Topic 6: Calculus Integration Markscheme 6.10 Area Under Curve Paper 2

Transcription

1 Topic 6: Calculus Integration Markscheme 6. Area Under Curve Paper. (a). N Standard Level (b) (i). N (ii).59 N (c) q p f ( ) = 9.96 N split into two regions, make the area below the -ais positive RR N [6]. (a) evidence of valid approach (M) e.g. f() =, graph a =.7, b =.7 ( a =, b = ) N (b) attempt to find ma (M) e.g. setting f () =, graph c =.5 (accept (.5,.)) N (c) attempt to substitute either limits or the function into formula M c e.g. V = [ ] [ ] f ( ), ln( ),.9... y V =.6 A N (d) valid approach recognizing regions (M) e.g. finding areas correct working e.g. f ( ) + f ( ); f ( ) f ( ) area =.7 (accept.6) A N []. (a) finding the limits =, = 5 integral epression e.g. 5 f ( ) area = 5. N

2 (b) evidence of using formula v = y correct epression 5 (M) e.g. volume = ( 5) volume = A N a (c) area is ( a ) a a = substituting limits (M) a a e.g. setting epression equal to area of R (M) correct equation a a e.g. = 5., a = 6 5., a = 6.79 N []. (a) (.) (accept (, ), (., )) N (b) (i) For using the product rule (M) f () = e cos + e sin = e (cos + sin ) N (ii) At B, f () = N (c) f () = e cos e sin + e sin + e cos = e cos AG N (d) (i) At A, f () = N (ii) Evidence of setting up their equation (may be seen in part (d)(i)) eg e cos =, cos = = ( =.57), y = e ( =.8) Coordinates are, e (.57,.8 ) N (e) (i) e sin or f ( ) A N (ii) Area =. A N [5]

3 5. (a) (i) p = N (ii) q = N (b) (i) f () = (M) = ( = ) = =, N (ii) Using V = b a y (limits not required) (M) V = A V =.5 N (c) (i) Evidence of appropriate method M eg Product or quotient rule Correct derivatives of and Correct substitution eg ( ) ( ) () ( ) f () = ( ) f () = ( + ) = ( ( + ) ) (ii) METHOD Evidence of using f () = at ma/min AG N (M) ( + ) = ( + = ) no (real) solution R Therefore, no maimum or minimum. AG N METHOD Evidence of using f () = at ma/min (M) Sketch of f () with good asymptotic behaviour Never crosses the -ais R Therefore, no maimum or minimum. AG N METHOD Evidence of using f () = at ma/min (M) Evidence of considering the sign of f () f () is an increasing function (f () >, always) R Therefore, no maimum or minimum. AG N

4 (d) For using integral (M) Area = a g ( ) a a + or f ( ) or ( ) Recognizing that a a g ( ) = f ( ) A Setting up equation (seen anywhere) (M) Correct equation eg a ( + ) a = a = =, a a [ ] =, a + a = a = N [] d 6. (a) cos N (b) Area of A = N (c) Evidence of attempting to find the area of B (M) eg y,. Evidence of recognising that area B is under the curve/integral is negative (M) eg y, cos, cos d Area of B =. accept Total Area = +. =. accept N [6]

5 7. (a) y (b) (i) N = (must be an equation) N (ii) f ( ) N (iii) Valid reason R N eg reference to area undefined or discontinuity Note: GDC reason not acceptable..5 (c) (i) V = f ( ) A N (ii) V = 5 (accept. ) A N (d) f () = e ( ) N (e) (i) =. (accept (., 7.9)) N (ii) p =, q = 7.9 (accept k < 7.9) N [7] 5

6 8. (a) (i) f () = + N (ii) For using the derivative to find the gradient of the tangent (M) f () = Using negative reciprocal to find the gradient of the normal M y = ( ) or y = + N (iii) Equating + + = + (or sketch of graph) M 8 = ( + )( ) = (accept, or =, = ) N = ( =.) (b) (i) Any completely correct epression (accept absence of ) A eg + +, + + N (ii) Area = (.5) 5 = (accept.) N (iii) Attempting to use the formula for the volume (M) eg, A N k k (c) f ( ) = + + Substituting Note: Award for, for, for. k + k + k + + (M) = k + k + k. 5 N [] 6

