Câu 18: Đáp án C
Kẻ $APbot Delta Rightarrow Pleft( t+2;1-2t;2t right)Rightarrow overrightarrow{AP}=left( t+3;-2t;2t-6 right)$
Ta có $overrightarrow{{{u}_{Delta }}}=left( 1;-2;2 right),APbot Delta Leftrightarrow overrightarrow{AP}.overrightarrow{{{u}_{Delta }}}=0Leftrightarrow left( t+3 right)+4t+2left( 2t-6 right)=0Leftrightarrow t=1Rightarrow Pleft( 3;-1;2 right)$
Câu 19: Đáp án D
Kẻ $BPbot ACRightarrow BPbot left( SAC right)Rightarrow widehat{left( SB;left( SAC right) right)}=widehat{BSP}$
$BP=dfrac{AB.BC}{AC}=dfrac{aasqrt{3}}{2a}=dfrac{asqrt{3}}{2}$
$SB=sqrt{S{{A}^{2}}+A{{B}^{2}}}=asqrt{3}$
$Rightarrow sin widehat{BSP}=dfrac{BP}{SB}=dfrac{1}{2}Rightarrow widehat{BSP}=30{}^circ $
Câu 20: Đáp án A
Ta có ${y}’=dfrac{1}{3}{{left( {{x}^{2}}+x+1 right)}^{dfrac{1}{3}-1}}.left( 2x+1 right)=dfrac{2x+1}{3sqrt[3]{{{left( {{x}^{2}}+x+1 right)}^{2}}}}$
Câu 21: Đáp án B
Kẻ $SHbot ABRightarrow SHbot left( ABCD right)$.
Ta có $AD//BCRightarrow ADbot left( SBC right)$
$Rightarrow dleft( AD,SC right)=dleft( A;left( SBC right) right)=2dleft( H;left( SBC right) right)=2HP.$
Trong đó $HPbot text{S}B.$
Cạnh $SH=sqrt{S{{A}^{2}}-A{{H}^{2}}}=sqrt{S{{A}^{2}}-{{left( dfrac{AB}{2} right)}^{2}}}=2a$
$Rightarrow HP=dfrac{HS.HB}{SB}=dfrac{2a.a}{asqrt{5}}Rightarrow dleft( AD;SC right)=dfrac{4a}{sqrt{5}}$.
Câu 22: Đáp án B
Ta có $I=dfrac{1}{2}.left. dfrac{{{3}^{2text{x}+1}}}{ln 3} right|_{0}^{1}=dfrac{12}{ln 3}$.
Câu 23: Đáp án C
Ta có $y’ = 2left( {{x^2} – x} right)left( {2{rm{x}} – 1} right) < 0 Leftrightarrow left[ {begin{array}{*{20}{l}}
{x < 0}\
{frac{1}{2} < x < 1}
end{array}} right.$
Câu 24: Đáp án A
Ta có $y = x + frac{4}{{x + 1}} Rightarrow y’ = 1 – frac{4}{{{{left( {x + 1} right)}^2}}};,left{ begin{array}{l}
x in left( {0;2} right)\
y’ = 0
end{array} right. Leftrightarrow x = 1$
Tính $yleft( 0 right) = 4;,yleft( 2 right) = frac{{10}}{3};,yleft( 1 right) = 3 Rightarrow left{ begin{array}{l}
a = 3\
A = 4
end{array} right. Rightarrow a + A = 7.$
Câu 25: Đáp án A
Ta có $left{ {begin{array}{*{20}{l}}
{{z_1} + {z_2} = 6}\
{{z_1}{z_2} = 13}
end{array}} right. Rightarrow {z^2} – 6z + 13 = 0.$
Câu 26: Đáp án A
Ta có $Fleft( x right)=-int{ln left( x+3 right)dleft( dfrac{1}{x} right)=-dfrac{ln left( x+3 right)}{x}+int{dfrac{1}{x}dleft[ ln left( x+3 right) right]}}$
$=-dfrac{ln left( x+3 right)}{x}+int{dfrac{1}{x}.dfrac{1}{x+3}dtext{x}=-dfrac{ln left( x+3 right)}{x}+dfrac{1}{3}int{left( dfrac{1}{x}-dfrac{1}{x+3} right)dtext{x}}}$
$=-dfrac{ln left( x+3 right)}{x}+dfrac{1}{3}ln left| dfrac{x}{x+3} right|+C.$
Mà $Fleft( -2 right)+Fleft( 1 right)=0Rightarrow left( dfrac{1}{3}ln 2+{{C}_{1}} right)+left( -ln 4+dfrac{1}{3}ln dfrac{1}{4}+{{C}_{2}} right)=0Rightarrow -dfrac{7}{3}ln 2+{{C}_{1}}+{{C}_{2}}=0$
$Rightarrow Fleft( -1 right)+Fleft( 2 right)=left( ln 2+dfrac{1}{3}ln dfrac{1}{2}+{{C}_{1}} right)+left( -dfrac{1}{2}ln 5+dfrac{1}{3}ln dfrac{2}{5}+{{C}_{2}} right)=dfrac{10}{3}ln 2-dfrac{5}{6}ln 5.$
