2014-2-MELAKA-SMK GajahBerang_MATHS QA

Section A [45 marks] Answer all questions in this section.

1. Find the limits of the following, if it exists :

(a) lim𝑥→3

𝑥2−2𝑥−3

|𝑥−3| [4 marks]

(b) lim𝑥→1

(𝑥+2)(√𝑥−1)

𝑥−1 [3 marks]

2. (a) Find the gradient of the normal to the curve y = ln 𝑥

𝑥2 at the point where x = e.

[3 marks]

(b) Evaluate ∫ 𝑒𝑥 sin 3𝑥 𝑑𝑥𝜋

60

, giving your answer in the exact form. [5 marks]

3. y y = √𝑥

P

y = 1

𝑥

0 x

The diagram above shows the curves y = √𝑥 and y = 1

𝑥 intersecting at the point P.

Find the coordinates of P. [2 marks] Calculate the area of the region bounded by the two curves, the y-axis and the line y = 2. [4 marks] Determine the volume of the solid generated when the region bounded by the curve

y = √𝑥, the x-axis and the line x = 1 is revolved 3600 about the x-axis. [3 marks]

4. Given that x<4, find ∫8

(4−𝑥)(8−𝑥)𝑑𝑥. [4 marks]

A chemical reaction takes place in a solution containing a substance S. At the beginning,

there are 2 grams of S in the solution and t hours later, there are x grams of S.

The rate of the reaction is such that x satisfies the differential equation

8𝑑𝑥

𝑑𝑡= (4 − 𝑥)(8 − 𝑥).

Solve this equation, giving t in terms of x. Then find to the nearest minute, the time taken

at which there are 3 grams of S present. [3 marks]

5. It is given that , at any point on the graph of y = f(x), 𝑑𝑦

𝑑𝑥= √(1 + 𝑦3) ,

(i) Show that 𝑑2𝑦

𝑑𝑥2 = 3

2 𝑦2 [2 marks]

(ii) Find expressions for 𝑑3𝑦

𝑑𝑥3 and 𝑑4𝑦

𝑑𝑥4 in terms of y and 𝑑𝑦

𝑑𝑥 . [3 marks]

(iii) The graph of y = f(x) passes through the origin. Show that the first two terms in the

Maclaurin’s series for y is x + 1

8𝑥4 [2 marks]

(iv) Hence, evaluate lim𝑥→0

𝑦

sin 2𝑥.

(You may use the Maclaurin series for sin 2x without proof) [3 marks]

6. Given y=ln (1+x2) and ∫ 𝑦𝑑𝑥 = 1.4373,2

0 𝑦𝑛 = 1.6094 𝑎𝑛𝑑 ∑ 𝑦𝑖

𝑛−11 = 4.9446.

Find the value of n. [4 marks]

Section B [ 15 marks] Answer any one question in this section.

7. Given the curve y = 3𝑥−9

𝑥2−𝑥−2 , find the intervals of x where the curve is a decreasing function.

Find also the stationary points and determine their nature. [6 marks]

Write the equations of the asymptotes and hence sketch the graph of y = 3𝑥−9

𝑥2−𝑥−2

Sketch on a separate diagram the graph of y = 𝑥2−𝑥−2

3𝑥−9 [9 marks]

8. Show that the equation 𝑒𝑥 = 3 − 2𝑥 has only one real root. [3 marks]

Show that the real root is between 0 and 1. [2 marks]

Given that x0 =0.6 as an initial approximation, use the Newton-Raphson method to find

the root correct to four decimal places. [5 marks]

Hence state the value of x, to four decimal places, which satisfies the equation

a) x=3−𝑒𝑥

2 [1 marks]

b) x=ln(3-2x) [1 marks]

Sketch the graphs of y=x and y=ln(3-2x) on the same axes, state clearly the coordinates of Intersection point. [3 marks]

SMKGB 2014 T2 TRIAL MARKSCHEME

1. Find the limits of the following, if it exists :

(a) lim𝑥→3

𝑥2−2𝑥−3

|𝑥−3|

lim𝑥→3−

𝑥2−2𝑥−3

|𝑥−3|= lim

𝑥→3−

(𝑥−3)(𝑥+1)

−(𝑥−3)= −4

lim𝑥→3+

𝑥2−2𝑥−3

|𝑥−3|= lim

𝑥→3+

(𝑥−3)(𝑥+1)

(𝑥−3)= 4 either one √ M1A1 lim

𝑥→3−

𝑥2−2𝑥−3

−(𝑥−3)≠

lim𝑥→3+

(𝑥−3)(𝑥+1)

(𝑥−3) ∴ lim

𝑥→3𝑓(𝑥) does not exit M1A1

(b) lim𝑥→1

(𝑥+2)(√𝑥−1)

𝑥−1= lim

𝑥→1

(𝑥+2)(√𝑥−1)

(√𝑥 −1)(√𝑥+1) = lim

𝑥→1

(𝑥+2)

(√𝑥+1) M1 lim

𝑥→1−

(𝑥+2)

(√𝑥+1) =

lim𝑥→1+

(𝑥+2)

(√𝑥+1) =

3

2 M1A1

2. (a) 𝑑𝑦

𝑑𝑥=

𝑥2(1

𝑥)−(ln 𝑥)(2𝑥)

