){\log }_{4}3\right]=0 \\ \iff \left\{\begin{array}{l}x=0\\ (x-1){\log }_{4}3=1\end{array}\right.\iff \left\{\begin{array}{l}x=0\\ x=\frac{1}{{\log }_{4}3}+1={\log }_{3}4+1\end{array}\right. \end{gathered}$$

Đáp án B

$$\begin{gathered} f\left(x\right)=\ln \left(\frac{x}{x+1}\right) \\ f\text{'}\left(x\right)=\left(\frac{x}{x+1}\right)\text{'}.\frac{1}{{\displaystyle \frac{x}{x+1}}}=\frac{1}{{(x+1)}^2}.\frac{x+1}{x}=\frac{1}{x(x+1)} \end{gathered}$$

Đáp án D

$$\begin{gathered} {\log }_6(a^2.\sqrt{b})={\log }_6{a}^2+{\log }_6\sqrt{b} \\ =2{\log }_{6}a+\frac{1}{2}{\log }_{6}b \\ =2\left({\log }_{6}a+\frac{1}{4}{\log }_{6}b\right) \\ =2\left({\log }_{6}a+{\log }_6\sqrt[4]{b}\right)=2{\log }_6\left(a.\sqrt[4]{b}\right) \end{gathered}$$

Đáp án C

$ĐK:\left\{\begin{array}{l} {\log }_{\frac{1}{2}}x>0 \\ x>0\end{array}\iff \left\{\begin{array}{l}x<1\\ x>0\end{array}\iff 0<x<1\right.\right.$

Đáp án D

$$\begin{gathered} 2^{{\sin }^2\mathrm{x}}+3^{{\cos }^2\mathrm{x}}=\mathrm{m}.3^{{\sin }^2\mathrm{x}}(*) \\ \mathrm{Đặt} {\sin }^2\mathrm{x}=\mathrm{t} (0\le \mathrm{t}\le 1) \\ \Rightarrow 2^{\mathrm{t}}+3^{1-\mathrm{t}}=\mathrm{m}.3^{\mathrm{t}} \\ \iff {\left(\frac{2}{3}\right)}^{\mathrm{t}}+3^{1-2\mathrm{t}}=\mathrm{m} \\ \mathrm{Đặt} \mathrm{f}\left(\mathrm{t}\right)={\left(\frac{2}{3}\right)}^{\mathrm{t}}+3^{1-2\mathrm{t}} \\ \mathrm{f}\text{'}\left(\mathrm{t}\right)={\left(\frac{2}{3}\right)}^{\mathrm{t}}\ln \left(\frac{2}{3}\right)-2.3^{1-2\mathrm{t}}\ln 3<0 \mathrm{với}\forall \mathrm{t}\in [0;1] \\ \Rightarrow \mathrm{Hàm} \mathrm{f}\left(\mathrm{t}\right) \mathrm{NB} \mathrm{trên} \mathrm{đoạn} [0;1] \\ \mathrm{Ta} \mathrm{có} \mathrm{f}\left(0\right)=4 \mathrm{và} \mathrm{f}\left(1\right)=1 \\ \mathrm{KL}: \mathrm{PT} (*) \mathrm{có} \mathrm{nghiệm} \mathrm{khi} 1\le \mathrm{m}\le 4 \end{gathered}$$

Đáp án D

$$\begin{gathered} ĐK:\left\{\begin{array}{l}1-3x\ne 1\\ 1-2x>0\end{array}\iff \left\{\begin{array}{l}x\ne 0\\ x<\frac{1}{2}\end{array}\right.\right. \\ {\log }_{1-3x}(1-2x)=-1 \\ \iff 1-2x={\left(1-3x\right)}^{-1} \\ \iff (1-2x)(1-3x)=1 \\ \iff 6x^2-5x=0 \\ \iff \left\{\begin{array}{l}x=0 \left(L\right)\\ x=\frac{5}{6}\left(L\right)\end{array}\right. \end{gathered}$$

Đáp án C

$$\begin{gathered} {\left(\frac{1}{5}\right)}^{x^2-4x}>1 \\ \iff x^2-4x<0 \\ \iff 0<x<4 \end{gathered}$$

