Last saved by: Kris-Anne Steele

SAT CLASS ANSWERS
CALCULUS 
1.	(a)	f ¢(x) = 3(2x + 5)^2 × 2	(M1)(A1)
Note: Award (M1) for an attempt to use the chain rule.
		 = 6(2x + 5)^2		(C2)
 
(b)	 + c	(A2)	(C2)
Note: Award (A1) for (2x + 5)^4 and (A1) for /8.
[4]

2.	(a)	y = 
	 (–4)	(A1)(A1)	(C2)
Note: Award (A1) for each element, to a maximum of [2 marks].
 
(b)	y = e^sin ^x
 = (cos x)(e^sin ^x)	(A1)(A1)	(C2)
Note: Award (A1) for each element.
[4]

3.	(a)	(x^2 + 1)^2
= 2(x^2 + 1) × (2x)	(M1)(M1)	(C2)
= 4x(x^2 + 1)
 
(b)	(ln(3x – 1))
= × (3)	(M1)(M1)	(C2)
= 
[4]

4.	f (1) = 1^2 – 3b + c + 2 = 0	(M1)
	f ¢(x) = 2x – 3b,
f ¢(3) = 6 – 3b = 0	(M1)
	3b = 6, b = 2	(A1)
	1 – 3(2) + c + 2 = 0, c = 3	(A1)
Note: In the event of no working shown, award (C2) for 1 correct answer.
[4]

5.	(a)	
	= 76.51 (or 76.5 to 3 sf)	(A1)	(C1)
 
(b)	 = f ¢(5)	(M1)
	= 3(5)^2	(A1)
	= 75	(A1)	(C3)
[4]

6.	Using 	(may be implied)	(M1)
	(A1)
= ln (k - 2) - ln1	(A1)(A1)
ln (k - 2) - ln1 = ln 7
	k - 2 = 7	(A1)
	k = 9		(A1)	(C6)
[6]

7.	f ¢(x) = 1 – x^2
	f (x) = 	(A1)
	f (3) = 0 Þ 3 – 9 + C = 0	(M1)
	Þ c = 6	(A1)
	f (x) = x –  + 6	(A1)
[4]

8.	f ¢(x) = –2x + 3
f (x) =  + 3x + c	(M1)
Notes: Award (M1) for an attempt to integrate. Do not penalize the omission of c here.
	1 = –1 + 3 + c	(A1)
c = –1	(A1)
f (x) = –x^2 + 3x – 1	(A1)	(C4)
[4]

9.	f ¢(x) = cos x Þ f (x) = sin x + C	(M1)
f   = –2 Þ –2= sin + C	(M1)
C = –3	(A1)
f (x) = sin x – 3	(A1)	(C4)
[4]

10.	y = dx	(M1)
=  – x + c	(A1)(A1)
Note: Award (A1) for first 3 terms, (A1) for “+ c”.
	13 =  + 4 – 2 + c	(M1)
c = 7	(A1)
y =  + x^2 – x – 7	(A1)	(C6)
[6]

11.	y = x^2 – x
 = 2x – 1 = gradient at any point.	(M1)
Line parallel to y = 5x
Þ 2x – 1 = 5	(M1)
x = 3	(A1)
y = 6	(A1)
Point (3, 6)	(C2)(C2)
[4]

12.	y = x^3 + 1
	 = 3x^2
	= Slope of tangent at any point
	Therefore at point where x = 1, slope = 3	(M1)
	Þ Slope of normal = –	(M1)(A1)
	Þ Equation of normal: y – 2 = –(x – 1)
	3y – 6 = –x + 1
	x + 3y – 7 = 0	(A1)	(C4)
Note: Accept equivalent forms eg y = –x + 2
[4]

13.	dx = 2	(M1)
Þ  = 2	(M1)
Þ ln a = 2	(A1)
Þ a = e^2	(A1)	(C4)
Note: If 7.39 given instead of e^2 then deduct [1 mark].
[4]

14.	(a)	(1 – x)dx	(A1)	(C1)
(b)	12dx
= 12	(M1)
= 12	(A1)
= 1	(A1)	(C3)
[4]

15.	(a)	(i)	f ¢(x) = –2e^–2x	(A1)
(ii)	f ¢(x) is always negative	(R1)	2
 
(b)	(i)	y = 1 +  (= 1 + e)	(A1)
(ii)	f ¢ (= –2e)	(A1)	2
Note: In part (b) the answers do not need to be simplified.
 
(c)	y – (1 + e) = –2e	(M1)
y = –2ex + 1 ( y = –5.44 x + 1)	(A1)(A1)	3
 
(d)

		(A1)(A1)(A1)
Notes: Award (A1) for each correct answer. Do not allow (ft) on an incorrect answer to part (i). The correct final diagram is shown below. Do not penalize if the horizontal asymptote is missing. Axes do not need to be labelled.

