Militar 2020 - Dominando

PROVA DE MATEMÁTICA (AMARELA) - ESCOLA NAVAL-2009/2010 

$\text { 1) Ao escrevermos } \quad \frac{x^{2}}{x^{4}+1}=\frac{A x+B}{a_{1} x^{2}+b_{1} x+c_{1}}+\frac{C x+D}{a_{2} x^{2}+b_{2} x+c_{2}}$
onde $a_{i}, b_{i}, c_{i}(1 \leq i \leq 2)$ e $A, B, C$ e $D$ säo constantes reais, podemos afirmar que $\quad A^{2}+C^{2}$ vale

(A) $\frac{3}{8}$
$(\mathrm{B}) \frac{1}{2}$
(c) $\frac{1}{4}$
(D) $\frac{1}{8}$
$(\mathrm{E}) 0$


RESPOSTA: C


Comentário

RESOLUÇÄO:


$\mathrm{x}^{4}+1=\mathrm{x}^{4}+2 \mathrm{x}^{2}+1-2 \mathrm{x}^{2}=\left(\mathrm{x}^{2}+1\right)^{2}-(\sqrt{2} \mathrm{x})^{2}=\left(\mathrm{x}^{2}+\sqrt{2} \mathrm{x}+1\right)\left(\mathrm{x}^{2}-\sqrt{2} \mathrm{x}+1\right)$

$\frac{x^{2}}{x^{4}+1}=\frac{A x+B}{x^{2}+\sqrt{2} x+1}+\frac{C x+D}{x^{2}-\sqrt{2} x+1}$

$\Leftrightarrow \mathrm{x}^{2}=(\mathrm{A}+\mathrm{C}) \mathrm{x}^{3}+(-\sqrt{2} \mathrm{A}+\mathrm{B}+\sqrt{2} \mathrm{C}+\mathrm{D}) \mathrm{x}^{2}+(\mathrm{A}+\mathrm{C}-\sqrt{2} \mathrm{B}+\sqrt{2 \mathrm{D}}) \mathrm{x}+(\mathrm{B}+\mathrm{D})$

$\Leftrightarrow\left\{\begin{array}{l}{\mathrm{A}+\mathrm{C}=0} \\ {-\sqrt{2} \mathrm{A}+\mathrm{B}+\sqrt{2} \mathrm{C}+\mathrm{D}=1 \Rightarrow-\sqrt{2} \mathrm{A}+\sqrt{2} \mathrm{C}=1 \Leftrightarrow \mathrm{C}-\mathrm{A}=\frac{\sqrt{2}}{2}} \\ {\mathrm{A}+\mathrm{C}-\sqrt{2} \mathrm{B}+\sqrt{2} \mathrm{D}=0} \\ {\mathrm{B}+\mathrm{D}=0}\end{array}\right.$


$\Rightarrow\left\{\begin{array}{l}{\mathrm{A}+\mathrm{C}=0} \\ {\mathrm{C}-\mathrm{A}=\frac{\sqrt{2}}{2}}\end{array} \Leftrightarrow \mathrm{C}=\frac{\sqrt{2}}{4}, \mathrm{A}=-\frac{\sqrt{2}}{4} \Rightarrow \mathrm{A}^{2}+\mathrm{C}^{2}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}\right.$



2) Sabendo que a equafäo $2 x=3 \sec \theta, \frac{\pi}{2}\langle\theta<\pi$ define implicitamente
$\theta$ como uma fungäo de $x$, considere a fungäo $f$ de variável real $x$ onde $f(x) \in \circ$ valor da expressäs $\frac{5}{2} \cos \sec \theta+\frac{2}{3} \operatorname{sen} 2 \theta \quad$ em termos de

$x .$ Qual o valor do produto $\left(x^{2} \sqrt{4 x^{2}-9}\right) f(x) ?$

(A) $5 x^{3}-4 x^{2}-9$
(B) $5 x^{3}+4 x^{2}-9$
(c) $-5 x^{3}-4 x^{2}+9$
(D) $5 x^{3}-4 x^{2}+9$
(E) $-5 x^{3}+4 x^{2}-9$

RESOLUÇÄO:
$2 \mathrm{x}=3 \sec \theta \Leftrightarrow \mathrm{x}=\frac{3}{2} \sec \theta$
$\mathrm{x}^{2} \sqrt{4 \mathrm{x}^{2}-9}=\left(\frac{3}{2} \sec \theta\right)^{2} \sqrt{4 \cdot\left(\frac{3}{2} \sec \theta\right)^{2}-9}=\frac{9}{4} \sec ^{2} \theta \sqrt{A \cdot \frac{9}{A} \sec ^{2} \theta-9}=\frac{9}{4} \sec ^{2} \theta \cdot 3 \sqrt{\sec ^{2} \theta-1}=$
$=\frac{27}{4} \sec ^{2} \theta \cdot \sqrt{\operatorname{tg}^{2} \theta}=\frac{27}{4} \sec ^{2} \theta \cdot|\operatorname{tg} \theta|$
$\frac{\pi}{2}<\theta<\pi \Rightarrow|\operatorname{tg} \theta|=-\operatorname{tg} \theta \Rightarrow x^{2} \sqrt{4 x^{2}-9}=-\frac{27}{4} \sec ^{2} \theta \cdot \operatorname{tg} \theta$
$\left(\mathrm{x}^{2} \sqrt{4 \mathrm{x}^{2}-9}\right) \cdot \mathrm{f}(\mathrm{x})=-\frac{27}{4} \sec ^{2} \theta \cdot \operatorname{tg} \theta \cdot\left(\frac{5}{2} \csc \theta+\frac{2}{3} \operatorname{sen} 2 \theta\right)=$
$=-\frac{135}{8} \sec ^{2} \theta \cdot \frac{\operatorname{sen} \theta}{\cos \theta} \cdot \frac{1}{\operatorname{sen} \theta}-\frac{9}{2} \sec ^{2} \theta \cdot \frac{\operatorname{sen} \theta}{\operatorname{ses} \theta} \cdot 2 \operatorname{sen} \theta \operatorname{ses} \theta=$
$=-\frac{135}{8} \sec ^{3} \theta-9 \sec ^{2} \theta \cdot\left(1-\cos ^{2} \theta\right)=-\frac{135}{8} \sec ^{3} \theta-9 \sec ^{2} \theta+9=$
$=-\frac{135}{8}\left(\frac{2}{3} x\right)^{3}-9\left(\frac{2}{3} x\right)^{2}-9=-5 x^{3}-4 x^{2}+9$


3) Sejam:

a) $f \quad$ uma funcäo real de variável $f(x)=\operatorname{arctg}\left(\frac{x^{3}}{3}-x\right), x>1$
b) $L$ a reta tangente ao gráfico da funcño $y=f^{-1}(x)$ no ponto $\left(0, f^{-1}(0)\right) .$ Quanto mede, em unidadades de área, a áxea do triângulo formado pela reta $L$ e os eixos coordenados?

$(\mathrm{A}) \frac{3}{2}$
(B) 3
$(\mathrm{C}) \quad 1$
(D) $\frac{2}{3}$

$(\mathrm{E}) \frac{4}{3}$

(A) $\frac{3}{2}$
$(B) 3$
(c) 1
(D) $\frac{2}{3}$
(E) $\frac{4}{3}$


RESPOSTA: B