Question

1. Determinati coordonatele punctului de extrem al functiei f:R->R f(x)=-x²+x+1
2. Rezolvati ecuatia $\frac{Pn}{2} = Pn-2$
3. Stiind ca sinA=$\frac{5}{13}$, calculati cos A, A ∈ (0, 90°)

Answer 1

$1.\text{Coeficientul lui a este negativ, deci functia admite punct de maxim.}\\\text{Deoarece avem de-a face cu o functie de gradul doi, punctul de maxim}\\\text{este chiar varful parabolei.}V(x_V,y_V)\\x_V=\dfrac{-1}{-2}=\dfrac{1}{2}\\\\y_V=\dfrac{-5}{-4}=\dfrac{5}{4}\\\Rightarrow \boxed{V\left(\dfrac{1}{2},\dfrac{5}{4}\right)}$

$2.\dfrac{P_n}{2}=P_{n-2}\\\dfrac{n!}{2}=(n-2)!\\n!=2(n-2)!\\(n-2)!\cdot(n-1)\cdot n=2(n-2)! |:(n-2)!\neq 0\\(n-1)\cdot n=2\\n^2-n-2=0\\\Delta=1+8=9\Rightarrow \sqrt{\Delta}=3\\n_1=\dfrac{1+3}{2}=\dfrac{4}{2}=2\\n_2=\dfrac{1-3}{2}=\dfrac{-2}{2}=-1\notin\mathbb{N}\\S:n=2$

$3.\sin^2A+\cos^2A=1\\\cos^2A=1-\sin^2A\\\cos^2A=1-\dfrac{25}{169}\\\cos^2A=\dfrac{144}{169}\\\\\cos A=\pm\dfrac{12}{13}\\\\A\in(0\textdegree,90\textdegree)\Rightarrow \cos A=\dfrac{12}{13}$