7 9. (a) METHOD Attempting to interchange and y (M) Correct epression = y 5 5 f ( ) = + N METHOD Attempting to solve for in terms of y (M) + 5 Correct epression = y 5 f ( ) = + N (b) For correct composition (g f) () = ( 5) + (g f) () = N + (c) = ( + = 9 9) = N 8 (d) (i) y = = N (ii) (Vertical asymptote) =, (Horizontal asymptote) y = N (Must be equations) (e) (i) + ln ( ) + C ( + ln + C) N (ii) [ + ln ( ) ] 5 (M) = (5 + ln ) (9 + ln) = 6 + ln N (f) Correct shading (see graph). N [8] 7

8 . (a) y P Q R (b) (i) Correctly finding derivative of + ie Correctly finding derivative of e ie e Evidence of using the product rule (M) f () = e + ( + )( e ) = ( )e AG N (ii) At Q, f () = (M) =.5, y = e.5 N Q is (.5, e.5 ) N (c) k < e.5 A N (d) Using f () = at the point of infleion M e ( + ) = This equation has only one root. R So f has only one point of infleion. AG N (e) At R, y = 7e (=.85...) 7e Gradient of (PR) is ( =.7) 7 e + Equation of (PR) is g () = + ( =.7 ) Evidence of appropriate method, involving subtraction of integrals or areas M Correct limits/endpoints eg ( ( ) g( ) ) f, area under curve area under PR e 7e Shaded area is ( + ) + =.59 N [] 8

9 . (a) (i) f () = + (ii) f () = (b) Gradient of tangent at y-intercept = f () = gradient of normal = (=.5) Finding y-intercept is.5 Therefore, equation of the normal is y.5 = ~( ) (y.5 =.5) M (y = (AG) (c) (i) EITHER solving = (M) = or = 5 y g () f () M Curves intersect at =, = 5 So solutions to f () = g () are =, = =.5( 5) = M = or = 5 (ii) Curve and normal intersect when = or = 5 (M) Other point is when = 5 y =.5(5) +.5 = (so other point (5, ) 5 5 (d) (i) Area = ( f ( ) g( )) or ( ) 5.5 Note: Award for the integral, for both correct limits on the integral, and for the difference. (ii) Area = Area under curve area under line (A = A A ) (M) = 5, A 5 = Area = 5 5 = 5 (or. (sf) [6] 9

10 . Note: There are many approaches possible. However, there must be some evidence of their method. Area = k sin (must be seen somewhere) Using area =.85 (must be seen somewhere) (M) EITHER Integrating cos = cos k + cos k Simplifying cos k +. 5 Equation cos k +. 5 =.85 (cos k =.7) Evidence of using trial and error on a GDC Eg sin =.5, too small etc (M) Using GDC and solver, starting with k sin.85 = (M) THEN k =.7 (A) (N) [6]. (a) y

11 (b) (i) -intercept = Accept,, = y-intercept = (Accept (, ), y = ) (ii) horizontal asymptote y = vertical asymptote = (c) (i) f () = ( ) = (A) ( ) (ii) no maimum / minimum points. since (R) ( ) (d) (i) + ln ( ) + c (accept ln ) f (M) (ii) A = ( ) Accept +, [ + ln ( ) ] (iii) A = [ + ln ( ) ] = (8 + ln ) ( + ln ) (M) = + ln (= 5., to sf) (N) 7 [6]. (a) (i) a = ( accept (, ) ) (ii) b = + ( accept ( +, ) ) (b) (i) h( )d h( ) (M). h( )d + h( ) (M). h( )d + h( ) (M). (ii) 5... ( ) = 5. (A) 5 (c) (i) y =.97 (ii). < k <.97 (A) []

