2014 2 johor smktinggikluang maths qa

7/22/2019 2014 2 JOHOR SMKTinggiKluang Maths QA

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2014-2-JOHOR-SMKTinggiKluang_MATHS QA byWu Yee Peng

Section A [45 marks]

Answer all questions in this section.

1. a) If f(x) = +||||

, evaluate

,

and

. [5 marks]

b) Given that

ax2+ 2 , - x 4 1 , otherwise

If g is continuous, find the constant value of a. [2 marks]

2. A curve has parametric equationsx= sin t, y = sin (t + ), < < .

Find an equation of the tangent to the curve at the point where t= [6 marks]

3.

Figure shows a sketch of the curve with equation y = x3ln (x2+ 2), x0. The finite region Rshown shaded in the figure, is bounded by the curve, the xaxis and the linex= 2. Use the

substitution u=x

2

+ 2 to show that the area of Ris

2 ln

. Hence, find the exact

area ofR. [10 marks]

4. Find the integrating factor for the differential equation x 2 = +. Hence, find thegeneral solution for the differential equation. Given that y = 1 when x= 1, solve the differential

equation. [9 marks]

g(x) =


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Marking Schemes

PRAPENTAKSIRAN STPM PENGGAL 2 2014

954/2 Mathematics (T) (Paper 2)

Q.

No.

Scheme Marks

1

SECTION A

a) + +

=

=

= 3 =

, does not exist.

b)4 1

= 2

4 1 = a(-)2+ 20 =

+ 2

a = - 8

1(either + or+

seen)

1 (Correct ans)

1 (Correct ans)1

1

5 marks

1

1 2 marks

2x = sin t , y = sin (t+

)

=c os , =cos =

cos 6cos Whent=

,

=

x= sin =

, y = sin (

) =

y - =

)

y =x+

1 (Both correct)

1(His )1(

= seen, can be

implied)

1 (or (, ) seen, can be

implied)

1 (yy1= m(xx1) )

1 6 marks

3R = ln 2 u =x

2

+ 2 du= 2xdxwhenx= 0, u = 2 ; whenx= 2, u = 4

1

11(Corresponding values

of u)


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ln 2 = ln 2 = 2 ln =

2 ln

y = ln u = = 2 v = 2 =

2

12 2 ln

= 12 ln2 2

2 2 (

1)

= ln

2 2

= ln 2 2 =

ln4

24

24

222 22=

[ 0 4 22423]

=

221

1(xeliminated)

1

1(Correct term to be y

and)

1(integration by parts)

1

1

1 10 marks

4x

2 =

+

=

+

Integrating factor, = ||=

- = +

=

1

=

1

=12 | 1|

=

ln|

1| + Cx2

When y = 1,x= 1

1 = ln |12 + 1| + C (1)2

1(His )1

1(Correct ans)

1

1

1

1

1


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C = 1 - ln 2

= ln |x2 + 1| + (1 ln 2)x2 1 9 marks5 = 1 ! +

!

= 1 2! cos3! =1 +xcosx+

! +

! (showed)

= 1 ! +

! + =1+x(1 !

! )+

! 1

!

!

! 1

!

! +

= 1 +x-

= 1 +x+ +Interval of convergence : (,

1

1

1

1

1

5 marks

6h =

y = 2x,

x= 1, y = 2 x= 1.25, y = 2.3784

x= 1.50, y = 2.8284 x= 1.75, y = 3.3636

x= 2, y = 4

2

= 12 0.25[2 4 22.37842.82843.3636]= 2.893 3 .

2 =

=

=

= 2.885 (3 d.p)

1 (At least 3 correct)

1(Using trapezium rule)

1(correct ans with 3 d.p)

1

1

1(correct ans)

1(Curve with correct

shape)

1 (Correct ans with

correct reason)

8 marks

Trapezium rule over-estimatethe

exact area under the curve because

the curve y = 2xconcave upwards.

y=2x


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7

SECTION B

a) = 1 8 8

= 4ex1 8

=

y

= 4

y +

= 4ex

y + 2= 4ex (Showed)

b) i) y = (x2)2ex = (x2)2ex+ e x 2(x 2)(1)

=xex(x2)

=0,xex(x2) = 0

x =0, x= 2

x = 0, y = 4 ;x= 2, y = 0

(0,4) and (2, 0) are turning points.

=x2ex+ ex(2x)[2xex+ ex(2)]= ex(x22)

At (0,4), = -2 (0) (2,0) is minimum point.

ii) = e

x(x22)

= 0, ex(x22) = 0ex 0,x22 = 0

(x 2)(x 2) = 0x = 2 = 2 2

= 2xe

x

+ e

x

(x

2

2)

Whenx= 2, = 22 2 2 0

1

1

2 marks

1(Product rule)

1(His = 0

1(Both correct)

1(Find )

1(Both correct)

5 marks

1(His = 0)

1(Correct ans)

1(His and show 0


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Whenx= -2, = 22 2 2 0

Whenx =2, y = 1.411 ;x= -2, y = 2.834

Points of inflexion = (

2, 1.411) and (-

2, 2.834)

iii) (x2)2exk = 0

(x2)2ex= k

Given that there are three distinct real roots, the graph y = (x2)2exand

y = k intersect at three points.

Set of values of k is {k: 0 < k < 4, k R}

1(Both correct)

1(Correct shape)

1(Correct points)

1 (All correct)

7 marks

1 1 mark

8a) i) y = +i

= -1(1+sin 2x)

-2(2 cos 2x)

= +i (Showed) =

+i[i][ +i+i

= i+

+

+i Whenx= 0,

=

+++

= 8 (Showed)

Whenx= 0, y = 1 f(0) = 1 = -2 f0 = -2

= 8

f(0) = 8

By Maclaurin theorem,

+i = f(0) + f(0)x+

! +

1(Quotient rule)

1

1

1

1(Both f(0) = 1 and

f(0)= -2 correct)

y

(2, 2.834)2,1.411)

(2,0)

(0,4)

x


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=1 +! + ! +

= 12x+ 4x2+

ii) +i.. = 1 2 4 .. = [ ] .. = [0.1(0.1)2+

0.1)][(-0.1)(-0.1)2+

0.13]

= 0.2027 (4 d.p)

b) f(x) = ln x + x3f (x) = 1

Xn+1=Xn

= Xn-+

+

=+ +

+

=

+ (Showed)

Xn+1=

+ ,let X0= 1X1=

+

= 2

Similarly,X2= 2.205

X3= 2.208

X4 = 2.208

The root is 2.21 (2 d.p)

1

1

1

1

1

10 marks

1(Using Newton-Raphson

formula)

1

1 (Subs.X0=1 to the

formula)

1(Until X is consistent)

1(Correct ans)5 marks

2014 2 johor smktinggikluang maths qa

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