web view2016 cchy mid-year exam 4e5n amaths 4047-2page 11 of 12. 2016 cchy 4e5n amaths p2 solution....

2016 CCHY 4E5N AMaths P2 Solution

S/N Solution Marks Remarks1(i)

tan =

8+5√24+3√2

=

8+5√24+3√2 x

4−3√24−3√2

=

32−24 √2+20√2−15 (2 )16−9 (2 )

=

2−4√2−2

= –1 + 2√2

M1

A1

1(ii) Method 1

sec2 = 1 + tan2

= 1 + (−1+2√2 )2

= 1 + 1 + 4(2) – 4√2 = 10 – 4√2

M1

A1

Seen or implied

Method 2AC2 = AB2 + BC2

= ( 4+3√2 )2 + (8+5√2 )2

= 16 + 9(2) + 24√2 + 64 + 25(2) + 80√2 = 148 +104√2

cos =

cos2 =

AB2

AC2

=

(4+3√2 )2

148+104√2

=

34+24√2148+104√2

=

2 (17+12√2 )2 (74+52√2 )

=

17+12√274+52√2

M1 M1 to find either AC2

or cos2

2016 CCHY Mid-Year Exam 4E5N AMaths 4047-2 of 12

sec2θ = ( 1cosθ )

2

=

74+52√217+12√2

=

74+52√217+12√2 x

17−12√217−12√2

=

74 (17 )−74 (12√2 )+52 (17√2 )−52 (12 ) (2 )172−144 (2 )

=

10−4√21

= 10 – 4√2A1

2 y = – x2 + 4x – 6

= – 2x + 4At x = 3, = – 2(3) + 4 = –2

y = – (3 )2 + 4(3) – 6 = –3

Equation of tangent : y = –2x + cAt ( 3, –3), –3 = –2(3) + c c = –3 + 6 = 3 equation of tangent is y = –2x + 3when x = 0, y = 3when y = 0, 2x = 3 x = 1 area of OAB = x x 3 units2

= 2 units2

A1

B1

M1

M1

A1

Recognise y-axis is x = 0 and x-axis is y = 0

A0 for omission of unit of measurement

3 f (x) = ax3+ bx2

+ 3x + 4

3 = a(1 )3 + b(1 )2 + 3(1) + 4 a + b = –4 ---------- (1)

6 = a(−1

2 )3

+ b(−1

2 )2

+ 3(−12 ) + 4

b = + ax4, b = 14 + a --------- (2)

M1

M1


subst (2) into (1), a + 14 + a = –4 a = –18 a = –18 x = –12subst a = –12 into (2), b = 14 + (–12) = 8Hence, a = –12 and b = 8

M1

M1

A1

[1] for both (1) & (2); accept equivalent form

solve a & b; accept elimination method

for both answers4(i) 4 x

– 2x+1

= 3

(22 )x – (2x )×2 – 3 = 0

(2x )2 – 2(2x ) – 3 = 0

M1A1

M1 : apply index rule correctly

Accept

22 x – 2(2x ) – 3 = 0

4(ii) (2x )2 – 2(2x ) – 3 = 0(2x−3 ) (2x+1 )=0 2

x= 3 or 2

x= – 1

(N.A.) x lg 2 = lg 3

x =

lg 3lg 2

= 158 (2dp)

M1

M1

A1

M0 if didn’t reject one of the answersor x ln 2 = ln 3

5(i) ∫ (10 e2 x+e−2 x ) dx = 10∫ e2 x

dx + ∫ e−2 x dx

= 10[ e2 x

2 ] +

[ e−2 x

−2 ] + c

= 5e2 x – e

−2 x + c

M1

A1

M1, apply correctly integration of exponential function; even if only 1 term is correct

A0 for omission of cAccept equivalent form

5(ii)∫

−k

k(10 e2 x+e−2 x )

dx = – 60

[5e2 x−12

e−2 x]−k

k

= – 60

5[e2 x ]−kk

– [e−2 x ]−kk

= – 60

5[e2 k−e−2 k ] – [e−2 k−e2 k ] = – 60

M1 substitution of limits


e2 k – e

−2 k = – 60

x2, 11e2 k– 11e−2 k

+ 120 = 0 (shown)

M1

A1

accept alternative method

6(i) sin (A – B) = sin A cos B – cos A sin B = – cos A sin B = cos A sin B = – =

M1

A1

M1 : apply formula correctly

A0 if not in simplest form

6(ii) sin ( A + B) = sin A cos B + cos A sin B = + =

M1

A1

M1 : apply formula correctly

6(iii) tan Atan B =

sin Acos A ÷

sin Bcos B

=

sin Acos A x

cos Bsin B

=

sin A cos Bcos A sin B

= ÷ = 2

M1

A1

M1 : change tan to sin and cos functions

Accept 25

7(a) 2cos2 x + 5 sin x + 1 = 0, 0 ≤ x ≤ 360

2( 1 – sin2 x ) + 5 sin x + 1 = 0

–2sin2 x + 5 sin x + 3 = 0( 2 sin x + 1)( – sin x + 3) = 0 sin x = – or sin x = 3 (N.A.)Basic angle, = 30x = 180 + 30, 360 – 30 = 210, 330

