X/7 N A T I O N A L Q U A L I F I C A T I O N S 9 T H U R S D A Y, M A Y. P M. P M MATHEMATICS ADVANCED HIGHER Read carefully. Calculators may be used in this paper.. Candidates should answer all questions. 3. Full credit will be given only where the solution contains appropriate working. L I X / 7 6 / 9 *X/7*
Answer all the questions. Marks. (a) Given f(x) = (x + )(x ) 3, obtain the values of x for which f (x) =. (b) Calculate the gradient of the curve defined by x x y 5 at the point y + = (3, ). 3 ( ) t + 3t. Given the matrix A =. 3 5 (a) Find A in terms of t when A is non-singular. (b) Write down the value of t such that A is singular. ( ) 6 3 (c) Given that the transpose of A is, find t. 6 5 3 3. Given that y dy x e = and y = when x =, find y in terms of x.. Prove by induction that, for all positive integers n, n =. r( r + ) n + r = 5 5. Show that ln ln 3 e e x x + e e x x = 9 ln. 5 6. Express z = ( + i) in the form a + ib where a and b are real numbers. 7 i Show z on an Argand diagram and evaluate z and arg (z). 6 [X/7] Page two
7. Use the substitution x = sin θ to obtain the exact value of (Note that cos A = sin A.) x x. Marks 6 8. (a) Write down the binomial expansion of ( + x) 5. (b) Hence show that. 9 5 is. 599. 9. Use integration by parts to obtain the exact value of x tan x. 5. Use the Euclidean algorithm to obtain the greatest common divisor of 36 and 65, expressing it in the form 36a + 65b, where a and b are integers.. The curve y = x x + dy is defined for x >. Obtain the values of y and at the point where x =. 5. The first two terms of a geometric sequence are a = p and a = p. Obtain expressions for S n and S n in terms of p, where Given that S n = 65S n show that p n = 6. = aj. Given also that a 3 = p and that p >, obtain the exact value of p and hence the value of n. S k k j=,, 3. The function f(x) is defined by f( x) = x + x ( x ± ). x Obtain equations for the asymptotes of the graph of f(x). Show that f(x) is a strictly decreasing function. Find the coordinates of the points where the graph of f(x) crosses (i) the x-axis and (ii) the horizontal asymptote. Sketch the graph of f(x), showing clearly all relevant features. 3 3 [Turn over for Questions to 6 on Page four [X/7] Page three
. Express x + 6x in partial fractions. ( x + ) ( x ) Hence, or otherwise, obtain the first three non-zero terms in the Maclaurin expansion of x + 6x. ( x + ) ( x ) Marks 5 5. (a) Solve the differential equation dy ( x + ) 3 y = ( x + ) (b) given that y = 6 when x =, expressing the answer in the form y = f(x). Hence find the area enclosed by the graphs of y = f(x), y = ( x) and the x-axis. 6 6. (a) Use Gaussian elimination to solve the following system of equations (b) x + y z = 6 x 3y + z = 5x + y z =. Show that the line of intersection, L, of the planes x + y z = 6 and x 3y + z = has parametric equations 5 x = λ y = λ z = 5λ. (c) Find the acute angle between line L and the plane 5x + y z =. [END OF QUESTION PAPER] [X/7] Page four
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Solutions for AH maths 9. (a) f (x) = (x + ) (x ) 3 f (x) = (x ) 3 + 3 (x + ) (x ) = (x ) ((x ) + 3 (x + )) = (x ) (x + ) = when x = and when x =. (b) Method Note: a candidate may obtain Method Mathod 3 x y + x = y 5 x + xy = y 5y x + x dy + y = ydy 5dy 6 + 3 dy = dy 5dy 5 = dy dy = dy = x + y y x 5 x y + x = y 5 xy x dy y 6 9 dy + = dy + = dy 6 9 dy + = dy dy = x y ( + x = y 5 x x y ( + x dy y) + = dy and then substitute. y) + x = y 5 6 9 dy + = dy dy = dy xy + y Note: a candidate may obtain = y (in and 3) and then + x substitute. E E E E E
. (a) t + 3t det ( 3 5 ) = 5 (t + ) 9t = t A 5 3t = t ( 3 t + ), (b) t = t = 5 (c) t + 3 ( ) ( ) = 6 3 t = 3t 5 6 5 3. e y x dy = e y dy = x y = when x = so ey dy = x e y = x + c = + c c = e y = x y = ln ( x). When n =, LHS =, RHS. So true when = = = n =. Assume true for n = k, k. r = r (r + ) = k + Consider n = k + k + r = r (r + ) = k r = = = r (r + ) + (k + ) (k + ) k + + (k + ) (k + ) k + (k + ) (k + ) = k + (k + ) ((k + ) + ) = ((k + ) + ) Thus, if true for n = k, statement is true for n = k +, and, since true for n =, true for all n.