Đáp án A

$f\text{'}\left(x\right)=\frac{(1-3x)\text{'}}{(1-3x)\ln 3}=\frac{-3}{(1-3x)\ln 3}=\frac{3}{(3x-1)\ln 3}$

Đáp án D

$$\begin{gathered} ĐK: {\log }_{\frac{1}{5}}(5x-1)\ge 0 \\ \iff 0<5x-1\le 1 \\ \iff \frac{1}{5}<x\le \frac{2}{5} \end{gathered}$$

Đáp án D

Đáp án D

$$\begin{gathered} {\log }_3{(5-2x)}^2=2x \\ \iff {(5-2x)}^2=3^{2x} \\ \iff \left|5-2x\right|=3^x \\ \iff \left|2x-5\right|=3^x \end{gathered}$$

Đáp án A

$$\begin{gathered} 3^{1-2x}=2 \\ \iff 1-2x={\log }_{3}2 \\ \iff x=\frac{1-{\log }_{3}2}{2} \end{gathered}$$

Đáp án D

$$\begin{gathered} ĐK: 4x-3>0\iff x>\frac{3}{4} \\ {\log }_{0,5}(4x-3)\ge -1 \\ \iff 4x-3\le 2 \\ \iff x\le \frac{5}{4} \\ KHĐK: \frac{3}{4}<x\le \frac{5}{4} \end{gathered}$$

$$\begin{gathered} 9^x-(m-3).3^x-3m=0 \\ \iff \left(3^x-m\right)(3^x+3)=0 \left(1\right) \end{gathered}$$

Vậy PT có nghiệm khi phương trình $3^x=m$ có nghiệm 

$\Rightarrow m>0$

Đáp án C

$$\begin{gathered} f\left(x\right)={\log }_5\left(\cos x\right) \\ \Rightarrow f\text{'}\left(x\right)=\frac{-\sin x}{\cos x.\ln 5}=\frac{-\tan x}{\ln 5} \end{gathered}$$

Đáp án A

$$\begin{gathered} 4^x.{\left(\sqrt{2}\right)}^{x-1}\le 1 \\ \iff 2^{2x}\le \frac{1}{2^{{\displaystyle \frac{1}{2}}(x-1)}} \\ \iff 2^{2x}\le 2^{\frac{-1}{2}(x-1)} \\ \iff 2x\le \frac{-1}{2}(x-1) \\ \iff 4x\le 1-x \\ \iff x\le \frac{1}{5} \end{gathered}$$

Đáp án D

$$\begin{gathered} Đk: x>\frac{1}{2} \\ f\text{'}\left(x\right)=\frac{2}{(2x-1)\ln \left({\displaystyle \frac{1}{4}}\right)}<0 với \forall x>\frac{1}{2} \\ \Rightarrow f\left(x\right) nghịch biến trên \left(\frac{1}{2};+\infty \right) \end{gathered}$$

$$\begin{gathered} NX: hàm y={(2-\sqrt{3})}^x là hàm nghịch biến trên \mathrm{ℝ} \\ Ta thấy khẳng định \left(1\right),\left(3\right),\left(4\right)đúng \\ \left(5\right) đúng vì: \\ {\left(2-\sqrt{3}\right)}^x>2+\sqrt{3}={\left(2-\sqrt{3}\right)}^{-1} \\ \iff x<-1 \\ \left(2\right) sai vì {\left(2-\sqrt{3}\right)}^x\ge 1\iff x\le 0 \end{gathered}$$

Đáp án A

$$\begin{gathered} {\log }_4\left(\frac{\sqrt{a}}{b^2}\right)={\log }_4\sqrt{a}-{\log }_4{b}^2 \\ =\frac{1}{2}{\log }_{4}a-2{\log }_{4}b=\frac{1}{2}{\log }_{4}a-\frac{1}{2}.4{\log }_{4}b \\ ={\log }_{4^2}a-4{\log }_{4^2}b={\log }_{16}a-4{\log }_{16}b \end{gathered}$$

Đáp án B

$$\begin{gathered} 3^{{\mathrm{x}}^2-6\mathrm{x}}={\left(\frac{1}{27}\right)}^3 \\ \iff 3^{{\mathrm{x}}^2-6\mathrm{x}}=3^{-3} \\ \iff {\mathrm{x}}^2-6\mathrm{x}+3=0 \\ \iff \mathrm{x}=3\pm \sqrt{6} \\ \Rightarrow \mathrm{Tổng} \mathrm{các} \mathrm{nghiệm} \mathrm{của} \mathrm{PT} \mathrm{là} 6 \end{gathered}$$

