Question

Va rog....URGENT zi repede

Answer 1

$\displaystyle \mathtt{Subiectul~1}\\\\\mathtt{1.~~~(2-3i)(2+3i)=4-9i^2=4-9\cdot(-1)=4+9=13}\\\\ \mathtt{2.~~~f(f(3))=?,~f:\mathbb{R}\rightarrow\mathbb{R},~f(x)=2x-1}\\\\ \mathtt{f(3)=2\cdot3-1=6-1=5}\\\\\mathtt{f(f(3))=2\cdot5-1=10-1=9}$

$\displaystyle \mathtt{3.~~~log_3\left(x^2+17\right)=log_381~~~~~~~~~~~~~~~~~~~~~C.E.~x^2+17\ \textgreater \ 0}\\\\\mathtt{x^2+17=81\Rightarrow x^2=81-17\Rightarrow x^2=64\Rightarrow x=8}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=-8}\\\\\mathtt{x=8~:~8^2+17\ \textgreater \ 0~A\Rightarrow x=8~este~solu\c{t}ie~a~ecua\c{t}iei}\\\\\mathtt{x=-8~:~(-8)^2+17\ \textgreater \ 0~A\Rightarrow x=-8~este~solu\c{t}ie~a~ecua\c{t}iei}\\\\\mathtt{S=\{-8;8\}}$

$\displaystyle \mathtt{4.~~~Cazuri~favorabile:10;~15;~20;~25;~30;...;95\Rightarrow18~cazuri}\\\\\mathtt{Cazuri~posibile:99-10+1=90}\\\\\mathtt{P= \frac{nr.~cazuri~favorabile}{nr.~cazuri~posibile}= \frac{18}{90}=\frac{1}{5} }$

$\displaystyle \mathtt{5.~~~A(1,a);~B(3,2);~C(2,1)~~~~~~~~~~~~~~~~~~~~~~~~A,~B,~C~-~coliniare}\\\\ \mathtt{Punctele~A(1,a);~B(3,2);~C(2,1)~sunt~coliniare\Leftrightarrow \left|\begin{array}{ccc}\mathtt1&\mathtt a&\mathtt1\\\mathtt3&\mathtt2&\mathtt1\\\mathtt2&\mathtt1&\mathtt1\end{array}\right|=0}\\\\\mathtt{1\cdot2\cdot1+1\cdot3\cdot1+a\cdot1\cdot2-1\cdot2\cdot2-a\cdot3\cdot1-1\cdot1\cdot1=0}\\\\\mathtt{2+3+2a-4-3a-1=0}\\ \\ \mathtt{-a=0}\\\\\mathtt{a=0}$

$\displaystyle \mathtt{6.~~~E(x)=sin \frac{x}{3}+cos \frac{x}{2} ~~~~~~~~~~~~~~~~~~~~~~~~~E\left( \frac{\pi}{2}\right)=\frac{1+\sqrt{2}}{2}}~\\\\\mathtt{E\left( \frac{\pi}{2}\right)=sin \frac{\frac{\pi}{2}}{3}+cos\frac{ \frac{\pi}{2} }{2}=sin\left(\frac{\pi}{2}\cdot \frac{1}{3}\right)+cos\left(\frac{\pi}{2}\cdot \frac{1}{2}\right)=}\\\\\mathtt{=sin \frac{\pi}{6}+cos\frac{\pi}{4}=\frac{1}{2}+\frac{\sqrt{2}}{2}=\frac{1+\sqrt{2} }{2}}$

$\displaystyle \mathtt{Subiectul~2}\\ \\ \mathtt{1.~~~A(a)=\left(\begin{array}{ccc}\mathtt1&\mathtt{2a}\\\mathtt{2a}&\mathtt4\end{array}\right)}\\\\ \mathtt{a)~A(1)+A(-1)=2A(0)}\\\\\mathtt{A(1)=\left(\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt2&\mathtt4\end{array}\right);~A(-1)=\left(\begin{array}{ccc}\mathtt1&\mathtt{-2}\\\mathtt{-2}&\mathtt4\end{array}\right);~A(0)=\left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt4\end{array}\right)}$