M1

B1

B1

A1

Apply correct formula

B0 if didn’t reject

B1 for

A1 for both answers

7(b) ( 1 + cot y ) tan y + 2 = 0, 0 ≤ y ≤ 2tan y + 1 + 2 = 0 tan y = –3basic angle, = 12490

M1

B1

Seen or implied


y = – 12490, 2 – 12490 ~ 189, 503

M1 Correct method to find the required ans; ISW (ignore subsequent working)

8ay = Abx

. ln y = ln A + x ln b

x 1 2 3 4 5 6y 36316 14765 5432 1999 813 299ln y 105 96 86 76 6

757

B1

P1 : correctly plotted the 6 points and joined the points by a straight line

8b(i) From the graph, ln A = 1145(accept range : 114 – 11.5)

A = e11⋅45

= 9390134605 ~ 90000 (1sf)

(accept :90 000, 100 000)

M1

A1

Correct reading from graph

8b(ii)

Gradient = –

4⋅54⋅7

(accept range : – 0975 to – 09375)

ln b = –

4⋅54⋅7

b = e−

4⋅54⋅7

= 038387 ~ 04 (1dp)

M1

A1

identify 2 points to calculate gradient; seen or implied

9a(i) ddx

(2 x−sin 2 x ) =

ddx

(2 x ) –

ddx

(sin 2 x )

= 2 – 2 cos 2x = 2 ( 1 – cos 2x ) = 2 (2 x ) = 4 x

M1

A1

differentiation of cosine function

Recall :cos 2x = 1 – 2 x

2 x = 1 – cos 2x

9a(ii)

x =

ddx

(2 x−sin 2 x )

∫π2

πsin2 x

dx

=

[ 2x−sin 2x ]π2

π M1


= [2π−sin 2 π−{2( π

2 )−2sin 2( π2 )}]

= [ 2π−0−π+0 ]

=

π4 A1

Accept 0785

9b(i) Method 1ddx

( x ln3 x ) = x

ddx

(ln 3 x ) + ln 3x

= x

ddx

(ln 3+ ln x ) + ln 3x

= x (1x ) + ln 3x

= 1 + ln 3x

M1

A1

Correct application of law of logarithm

Method 2ddx

( x ln3 x ) = x

ddx

(ln 3 x ) + ln 3x

= x(3

3 x ) + ln 3x = 1 + ln 3x

M1

A1

Correct application of differentiate rule for ln function

9b(ii)

ln 3x =

ddx

( x ln3 x ) – 1

∫1

e

ln3 x dx =

[ x ln3 x ]1

e

– ∫

1

e

x0

dx

= eln3e – ln 3 – [ x ]

1

e

= eln3e – ln 3 – e + 1 = 2887725532 ~ 289 (3sf)

M1

A1

10(a)

2(n3 )

= (n+1

3 )2 x

n (n−1 ) (n−2 )1×2×3 =

(n+1 ) n (n−1 )1×2×3

2 x (n – 2) = n + 12n – 4 = n + 1n = 5

M1

A1

10b(i) (2+ p )5


= (50 )25 p0

+ (51 )24 p1

+ (52 )23 p2

+ …

= 32 + 80p + 80 p2+ ….

A2 [–1] for each error, max 2 errors

10b(ii)(a) Let p = x – 2x2

(2+x−2 x2)5

= 32 + 80(x – 2x2) + 80(x – 2x2

)2 + …= –160x2

+ 80x2+ …

= –80x2+ …

coefficient of x2is –80

M1

A1

10b(ii)(b) 19995 = (2−0⋅001 )5

~ 32 + 80(–0001)+80(–0001)2

= 3192008 ~ 31920 (correct to 3 dp)

M1

A1

identify p = – 0001

11(i) x2 + y2

+ 2kx – 6ky = 10(10 – k2)

At (4, –2),42

+ (−2 )2 + 2k(4)– 6k(–2)= 10(10 – k2)

20 + 20k = 100 – 10k 2

10k 2 + 20k – 80 = 0

÷10, k2+ 2k – 8 = 0

( k + 4)(k – 2) = 0 k = – 4 or k = 2 (N.A.)Hence, k = 2

M1

M1

A1

subst correct x and y values into eqn to form an eqn in k only

A0 if didn’t reject –ve value

11(ii) x2 + y2

+ 4x – 12y = 60( x+2 )2 + ( y−6 )2 = 60 + 4 + 36 = 102

centre = (–2,6) radius = 10 units

M1

A1A1

accept alternative method

A0 for omission of units

11(iii) Let Q = (x,y)Centre of circle = mid-point of PQ

(–2,6) = (x+4

2, y+(−2 )

2 )x+4

2 = –2

y−22 = 6

x + 4 = –4 y – 2 = 12 x = –8 y = 14

M1

A1


Q = (–8, 14)