5. Method ln e x + e x ln 3 e x e x Let u = e x e x, then du = (e x + e x ). When x = ln 3, u = 3 3 = 6 5 and when x = ln, u = = 3. ln ln 3 e x + e x e x e = 3/ du x 5/6 u = [ln u] 3/ 5/6 = ln 3 ln 5 6 = ln 9 5 Method ln ln 3 e x + e x e x e x = [ln (ex e x )] ln ln 3, = ln ( ) ln ( 3 3) = ln 3 ln 5 6 = ln 9 5 6. ( + i) 7 i = + i 7 i 3 + i = 7 + i 7 i 7 + i ( 3 + i) (7 + i) = 5 = + i z = + = arg z = tan = tan ( ) = 3π (or 35 ).
7. x = sin θ = cos θ dθ x = θ = ; x = sin θ = θ = π x x = π/ = π/ sin θ ( cos θ) dθ sin θ sin θ cos θ = π/ ( sin θ) dθ ( cos θ) dθ = π/ ( cos θ) dθ = θ sin θ π/ = { π } = π 8. (a) ( + x) 5 = + 5x + x + x 3 + 5x + x 5 (b) Let x =, then 9 5 = ( + ( )) 5 = 5 + + 5 = 5 + 9 + 9 = 599 9. x tan x = tan x x = x tan x = x tan x x x, + x x 3 + x ln ( + ) x = tan ln ln = π 8 ln,
. 65 = 36 + 68 36 = 9 68 + 3 68 = 3 3 = 36 9 68 = 36 9 (65 36) = 36 9 65. When x =, y =. y = x x + ln y = ln (x x + ) = (x + ) ln x dy y = x + x + x ln x, Hence, when x =, y = and dy = 3 + = 3.. a j = p j S k = p + p + + p k = p (pk ) p S n = p (pn ) p S n = p (pn ) p p (p n ) p = 65p (pn ) p (p n + ) (p n ) = 65 (p n ) p n + = 65 p n = 6 a = p a 3 = p 3 but a 3 = p so p 3 = p p = p = since p >. p n = 6 = 6 = ( ) n =
3. f (x) = x + x x = x + x (x ) (x + ) Hence there are vertical asymptotes at x = and x =. f (x) = x + x x = + x x x = + x x as x. So y = is a horizontal asymptote. Hence f (x) f (x) = x + x x f (x) = (x + ) (x ) (x + x) x (x ) = x3 x + x x 3 x (x ) = (x + x + ) (x ) = ((x + ) + 3 ) < (x ) is a strictly decreasing function. f (x) = x + x x = x = or x = f (x) = x + x x = x + x = x x = for shape for position Alternatively for the horizontal asymptote: x x + x x x + f (x) = + x + x as x
. x + 6x (x + ) (x ) = A (x + ) + B x + + C x x + 6x = A(x ) + B(x + ) (x ) + C (x + ) Let x = then = 6A A =. Let x = then 6 + = 36C C =. Let x = then = A 8B + C = 8 8B + B =. Thus x + 6x (x + ) (x ) = (x + ) + x. Let f (x) = (x + ) + (x ) then f (x) = (x + ) + (x ) f () = = f (x) = (x + ) 3 (x ) f () = 6 = 6 9 f (x) = (x + ) + (x ) 3 f () = 3 3 = 3 M 3 Thus x + 6x (x + ) (x ) = 9x 6 + 3x 6 + E
5. (a) (x + ) dy 3y = (x + ) dy 3 x + y = (x + )3 Integrating factor: since 3 = 3 ln (x + ). x + Hence the integrating factor is (x + ) 3. dy (x + ) 3 3 (x + ) y = d ((x + ) 3 y) = y (x + ) 3 = = x + c y = 6 when x =, so = + c c =. Hence y = (x + ) (b) (x + ) = ( x) x + = x x = or x + = + x which has no solutions. y = f (x) y = ( x) Area = (x + ) + ( x) M = (x + ) = 5 [(x + )5 ] = 5 = 5
6. (a) x + y z = 6 x 3y + z = 6 3 5 5x + y z = 6 5 7 9 3 6 5 7 5 7,, z = 7 ( 7 5 ) = 5 5y = y = x + 5 = 6 x = 3 (b) Let x = λ. Method In first plane: x + y z = 6. λ + (λ ) (5λ ) = 5λ 5λ + 6 = 6. In the second plane: x 3y + z = λ 3(λ ) + (5λ ) = 5λ 5λ + =. Method Method y z = 6 λ y = 6 + z λ 3y + z = λ ( 8 3z + 3λ) + z = λ z = 5λ z = 5λ and y = λ x + y z = 6 () x 3y + z = () 5x z = () + 3 () x y = () + () y = x z = 5x x = λ, y = λ, z = 5λ (c) Direction of L is i + j + 5k, direction of normal to the plane is 5i + j k. Letting θ be the angle between these then 5 + 8 cos θ = M, 5 = 7 3 This gives a value of 3. which leads to the angle 3. 9 = 3..,