$\displaystyle \mathtt{A(1)+A(-1)=\left(\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt2&\mathtt4\end{array}\right)+\left(\begin{array}{ccc}\mathtt1&\mathtt{-2}\\\mathtt{-2}&\mathtt4\end{array}\right)=}\\\\ \mathtt{=\left(\begin{array}{ccc}\mathtt{1+1}&\mathtt{2+(-2)}\\\mathtt{2+(-2)}&\mathtt{4+4}\end{array}\right)=\left(\begin{array}{ccc}\mathtt2&\mathtt0\\\mathtt0&\mathtt8\end{array}\right)=2\cdot\left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt4\end{array}\right)=2A(0)}$

$\displaystyle \mathtt{b)~det(A(a))=0}\\\\\mathtt{\left|\begin{array}{ccc}\mathtt1&\mathtt{2a}\\\mathtt{2a}&\mathtt4\end{array}\right|=0\Rightarrow1\cdot4-2a\cdot2a=0\Rightarrow4-4a^2=0\Rightarrow -4a^2=-4\Rightarrow}\\\\\mathtt{\Rightarrow4a^2=4\Rightarrow a^2=1\Rightarrow a=1}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightrrow a=-1}$

$\displaystyle \mathtt{A(2)\cdot X=A(8)\Rightarrow X=(A(2))^{-1}A(8)}\\\\ \mathtt{A(2)=\left(\begin{array}{ccc}\mathtt1&\mathtt4\\\mathtt4&\mathtt4\end{array}\right);~A(8)=\left(\begin{array}{ccc}\mathtt1&\mathtt{16}\\\mathtt{16}&\mathtt4\end{array}\right)}\\\\ \mathtt{det(A(2))=\left|\begin{array}{ccc}\mathtt1&\mathtt4\\\mathtt4&\mathtt4\end{array}\right|=1\cdot4-4\cdot4=4-16=-12}$

$\displaystyle \mathtt{A(2)=\left(\begin{array}{ccc}\mathtt1&\mathtt4\\\mathtt4&\mathtt4\end{array}\right)\Rightarrow(A(2))^{tr}=\left(\begin{array}{ccc}\mathtt1&\mathtt4\\\mathtt4&\mathtt4\end{array}\right)} \\ \\ \mathtt{D_{11}=(-1)^{1+1}\cdot4=1\cdot4=4}\\\\\mathtt{D_{12}=(-1)^{1+2}\cdot4=(-1)\cdot1=-4}\\ \\ \mathtt{D_{21}=(-1)^{2+1}\cdot4=(-1)\cdot4=-4}\\ \\ \mathtt{D_{22}=(-1)^{2+2}\cdot1=1\cdot1=1}$

$\displaystyle \mathtt{(A(2))^*=\left(\begin{array}{ccc}\mathtt4&\mathtt{-4}\\\mathtt{-4}&\mathtt1\end{array}\right)}\\ \\ \mathtt{(A(2))^{-1}=-\frac{1}{12}\cdot \left(\begin{array}{ccc}\mathtt4&\mathtt{-4}\\\mathtt{-4}&\mathtt1\end{array}\right)=\left(\begin{array}{ccc}\mathtt{-\frac{1}{3}}&\mathtt{\frac{1}{3}}\\\mathtt{\frac{1}{3} }&\mathtt{-\frac{1}{12}}\end{array}\right)}$

$\displaystyle \mathtt{(A(2))^{-1}A(8)=\left(\begin{array}{ccc}\mathtt{- \frac{1}{3} }&\mathtt{ \frac{1}{3}}\\\mathtt{ \frac{1}{3} }&\mathtt{- \frac{1}{12} }\end{array}\right)\cdot\left(\begin{array}{ccc}\mathtt1&\mathtt{16}\\\mathtt{16}&\mathtt4\end{array}\right)=\left(\begin{array}{ccc}\mathtt5&{-4}\\\mathtt{-1}&\mathtt5\end{array}\right)} \\ \\ \mathtt{X=\left(\begin{array}{ccc}\mathtt5&{-4}\\\mathtt{-1}&\mathtt5\end{array}\right)}$