Đáp án D

$$\begin{gathered} A và C sai vì 2^{x^3}\ne {\left(2^x\right)}^3=8^x \\ B sai vì \frac{2^{x^4}}{2^x}=2^{x^4-x}\ne 2^{x^3} \\ D đúng vì 8^x.{\left(2^x\right)}^{x^2-3}=2^{3x}.2^{x^3-3x}=2^{3x+x^3-3x}=2^{x^3} \end{gathered}$$

Đáp án B

$$\begin{gathered} {\log }_6\left(\cos x\right)=m. ĐK: \cos x>0 \\ \iff \cos x=6^m \\ PT có nghiệm \iff 0<6^m\le 1 \iff m\le 0 \end{gathered}$$

Đáp án D

$$\begin{gathered} y={\left(2-\sqrt{3}\right)}^{x^2-1} \\ TXĐ: D=\mathrm{ℝ} \\ y\text{'}=2x.{\left(2-\sqrt{3}\right)}^{x^2-1}.\ln \left(2-\sqrt{3}\right) \\ y\text{'}=0\iff x=0 \\ Dễ thấy 2.{\left(2-\sqrt{3}\right)}^{x^2-1}.\ln \left(2-\sqrt{3}\right)<0 với mọi x \\ \Rightarrow y\text{'}>0 khi x<0 và y\text{'}<0 khi x>0 \\ \Rightarrow y\uparrow /(-\infty ;0) và y\downarrow /(0;+\infty ) \end{gathered}$$

$$\begin{gathered} Xét đáp án A: \\ {2}^{x^2}=\frac{1}{3}\iff x^2={\log }_2\left(\frac{1}{3}\right)=-{\log }_{2}3 \\ Ta thấy -{\log }_{2}3<0 nên PT x^2=-{\log }_{2}3 VN \\ Xét đáp án B: \\ {\log }_4(3x+1)=2-\sqrt{3}. ĐK: x>\frac{-1}{3} \\ \iff 3x+1=4^{2-\sqrt{3}} \\ \iff x=\frac{4^{2-\sqrt{3}}-1}{3}\left(TMĐK\right) \\ Xét đáp án C: \\ {\log }_3\left(4+\left|x\right|\right)=0,7 \\ \iff 4+\left|x\right|=3^{0,7} \\ \iff \left|x\right|=3^{0,7}-4(*) \\ Vì 3^{0,7}-4 <0 nên (*) VN \\ Xét đáp án D: \\ {5}^x={\log }_5\left(1-x^2\right). ĐK: 1-x^2>0\iff -1<x<1 \\ Ta dễ thấy PTVN \end{gathered}$$

Đáp án A

$$\begin{gathered} {\log }_{15}6=\frac{{\log }_{2}6}{{\log }_{2}15}=\frac{1+{\log }_{2}3}{{\log }_{2}3+{\log }_{2}5}(*) \\ \mathrm{Ta} \mathrm{có} \mathrm{b}={\log }_{5}6 \\ \iff \mathrm{b}={\log }_{5}3+{\log }_{5}2 \\ \iff \mathrm{b}=\frac{{\log }_{2}3}{{\log }_{2}5}+{\log }_{5}2 \\ \iff \mathrm{b}=\frac{\mathrm{a}}{{\log }_{2}5}+\frac{1}{{\log }_{2}5}=\frac{\mathrm{a}+1}{{\log }_{2}5} \\ \Rightarrow {\log }_{2}5=\frac{\mathrm{a}+1}{\mathrm{b}} \mathrm{thay} \mathrm{vào} (*) \\ \Rightarrow {\log }_{15}6=\frac{1+\mathrm{a}}{\mathrm{a}+{\displaystyle \frac{\mathrm{a}+1}{\mathrm{b}}}}=\frac{\mathrm{b}(1+\mathrm{a})}{\mathrm{ab}+\mathrm{a}+1}=\frac{\mathrm{b}(1+\mathrm{a})}{\mathrm{a}(\mathrm{b}+1)+1} \end{gathered}$$