(i)(ii)(iii)

(iv)	Area =  (or equivalent)	(M1)(M1)
Notes: Award (M1) for the limits, (M1) for the function.
Accept difference of integrals as well as integral of difference. Area below line may be calculated geometrically.
	Area = 
	 = 	(A1)
	 = 0.1795 …= 0.180 (3 sf)	(A1)
	OR
	Area = 0.180	(G2)	7
[14]

16.	(a)
	(A1)(A1)	2
Note: Award (A1) for a second branch in approximately the correct position, and (A1) for the second branch having positive x and y intercepts. Asymptotes need not be drawn.
 
(b)	(i)	x-intercept = 	(A1)
y-intercept = 1 (Accept (0, 1), y = 1)	(A1)
 
(ii)	horizontal asymptote y = 2	(A1)
vertical asymptote x = 1	(A1)	4
 
(c)	(i)	f ¢(x) = 0 – (x – 1)^–2 	(A2)
(ii)	no maximum / minimum points.
since  ¹ 0	(R1)	3
 
(d)	(i)	2x + ln (x – 1) + c (accept lnïx – 1ï)	(A1)(A1)(A1)
(ii)	A = 	(M1)(A1)
Notes: Award (A1) for both correct limits.
Award (M0)(A0) for an incorrect function.

(iii)	A = 
	 = (8 + ln 3) – (4 + ln 1)	(M1)
	 = 4 + ln 3(= 5.10, to 3 sf)                                                    (A1)	(N2)	7
[16]

17.	(a)	y = ln x Þ 	(A1)
when x = e, 
tangent line: y = (x – e) + 1	(M1)
y = (x) – 1 + 1 = 	(A1)
x = 0 Þ y =  = 0	(M1)
(0, 0) is on line	(AG)	4
 
(b)	(x ln x – x) = (1) × ln x + x ×  – 1 = ln x	(M1)(A1)(AG)	2
Note: Award (M1) for applying the product rule, and (A1) for (1) × ln x + x × .
 
(c)	Area = area of triangle – area under curve	(M1)
= 	(A1)
= 	(A1)
=  – {(e ln e – 1 ln1) – (e – 1)}	(A1)
=  – {e – 0 – e + 1}
= e – 1.	(AG)	4
[10]

18.	(a)	(i)	  (or )	(A1)(A1)	(N1)(N1)
 
(ii)	
	(M1)

		(A1)	(N1)
 
(iii)	

	(A1)	(N1)	5
 
(b)	(i)		(A1)	(N1)
(ii)		(M1)
	(A1)(A1)	(N2)	4
 
(c)	(i)	when x = 4, gradient of tangent is 4 – 1 = 3 (may be implied)	(A1)
gradient of normal is 	(A1)
	(A1)	(N3)
 
(ii)	 (or sketch/graph)	(M1)

	 (may be implied) 	(A1)

	(A1)	(N2)	6
[15]

19.	(a)	(i)	f (x) = 
= 2 +  by division or otherwise	(M1)
Therefore as êx ê® ¥ f (x) ® 2	(A1)
Þ y = 2 is an asymptote	(AG)
	OR  = 2	(M1)(A1)
Þ y = 2 is an asymptote	(AG)
	OR make x the subject
yx – 3y = 2x + 1
x(y – 2) = 1 + 3y	(M1)
x = 	(A1)
Þ y = 2 is an asymptote	(AG)
Note: Accept inexact methods based on the ratio of the coefficients of x.
 
(ii)	Asymptote at x = 3	(A1)
(iii)	P(3, 2)	(A1)	4
 
(b)	f (x) = 0 Þ x = –	(M1)(A1)
x = 0 Þ f (x) = –	(M1)(A1)	4
Note: These do not have to be in coordinate form.
 
(c)
	(A4)	4
Note:	Asymptotes (A1)
	Intercepts (A1)
	“Shape” (A2).
 
