x + y + z = 1 x + 2y + mz = 2 x + 4y + m 2 z = k (a) We use Gausian elimination to transform the augmented matrix of the system

Transcripción

1 UC3M Matemáticas para la Economía Examen Final, 20/01/2015 SOLUCIONES 1 Se considera el siguiente sistema lineal dependiente de los parámetros k, m R x + y + z = 1 x + 2y + mz = 2 x + 4y + m 2 z = k (a) (8 points) Discutir el sistema según los valores de los parámetros k, m. (b) (2 points) Resolver el sistema para m = 1 y k = 7, si es posible. (a) We use Gausian elimination to transform the augmented matrix of the system m 2, 1 4 m 2 k into an equivalent echelon matrix. Adding the first row to the second one and subtracting it to the third one we get 1 m m 2 1 k 1 Subtracting the second row to the third row we get finally m m 2 m 2 k 4 The rank of the system s matrix A is 3 iff m 2 m 2 0. In this case the rank of (A B) is also 3. The solutions of m 2 m 2 = 0 are m = and m = 2. Cases: m and m 2. A unique solution exists; m = or m = 2 and k 4. The bottom row of A is null, but of (A b) is not. No solution exists in this case. m = or m = 2 and k = 4. The bottom row of A is null, as well as of (A b). The rank of A is 2 < 3, thus the system admits infinitely many solutions. (b) When m = 1 and k = 7 the system admits a unique solution. The equivalent echelon system is x + y + z = 1 3y + 2z = 3. 2z = 3 The solution is (x = 1/2, y = 2, z = 3/2).

2 Examen Final, 20/01/2015 Página 2 de 10 2 (a) (5 points) Calcular el determinante D = (b) (5 points) Se considera una matriz cuadrada A y su traspuesta, A t, que verifican ( ) AA t 4 3 =. 3 4 Calcular el determinante de A y estudiar si A admite inversa. (a) Multiplying the first row by 2, 3 and 4 and subtracting to the second, third and fourth rows respectively, the resulting determinant has 3 proportional rows, thus the determinant is = = (b) Note that by the properties of the determinant, AA t = A A t = A 2 = , thus A = ±5. Yes, A is invertible since A 0.

3 Examen Final, 20/01/2015 Página 3 de 10 3 Se considera la función f(x) = e x/(1 x). (a) (7 points) Encontrar el dominio y calcular las asíntotas de f. Estudiar la continuidad y los intervalos de crecimiento/decrecimiento de f. Esbozar la gráfica de f. (b) (3 points) Discutir si f admite inversa en el intervalo (1, ) y hallarla, si es posible. (a) The function is not defined at 1, hence it is not continuous at 1. The domain is R {1}. thus, x = 1 is a vertical asymptote. lim f(x) = x 1 e = 0, + lim f(x) = e = x 1 lim f(x) = lim ) = e, x x e1/(x lim f(x) = lim ) = e x x e1/(x thus, y = e is an horizontal asymptote both at x = and at x =. There are no slanted asymptotes. (b) We will use the derivative of f to study monotonicity. The function is not continuous at x = 1, so it is not differentiable at this point. However, it is differentiable at any other point. The derivative is f (x) = ( ) x e x/(1 x) 1 = 1 x (1 x) 2 ex/(1 x) > 0 for any x in the domain of f, thus f is strictly increasing in its domain. The inverse of f exists in its domain, so in particular in the interval (1, ). This is because any horizontal line intercepts the graph of f at a point at most, thus f is one-to-one. See the graph below. To find the inverse, we must solve e x/(1 x) = y for x 1. Taking logarithms in both sides we find x/(1 x) = ln y and solving we have x + x ln y = ln y or x(1 + ln y) = ln y, that is, x = ln y/(1 + ln y). Hence f (y) = ln y 1 + ln y.

4 Examen Final, 20/01/2015 Página 4 de 10 4 (a) (6 points) Se considera la función f(x) = Tiene f algún punto de inflexión? 2x2. Estudiar los intervalos en los que f es cóncava o convexa. x 1 (b) (4 points) Enunciar el Teorema de Weierstrass. Se considera la función g(x) = x 4 2x 2. Encontrar los extremos locales y globales de g en el intervalo I = [0, 2]. (a) The domain of f is R {1}. f (x) = 2x2 4x (x 1) 2, f (x) = 4 (x 1) 3. Note that f < 0 in (, 1) and f > 0 in (1, ), thus f is concave in (, 1) and convex in (1, ). However, there is no inflection point, as 1 is not in the domain of f. (b) Theorem of Weierstrass: if f is continuous on a bounded and closed interval, then f has global maximum and global minimum in the interval. g (x) = 4x 3 4x, g (x) = 12x 2 4. g is null at x = 0 and x = ±1, it is negative in (, ) (0, 1) and positive in (, 0) (1, ). Hence 0 is a local maximum and 1 is a local minimum, which are points that belong to the interval I, hence they are also local extremum of f in I. The Theorem of Weierstrass guarantees that f attains global extremum in I. To find them, we evaluate the function at the candidates: f(0) = 0, f(1) =, f(2) = 8. We conclude that x = 1 is the global minimum of f in I and that x = 2 is the global maximum. Hence, x = 0 remains a local maximum of f in I.

5 Examen Final, 20/01/2015 Página 5 de 10 5 (a) (5 points) Dados a, b, c R, c 0, calcular (b) (5 points) Calcular (Sugerencia: Cambio de variable). e ax e bx lim x 0 tan cx. (e x 1) 1/2 e 2x dx. (a) The limit is an indetermination 0/0. Applying L Hôspital rule we get lim x 0 ae ax be ( ) bx = a b. c c cos 2 cx (b) Let t = e x 1, so dt = e x dx. Then the integral is transformed into e 2x (e x 1) 1/2 dx = (e x 1) 1/2 e x e x dx = (t 1/2 + t 3/2 )dt where C is an arbitrary constant. = 2 3 t3/ t5/2 + C = 2 3 (ex 1) 3/ (ex 1) 5/2 + C,

6 Examen Final, 20/01/2015 Página 6 de 10 6 (a) (5 points) Encontrar el área de la región delimitada por la gráfica de la función f(x) = x x 3 y el eje horizontal, entre x = y x = 1 2. (b) (5 points) Calcular la recta tangente a la función g(x) = cos x sen x en el punto x = π/4. (Observación: sen (π/4) = cos (π/4) = 2/2). (a) Function f meets the horizontal axis at, 0 and 1. We have to check the sign of f in (, 0) and (0, 1/2), since the area is defined as A = f(x) dx. The function is negative in the first interval and positive in the second one, thus A = 0 f(x)dx+ 0 f(x)dx = Note that the defined integral is not an area! f(x)dx = (b) The equation of the tangent line is 0 x 3 xdx+ x x 3 dx = x2 2 x /2 x x 3 dx = x4 4 x2 2 y g(π/4) = g (π/4)(x π/4). 0 + x2 2 x4 4 1/2 0 = ( ) = = 9 64 < 0 Note g(π/4) = 1/2 and g (x) = cos 2 x sen 2 x, thus g (π/4) = 0 and y = 1/2 is the tangent line. = 1 4 +( ) =