Solutionbank Edexcel AS and A Level Modular Mathematics

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Vernon Summers
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1 Page of Exercise A, Question The curve C, with equation y = x ln x, x > 0, has a stationary point P. Find, in terms of e, the coordinates of P. (7) y = x ln x, x > 0 Differentiate as a product: = x + x ln x = x + x ln x = x ( + ln x ) x = 0 + ln x = 0 as x > 0 ln x = ln x = x = e Substituting x = e, in y = x ln x y = ( e ) ln e = e So coordinates are ( e, e Pearson Education Ltd 008

2 Page of Exercise A, Question f ( x ) = e x, x 0 The curve C with equation y = f ( x ) meets the y-axis at P. The tangent to C at P crosses the x-axis at Q. (a) Find, to 3 decimal places, the area of triangle POQ, where O is the origin. (5) The line y = intersects C at the point R. (b) Find the exact value of the x-coordinate of R. (3) (a) C meets y-axis where x = 0 y = e Find gradient of curve at P. = e x At x = 0, = e Equation of tangent is y e = e x This meets x-axis at Q, where y = 0 Q, 0 Area of POQ = e = e = 0.09 (b) At R, y = = e x x = ln x = + ln x = ( + ln ) Pearson Education Ltd 008

3 Page of Exercise A, Question 3 3x f ( x ) =, x > x + x + 7 x (a) Show that f ( x ) = 3, x >. (5) (b) Find f ( x ). () (c) Write down the domain of f ( x ). () x (a), x > 3x x + x + 7 ( x + ) ( x ) 3x ( x ) ( x + 7 ) ( x + ) ( x ) 3x x 7 ( x + ) ( x ) 3 ( 3x 7 ) ( x + ) ( x + ) ( x ) 3x 7 x 3 ( x ) x (b) Let y = 3 x x x = x = x = 3 y 3 y 3 y

4 Page of x = + 3 y So f ( x ) = + or 3 x 7 y 3 y or 7 x 3 x (c) Domain of f ( x ) is the range of f(x). x > > 0 f ( x ) = 3 < 3 x So the domain of f ( x ) is x < 3 Pearson Education Ltd 008 x

5 Page of Exercise A, Question (a) Sketch, on the same set of axes, for x > 0, the graphs of y = + ln 3x and y = () The curves intersect at the point P whose x-coordinate is p. Show that (b) p satisfies the equation p ln 3p p = 0 () (c) < p < () The iterative formula x n + = e + x n, x 0 = 3 is used to find an approximation for p. (d) Write down the values of x, x, x 3 and x giving your answers to significant figures. (3) x (e) Prove that p =.66 correct to 3 significant figures. () (a) (b) At P, + ln 3p = p + p ln 3p = p ln 3p p = 0 (c) Let f ( p ) p ln 3p p f ( ) = ln 3 = f ( ) = ln 6 3 = Sign change implies root between and, so < p <. (d) x n + = e + x n, x 0 = 3 p

6 Page of x = e 3 x =.770 ( s.f.) x 3 =.59 ( s.f.) x =.697 ( s.f.) 3 =.9 ( s.f.) (e) f (.665 ) = f (.655 ) = root between.655 and.665 So p =.66 (3 s.f.) Pearson Education Ltd 008

7 Page of Exercise A, Question 5 The curve C has equation y = cos x sin x The curve C has equation y = sin x (a) Show that the x-coordinates of the points of intersection of C and C satisfy the equation cos x sin x = (3) (b) Express cos x sin x in the form R cos ( x +α), where R > 0 and π 0 <α<, giving the exact value of R and giving α in radians to 3 decimal places. () (c) Find the x-coordinates of the points of intersection of C and C in the interval 0 x <π, giving your answers in radians to decimal places. (5) (a) Where C and C meet cos x sin x = sin x Using cos x sin x sin x cos x So cos x + ( cos x ) = sin x cos x sin x = (b) Let cos x sin x R cos ( x +α) R cos x cos α R sin x sin α Compare: R cos α =, R sin α = Divide: tan α = α=0.6 (3 d.p.) Square and add: R ( cos α + sin α ) = + = 5 R = 5 So cos x sin x 5 cos ( x ) (c) cos x sin x =

8 Page of 5 cos ( x ) = cos x = 5 x =.07, x < 6.77 x = 0.63,.7 x = 0.3,.36 ( d.p.) Pearson Education Ltd 008

9 Page of Exercise A, Question 6 (a) Given that y = ln sec x, otherwise, to show that = tan x. (3) < x 0, use the substitution u = sec x, or The curve C has equation y = tan x + ln sec x, < x 0. At the point P on C, whose x-coordinate is p, the gradient is 3. (b) Show that tan p =. (6) π (c) Find the exact value of sec p, showing your working clearly. () (d) Find the y-coordinate of P, in the form a + k ln b, where a, k and b are rational numbers. () (a) y = ln sec x Let u = sec x so y = ln u = Using = (b) = sec x tan x = sec x tan x = sec x tan x = tan x = sec x + tan x So sec p + tan p = 3 + tan p + tan p = 3 As u tan p + tan p = 0 ( tan p ) ( tan p + ) = 0 π du So tan p = du du du u sec x < x 0 (th quadrant), tan p is negative π

10 Page of (c) sec p = + tan p = 5 sec p = + 5 (th quadrant) (d) y = ln 5+ ( ) = + ln 5 Pearson Education Ltd 008

11 Page of 3 Exercise A, Question 7 The diagram shows a sketch of part of the curve with equation y = f ( x ). The curve has no further turning points. On separate diagrams show a sketch of the curve with equation (a) y = f ( x ) (3) (b) y = f ( x ) (3) In each case show the coordinates of points in which the curve meets the coordinate axes. The function g is given by g : x x + k, x R, k > (c) Sketch the graph of g, showing, in terms of k, the y-coordinate of the point of intersection of the graph with the y-axis. (3) Find, in terms of k, (d) the range of g(x) () (e) gf(0) ()

12 Page of 3 (f) the solution of g ( x ) = x (3) (a) (b)

13 Page 3 of 3 (c) (d) g ( x ) k (e) gf ( 0 ) = g ( ) = k = k (f) y = x meets y = x + k where x = ( x + ) k x = ( + k ) x = + k Pearson Education Ltd 008

(a) Show that there is a root α of f (x) = 0 in the interval [1.2, 1.3]. (2)

. f() = 4 cosec 4 +, where is in radians. (a) Show that there is a root α of f () = 0 in the interval [.,.3]. Show that the equation f() = 0 can be written in the form = + sin 4 Use the iterative formula