(d)	f ¢(x) = 	(M1)
= 	(A1)
= Slope at any point
	Therefore slope when x = 4 is –7	(A1)
And f (4) = 9      ie S(4, 9)	(A1)
Þ Equation of tangent: y – 9 = –7(x – 4)	(M1)
	7x + y – 37 = 0	(A1)	6
 
(e)	at T,  = –7	(M1)
Þ (x – 3)^2 = 1	(A1)
x – 3 = ±l	(A1)
	(A1)(A1)	5
 
(f)	Midpoint [ST] = 
= (3, 2)
= point P	(A1)	1
[24]

20.	(a)	f (1) = 3	f (5) = 3	(A1)(A1)	2
 
(b)	EITHER	distance between successive maxima = period	(M1)
		 = 5 – 1	(A1)
		 = 4	(AG)
	OR	Period of sin kx = ;	(M1)
	so period = 	(A1)
	= 4	(AG)	2

(c)	EITHER A sin + B = 3 and A sin+ B = –1	(M1) (M1)
Û A + B = 3, – A + B = –1	(A1)(A1)
Û A = 2, B = 1	(AG)(A1)
OR Amplitude = A	(M1)
	A = 	(M1)
	A = 2	(AG)
Midpoint value = B	(M1)
	B = 	(M1)
	B = 1	(A1)	5
Note: As the values of A = 2 and B = 1 are likely to be quite obvious to a bright student, do not insist on too detailed a proof.
 
(d)	f (x) = 2 sin + 1
f ¢(x) = + 0	(M1)(A2)
Note: Award (M1) for the chain rule, (A1) for , (A1) for 2 cos.
	= p cos	(A1)	4
Notes: Since the result is given, make sure that reasoning is valid. In particular, the final (A1) is for simplifying the result of the chain rule calculation. If the preceding steps are not valid, this final mark should not be given. Beware of “fudged” results.
 
(e)	(i)	y = k – px is a tangent Þ –p = p cos	(M1)
	Þ –1 = cos	(A1)
	Þ x = p or 3p or ...
	Þ x = 2 or 6 ...	(A1)
	Since 0 £ x £ 5, we take x = 2, so the point is (2, 1)	(A1)

(ii)	Tangent line is: y = –p(x – 2) + 1	(M1)
	y = (2p + 1) – px
	k = 2p + 1	(A1)	6
 
(f)	f (x) = 2 Þ 2 sin + 1 = 2	(A1)
	Þ sin	(A1)
	Þ 
	x = 	(A1)(A1)(A1)	5
[24]

21.	(a)	p = 3	(A1)	(C1)
(b)	Area = 	(M1)
= 	(A1)
= 3 square units	(A1)	(C3)
[4]

22.	(a)		f ²(x)	= 2x – 2
Þ	f ¢(x)	= x^2 – 2x + c	(M1)(M1)
		= 0 when x = 3
Þ	0	= 9 – 6 + c
	c	= –3	(A1)
	f ¢(x)	= x^2 – 2x – 3	(AG)
	f (x)	=  – x^2 – 3x + d	(M1)
	When x = 3,	f (x)	= –7
Þ	–7	= 9 – 9 – 9 + d	(M1)
Þ	d	= 2	(A1)	6
Þ	f (x)	=  – x^2 – 3x + 2

(b)	f (0) = 2	(A1)
f (–1) = – – 1 + 3 + 2
= 3	(A1)
f ¢(–1) = 1 + 2 – 3
= 0	(A1)	3
 
(c)	f ¢(–1) = 0 Þ  is a stationary point
		(A4)	4
Note: Award (A1) for maximum, (A1) for (0, 2)
(A1) for (3, –7), (A1) for cubic.
[13]

23.	(a)	y = x(x – 4)^2
(i)	y = 0 Û x = 0 or x = 4	(A1)
(ii)	 = 1(x – 4)^2 + x × 2(x – 4) = (x – 4)(x – 4 + 2x)
	 = (x – 4)(3x – 4)	(A1)
 = 0 Þ x = 4 or x = 		(A1)
 is a maximum	(R1)
Note: A second derivative test may be used.
	x =  Þ y = 
	(A1)
Note: Proving that  is a maximum is not necessary to receive full credit of [4 marks] for this part.
 
(iii)	(3x^2 – 16x + 16) = 6x – 16	(A1)
 = 0 Û 6x – 16 = 0	(M1)
Û x = 	(A1)
x =  Þ y = 
	(A1)	9
Note: GDC use is likely to give the answer (1.33, 9.48). If this answer is given with no explanation, award (A2), If the answer is given with the explanation “used GDC” or equivalent, award full credit.

(b)
	(A3)	3
Note: Award (A1) for intercepts, (A1) for maximum and (A1) for point of inflexion.
 
(c)	(i)	See diagram above	(A1)
(ii)	0 < y < 10 for 0 £ x £ 4	(R1)
So 	(R1)	3
[15]

24.	(a)	y = e^x/2 at x = 0 y = e^0 = 1 P(0, 1)	(A1)(A1)	2
 
(b)	V = p	(A4)	4
Notes:	Award (A1) for p
(A1) for each limit
(A1) for (e^x/2)^2.
 
