Editorial - Matlan - Soal TKA 2

Problem 1

$ƒ\left(x\right)=m{x}^3+4x^2-nx+2$

$\mathrm{mod}(ƒ\left(x\right),x+2)=12$

$\mathrm{mod}(ƒ\left(x\right),x-1)=3$

$\mathrm{mod}(ƒ\left(x\right),2x^2-2x+1)=$?

$ƒ\left(1\right)=3=m+4-n+2$

$m-n=-3$

$ƒ(-2)=12=-8m+16+2n+2$

$8m-2n=6$

$4m-n=3$

$3m=6$

$m=2$

$-n=-3-2=-5$

$n=5$

$ƒ\left(x\right)=2x^3+4x^2-5x+2$

Divide $ƒ\left(x\right)$ by $2x^2-2x+1$

We find that the remainder is $-1$

Problem 5

$P=\left[\begin{array}{cc}3& -2\\ n+1& 3\end{array}\right]$

$Q=\left[\begin{array}{cc}1& m-2\\ 3& 4\end{array}\right]$

$R=\left[\begin{array}{cc}4& 5\\ 6& 7\end{array}\right]$

$P+Q=R$

$T=P^{\text{t}}+Q^2-R$

$\det \left(T\right)=$?

$n+1+3=6$

$n=2$

$-2+m-2=5$

$m=9$

$T=\left[\begin{array}{cc}& \\ & \end{array}\right]$

Problem 6

$A=\left[\begin{array}{cc}3& -1\\ 4& 2\end{array}\right]$

$B^{\text{t}}=\left[\begin{array}{cc}5& 4\\ 2& 2\end{array}\right]$

$\det \left({(A\cdot B)}^{-1}\right)=$?

$B=\left[\begin{array}{cc}5& 2\\ 4& 2\end{array}\right]$

$\det \left({(A\cdot B)}^{-1}\right)=\frac{1}{\det (A\cdot B)}$

$A\cdot B=\left[\begin{array}{cc}3& -1\\ 4& 2\end{array}\right]\left[\begin{array}{cc}5& 2\\ 4& 2\end{array}\right]=\left[\begin{array}{cc}11& 4\\ 28& 12\end{array}\right]$

$\det (A\cdot B)=\frac{1}{11\cdot 12-28\cdot 4}=\frac{1}{20}$

$\det \left({(A\cdot B)}^{-1}\right)=20$

Problem 9

$y=a{(x-4)}^2+10$

$-6=16a+10$

$16a=-16$

$a=-1$

$y={(X-4)}^2+10$

Scale by $2$, centered on origin

$\frac{1}{2}y=-{(\frac{x}{2}-4)}^2+10$

$y=-2{(\frac{x}{2}-4)}^2+20$

$y=-2\cdot \frac{x^2}{4}+8x-32+20$

$y=-12+8x+\frac{-x^2}{2}$

Problem 10

$ƒ\left(x\right)=a{b}^x+c$

$a+c=2$

$ab+c=6$

$a{b}^2+c=18$

Just by observation, $b=3$

$a+c=2$

$3a+c=6$

$2a=4$

$a=2$

$c=0$

$\therefore a+b+c=2+3+0=5$

Problem 11

$ƒ\left(x\right)=3^{x-2}$

Translated by $\left[\begin{array}{c}-2\\ 3\end{array}\right]$

$\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]+\left[\begin{array}{c}-2\\ 3\end{array}\right]$

Substitute.

$y-3=3^x$

$y=3^x+3=g\left(x\right)$

Rapidfire some subproblems.

$g\left(1\right)=6$ True

$g\left(0\right)=3$ False

$g(-1)$ $+g\left(2\right)=\frac{1}{3}+3+9+3\ne 15$ False

Subproblem 11.4

$g(ƒ\left(2\right))=12$

$g(ƒ\left(2\right))=g\left(1\right)=6$

The statement is false.

Subproblem 11.5

$ƒ(g\left(1\right))=81$

$g\left(1\right)=6$

$ƒ\left(6\right)=3^4=81$

The statement is true.

Problem 18

Subproblem 18.1

$\overrightarrow{\mathrm{DE}}=\overrightarrow{a}+\overrightarrow{c}$

$\overrightarrow{\mathrm{DE}}=\overrightarrow{\mathrm{DA}}+\overrightarrow{\mathrm{DH}}$

True.

Subproblem 18.2

$\overrightarrow{\mathrm{DE}}=2\sqrt{2}$

True, just by pythagoras.

Subproblem 18.3

$\overrightarrow{\mathrm{DB}}+\overrightarrow{\mathrm{BF}}=2\sqrt{3}$

$\overrightarrow{\mathrm{DB}}=(2,2,0)$

$\overrightarrow{\mathrm{BF}}=(0,0,2)$

$\overrightarrow{\mathrm{DB}}+\overrightarrow{\mathrm{BF}}=(2,2,2)=\sqrt{4+4+4}=2\sqrt{3}$

True!

Subproblem 18.4

$\overrightarrow{\mathrm{DC}}+\overrightarrow{\mathrm{BF}}+\overrightarrow{\mathrm{BA}}=\overrightarrow{c}-\overrightarrow{a}=\overrightarrow{\mathrm{DH}}-\overrightarrow{\mathrm{DA}}$

$\left[\begin{array}{c}0\\ 2\\ 0\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 2\end{array}\right]+\left[\begin{array}{c}-2\\ 0\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 2\end{array}\right]-\left[\begin{array}{c}\\ \\ \end{array}\right]$

IDK ABOUT VECTORS BRO

Problem 19

$\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{PC}}+\overrightarrow{\mathrm{CQ}}=\frac{\overrightarrow{\mathrm{CA}}}{2}+\frac{2\overrightarrow{\mathrm{CB}}}{3}$

$\frac{1}{2}\left[\begin{array}{c}-2\\ -8\\ 6\end{array}\right]+\frac{2}{3}\left[\begin{array}{c}-15\\ 3\\ 9\end{array}\right]=\left[\begin{array}{c}-11\\ -2\\ 9\end{array}\right]$

$\left|\overrightarrow{\mathrm{PQ}}\right|=\sqrt{121+4+81}=\sqrt{206}$