11(iv) Method 1At R(8,1),82

+ 12 + 4(8) – 12(1) = 85 > 60Hence, R(8,1) lies outside the circle

R1A1

Method 2Let M = centre of circle

MR = √ (8−(−2 ) )2+ (1−6 )2

= √125 ~ 111803Since MR > radius of circle, hence, R is outside the circle

R1

A1

12(i) V = base area x heightV = x

2h -------------- (1)

Surface area, A = x2

+ 4xh

x2

+ 4xh = 75 4xh = 75 – x

2

h = 75−x2

4 x ---------- (2) subst (2) into (1),

V = x2

x 75−x2

4 x

=

14

(75 x−x3) (shown)

M1

A1

12(ii)

V =

14

(75 x−x3)

= [75−3 x2 ]At stationary point, = 075 – 3x2 = 0 x2 = 25 x = 5 or x = –5 （N.A.）Hence, x = 5

M1

M1

A1

Seen or implied

12(iii)(a)x 49 5 51

+ve 0 -veM1

- No description required

- either indicate


slope

Hence, V has a maximum value A1

actual values or +ve / – ve sign

12(iii)(b)

V =

14

(75 x−x3)

=

14

(75×5−53 )

= 625Hence, volume of box = 625 cm3

M1

A1 A0 for omission of units of measurement

13a(i) 2x2 + 11x + 5 = 0

Sum of roots, + = – -------- (1)Product of roots, = -------- (2)(α +β )2= α

2+β2

+ 2

(−11

2 )2

= α2

+β2+ 2(

52 )

α2

+β2 = – 5

= 25

B1

M1

A1

for both (1) & (2)

13a(ii) (α−β )2 = α2

+β2– 2

= 25 – 2(52 )

= – = or – = – (N.A.) α 3

–β3= ( – )(α

2+β2

+ )

=(101

4+ 5

2 ) = = 124.875

M1

A1

M1

A1

Apply formula correctly

13b Sum of new roots :3+ + +3 = 4( + )

= 4(−112 )

= – 22B1


Tan Mui Mui, 03/06/16,

Product of new roots :

(3+ )( +3) = 3α 2+ 10 + 3β2

= 3(α2

+β2) + 10

= 3(101

4 ) + 10(52 )

= new equation is x

2 + 22x + = 0

Or 4x2 + 88x + 403 = 0

B1

A1

14(i) Let f(x) = 3x3– 14x2

– 7x + 10Try x – 1

f(1) = 3(1 )3 – 14(1 )2 – 7(1) + 10 = –8 x – 1 is not a factor

Try x + 1

f(-1) = 3(−1 )3 – 14(−1 )2 – 7(–1) + 10 = 0 x + 1 is a factor

Let 3x3– 14x2

– 7x + 10 = ( x + 1)(3x2

+ bx +10)= (10 + b)x + ( b + 3)x

2 + ….

Comparing x term, 10 + b = – 7 b = – 17

3x3– 14x2

– 7x + 10 = ( x + 1)(3x2

– 17x + 10)= ( x + 1)(x – 5)(3x – 2)

M1

M1

A2

M1 for attempting to find factor

Accept alternative method

[–1] for each error, max 2 errors

14(ii) Let ey= x

3x3– 14x2

– 7x + 10 = 0 has 2 positive roots, since e

y is always positive,

y has only 2 solutions

R1

A1

14(iii) 4−5x−8x2

3 x3−14 x−7 x+10 =

4−5 x−8 x2

( x+1 ) ( x−5 ) (3 x−2 )

Let

4−5 x−8 x2

( x+1 ) ( x−5 ) (3 x−2 )

=

Ax+1 +

Bx−5 +

C3 x−2 M1


3x -2 -2xx -5 -15x3x2 10 -17x

Tan Mui Mui, 03/06/16,

4 – 5x – 8x2

= A(x– 5)(3x – 2) + B(x+1)(3x–2) + C(x+1)(x–5)

when x = –1,

4 – 5(–1) – 8(−1 )2 = A(–1–5)( –3– 2) 1 = 30A A =

when x = 5,

4 – 5(5) – 8(5 )2 = B(5+1)(15 –2) –221 = 78B B = – = –when x = ,

4 – 5(23 ) – 8

( 23 )

2

= C( 23+1)( 2

3−5)

– = –C C = =

4−5x−8x2

3 x3−14 x−7 x+10

=

130 ( x+1 ) –

176 ( x−5 ) +

25 (3 x−2 )

M1

A2

Accept alternative method; award M1 even if only attempts to solve for only 1 unknown

[–1] for each error, max 2 errors

14(iv) ∫ 4−5x−8x2

3 x3−14 x2−7 x+10 dx

= ∫ 1

x+1 dx – ∫ 1

x−5 dx +

∫ 1

3 x−2 dx= ln (x+1) – ln (x-5) + xln (3x–2) + c= ln (x+1) – ln (x-5) +ln (3x–2) + c

A2 [–1] for each error, max 2 errors


web view2016 cchy mid-year exam 4e5n amaths 4047-2page 11 of 12. 2016 cchy 4e5n amaths p2 solution....

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