(c)	V = 	(A1)
= p	(A1)
= p[e^ln2 – e^0]	(A1)
= p[2 – 1] = p	(A1)(A1)
= p	(AG)	5
[11]

25.	Using V = 	(M1)
Correctly integrating 	A1
V = p	A1
	= 	(A1)
Setting up their equation 	M1
a^2 = 1.69
a = 1.3	A1	N2
[6]

26.	(a)	Evidence of using 	(M1)
eg 3e^3t ^- ^2
a(1) = 3e	(= 8.15)	A1	N2
 
(b)	Attempt to solve e^3t ^- ^2 = 22.3	(M1)
eg (3t - 2) (ln e) = ln 22.3, sketch
t = 1.70	A1	N2
 
(c)	Evidence of using s =  (limits not required)	M1

	A1	N1
[6]

Ÿ 827.	(a)	d = dt	(M1)(A1)(A1)	(C3)
Note: Award (M1) for ò, (A1) for both limits, (A1) for 4t + 5 – 5e^–t
 
(b)	d = 	(A1)(A1)
Note: Award (A1) for 2t^2 + 5t, (A1) for 5e^–t.
	   = (32 + 20 + 5e^–4) – (5)
   = 47 + 5e^–4 (47.1, 3sf )	(A1)	(C3)
[6]

28.	METHOD 1
l + 2w = 60	(M1)
l = 60 - 2w	(A1)
A = w(60 - 2w)	(= 60w - 2w^2)	(A1)
 = 60 - 4w		(A1)
Using  = 0	(60 - 4w = 0)	(M1)
w = 15	(A1)	(C6)
METHOD 2
w + 2l = 60	(A1)
w = 60 - 2l	(A1)
A = l(60 - 2l)	(= 60l - 2l^2)	(A1)
 = 60 - 4l		(A1)
Using  = 0	(60 - 4l = 0)	(M1)
l = 15
w = 30	(A1)	(C6)
[6]

29.	(a)	x = 1	(A1)	1
(b)	(i)	f (–1000) = 2.01	(A1)
(ii)	y = 2	(A1)	2
 
(c)	f ¢(x) = 	(A1)(A1)
	= 	(A1)
	= 	(AG)	3
Notes:	 Award (M1) for the correct use of the quotient rule, the first (A1) for the placement of the correct expressions into the quotient rule.
Award the second (A1) for doing sufficient simplification to make the given answer reasonably obvious.
 
(d)	f ′(3) = 0	Þ stationary (or turning) point	(R1)
f ²(3) =  > 0	Þ minimum	(R1)	2
 
(e)	Point of inflexion	Þ	f ²(x) = 0 Þ x = 4	(A1)
x = 4 Þ y = 0	Þ	Point of inflexion = (4, 0)	(A1)
	OR
	Point of inflexion = (4, 0)	(G2)	2
[10]

30.	(a)	(i)	p = (10x + 2) – (1 + e^2x)	A2	2
Note: Award (A1) for (l + e^2x) – (10x + 2).
(ii)	= 10 – 2e^2x	A1A1
	 = 0  (10 – 2e^2x = 0)	M1
	x =  (= 0.805)	A1	4

(b)	(i)	METHOD 1
	x = 1 + e^2x	M1
1n(x – 1) = 2y	A1
	f ^–1(x) = 	A1	3
	METHOD 2
	y – 1 = e^2x	A1
 = x	M1
f ^–1(x) = 	A1	3
 
(ii)	a = 	M1
	=  × 21n2	A1
= 1n 2	AG	2
 
(c)	Using V = 	(M1)
	Volume = 	A2	3
[14]

31.	(a)	From graph, period = 2π	(A1)	1
(b)	Range = {y |–0.4 < y < 0.4}	(A1)	1
 
(c)	(i)	f ¢(x) = {cos x (sin x)^2}
= cos x (2 sin x cos x) – sin x (sin x)^2 or –3 sin^3 x + 2 sin x	(M1)(A1)(A1)
Note: Award (M1) for using the product rule and (A1) for each part.

(ii)	f ¢(x) = 0	(M1)
Þ sin x{2 cos x – sin^2 x} = 0 or sin x{3 cos x – 1} = 0	(A1)
Þ 3 cos^2 x – 1 = 0
Þ cos x = ±	(A1)
At A, f (x) > 0, hence cos x = 	(R1)(AG)
(iii)	f (x) = 	(M1)
       = 	(A1)	9
 
(d)	x = 	(A1)	1
 
(e)	(i)		(M1)(A1)
 
(ii)	Area = 	(M1)
	= 	(A1)	4
 
(f)	At C f ²(x) = 0	(M1)
Û 9 cos^3 x – 7 cos x = 0
Û cos x(9 cos^2 x – 7) = 0	(M1)
Þ x =  (reject) or x = arccos = 0.491 (3 sf)	(A1)(A1)	4
[20]