12 5. (a) (i) cos =, sin = therefore cos + sin = (AG) (ii) cos + sin = + tan = tan = l = Note: Award (A) for.6. = (M) (G) (b) y = e (cos + sin ) dy = e (cos + sin ) + e ( sin + cos ) (M) = e cos (c) dy = for a turning point e cos = (M) cos = = a = y = e (cos + sin ) = e b = e d y (d) At D, = e cos e sin = e (cos sin ) = cos sin = = Note: Award (M)(A)(A) for a =.57, b =.8. y = e (cos + sin ) = e (M) (AG) 5 (e) Required area = (cos + sin ) (M) = 7.6 sq units (G) Αrea = 7.6 sq units (G) Note: Award (M)(G) for the answer 9.8 obtained if the calculator is in degree mode. [7]

13 6. (a) ( + sin ( + )) = cos ( + ) + c (C) (b) + sin ( + ) = (M) sin ( + ) = + =.98, +.98,... =.98,.8,... Required value of =.8 7. (a) (i) f () = e (C) (ii) f () is always negative (R) [6] (b) (i) y = + e (= + e) (ii) f = e (= e) (c) y ( + e) = e + (M) y = e + ( y = 5. + ) (d) (i) (ii) (iii) P (i)(ii)(iii) 8 y 6 P (iv) Area = [( + e ) ( e + )] (or equivalent) (M)(M) Area = [( e + e) = e + e =.795 =.8 ( sf) []

14 8. (a) At A, = => y = sin (e ) = sin () (M) => coordinates of A = (,.8) A(,.8) (G) (b) sin (e ) = => e = (M) => = ln (or k = ) = ln (or k = ) (A) (c) (i) Maimum value of sin function = (ii) dy = e cos (e ) Note: Award for cos (e ) and for e. (iii) d y = at a maimum (R) e cos (e ) = => e = (impossible) or cos (e ) = (M) => e = => = ln (AG) 6 (d) (i) Area = ln sin (e ) Note: Award for, for ln, for sin (e ). (ii) Integral =.9585 =.96 ( sf) (G) 5 (e) y p = (M) At P, = =.877 ( sf) (G) [8]

15 9. (a) (i) a = (ii) b = 5 (b) (i) f () = (A) (ii) = ( + 5)( ) = = 5 or = (M) = 5 or = (G) (iii) = f () = + ( ) + 5() (M) = =6 f () = 6 (G) 7 (c) (i) f () = 5 at = (M) Line through (, ) of gradient 5 y = 5 y = 5 (G) (ii) = 5 (M) + = ( ) = = = (G) (d) Area =5 ( sf) (G) Area = ( + + 5) = (M) 6 5 = 75 = 5 ( sf) [5] 5

16 . (a) y {.5< <.5< y< MAXIMUM POINT integers on ais { 5 LEFT INTERCEPT < <.5.5< <.< <.6 MINIMUM.< y < POINT { RIGHT INTERCEPT 5 (b) is a solution if and only if + cos =. (M) Now + cos = + ( ) = (c) By using appropriate calculator functions = (M) =.6967 (6sf) (d) See graph: ( + cos ) (e) EITHER ( + cos ) = (6 sf) (A) Note: This answer assumes appropriate use of a calculator eg fnint( Y, X,, ) = fnint : withy = + cos ( + cos ) = [ + sin + cos ] = ( ) + ( sin sin ) + (cos cos ) = + + = (6 sf) [5] 6

17 . (a) y = ( ) (i) y = = or = (ii) dy = ( ) + ( ) = ( )( + ) = ( )( ) dy = = or = dy = = ( )( ) = > is a maimum (R) dy = = ( )() = < Note: A second derivative test may be used = y = = = = , 7 d y d d = = ( 6 + 6) = 6 6 d y = 6 6 = (M) 8 = (iii) (( )( ) ) = y = = = = , 7 9 (b) y ma pt. pt. of infleion 5 intercepts (A) (c) (i) See diagram above (ii) < y < for (R) So < y < < y < (R) [5] 7