QUIZ MATHS Tentukan nilai terkecil dari k bilangan bulat sehingga (k-10)^4026 >=(2013)^2013 Note: >.... Question from @46jekkoi

November 2019 1 76 Report

QUIZ MATHS

Tentukan nilai terkecil dari k bilangan bulat sehingga (k-10)^4026 >=(2013)^2013
Note: >= dibaca lebih besar atau sama dengan

MRizky099

$\displaystyle (k-10)^no numeric noise key 1088 \geq (2013)^no numeric noise key 1087 \\ \left((k-10)^2 \right)^no numeric noise key 1086 1085 (k^2-20k+100)^no numeric noise key 1084 1083 k^2-20k+100 2013 k^2-20k-1913 0 \\$

Pengubah Nol

$\displaystyle k^2-20k-1913 = 0 \to a=1; \: b=-20; \: c = -1913\\\\ \begin{aligned} k_{1,2} &= \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ &= \frac{-(-20)\pm \sqrt{(-20)^2-4(1)(-1913)}}{2(1)} \\ &= \frac{20\pm \sqrt{400+4\cdot 1913}}{2} \\ &= \frac{20\pm\sqrt{4(100+1913)}}{2} \\ &= \frac{20\pm2\sqrt{2013}}{2} \\ &= 10\pm\sqrt{2013} \end{aligned}\\ \begin{array}{lcl} k = 10-\sqrt{2013} & atau & k = 10+\sqrt{2013}\end{array}$

Uji titik

$\displaystyle \begin{aligned} \{k \leq 10-\sqrt{2013}\} &: k = -35 & \to (-35)^2-20(-35)-1913 &= (+) > 0 \\ \{10-\sqrt{2013} \leq k \leq 10+\sqrt{2013}\} &: k = 0 & \to (0)^2-20(0)-1913 &= (-) < 0 \\ \{k \geq 10+\sqrt{2013}\} &: k = 55 & \to (55)^2-20(55)-1913 &= (+) > 0\end{aligned}\\ \therefore HP_k = \{k \leq 10-\sqrt{2013} \bigvee k \geq 10+\sqrt{2013} \}$