𝑥4 quotient rule M1

𝑑𝑦

𝑑𝑥=

1−2

𝑒3 = −1

𝑒3 substitute x = e M1Gradient of

normal = e3 A1

(b) ∫ 𝑒𝑥 sin 3𝑥 𝑑𝑥 = [𝑒𝑥 (− cos 3𝑥)

3] − ∫ 𝑒𝑥 (− cos 3𝑥)

3𝑑𝑥 by parts M1

∫ 𝑒𝑥 sin 3𝑥 𝑑𝑥 = [𝑒𝑥 (− cos 3𝑥)

3] +

1

3[𝑒𝑥 (sin 3𝑥)

3− ∫ 𝑒𝑥 (sin 3𝑥)

3𝑑𝑥] + c M1

10

9∫ 𝑒𝑥 sin 3𝑥 𝑑𝑥 = [𝑒𝑥 (− cos 3𝑥)

3] +

1

9𝑒𝑥 sin 3𝑥+ c A1

∫ 𝑒𝑥 sin 3𝑥 𝑑𝑥 =9

10[𝑒𝑥 sin 3𝑥

9− 𝑒𝑥 cos 3𝑥

3]

𝜋6⁄

0

𝜋6⁄

0 M1

= 1

10(𝑒

𝜋6⁄ − 0 − 0 + 3𝑒0)

= 1

10(𝑒

𝜋6⁄ + 3) A1

3. √𝑥 = 1/𝑥 simultaneous M1

x = 1, y = 1 P(1, 1) A1

Area = ∫ (√𝑦 )2𝑑𝑦 + ∫1

𝑦 𝑑𝑦

2

1

1

0 limits & add M1 M1

= [ 2

3𝑦

32⁄ ]

10

+ [ ln y] 21

A1

= 1/3 + ln 2 A1

Volume = 𝜋 ∫ 𝑥1

0 𝑑𝑥 M1

= 𝜋 [ 1

2𝑥2]

10

A1

= ½ π A1

4. ∫8

(4−𝑥)(8−𝑥)𝑑𝑥. 𝑙𝑒𝑡

8

(4−𝑥)(8−𝑥) ≡

𝐴

4−𝑥+

𝐵

8−𝑥 M1

𝐴 = 2 , 𝐵 = −2 A1

∫8

(4−𝑥)(8−𝑥)𝑑𝑥 = -2 ln (4-x) + 2ln (8-x) + c M1A1

t= -2 ln (4-x) + 2ln (8-x) + c

when x=2, t=0, c= -2ln3 M1A1

t= -2 ln (4-x) + 2ln (8-x) -2 ln 3

when x=3, t=2 ln 5

3 = 1.022 =61minutes A1

5. (i) 𝑑2𝑦

𝑑𝑥2 = 1

2(1 + 𝑦3)−

1

2 (3𝑦2)𝑑𝑦

𝑑𝑥 M1

= 3

2𝑦2 substsitute dy/dx A1

(ii) 𝑑3𝑦

𝑑𝑥3 = 3𝑦𝑑𝑦

𝑑𝑥 A1

𝑑4𝑦

𝑑𝑥4 = 3(𝑑𝑦

𝑑𝑥)2 + 3y

𝑑2𝑦

𝑑𝑥2 product rule M1

= 3(𝑑𝑦

𝑑𝑥)2 + 3y(

3

2𝑦2 )

= 3(𝑑𝑦

𝑑𝑥)2 +

9

2𝑦3 or 3 +

15

2𝑦3 A1

(iii) x =0 , y = 0

𝑑𝑦

𝑑𝑥 = 1,

𝑑2𝑦

𝑑𝑥2 = 0 ,

𝑑3𝑦

𝑑𝑥3= 0 , 𝑑4𝑦

𝑑𝑥4 = 3

y = 0 + x + 0x2 + 0 x3 + 3 (𝑥4

4! ) + … M1

= x + 1

8𝑥4 + … A1

(iv) lim𝑥→0

𝑦

sin 2𝑥 = lim

𝑥→0

𝑥+ 1

8𝑥4

2𝑥− (2𝑥)3

6 +

(2𝑥)

120

5+⋯

A1 for sin 2x

= lim𝑥→0

1+ 1

8𝑥3

2− 4𝑥2

3+

4𝑥4

15+⋯

M1

= ½ A1

6. x=0, y = ln (1+0) = 0 B1

Area = 1

2 .

2

𝑛 ((𝑦𝑜 + 𝑦𝑛) + 2( 𝑦1 + 𝑦2 + 𝑦3 + ⋯ + 𝑦𝑛 ) B1for

2

𝑛 ,

1

2

1

2 .

2

𝑛 ((0 + 1.6094) + 2(4.9446 ) = 1.4373 M1

n = 8 A1

7. 𝑑𝑦

𝑑𝑥=

3(𝑥2−𝑥−2)−(3𝑥−9)(2𝑥−1)

(𝑥2−𝑥−2)2 M1

= −3(𝑥−1)(𝑥−5)

(𝑥2−𝑥−2)2

For decreasing function, −3(𝑥−1)(𝑥−5)

(𝑥2−𝑥−2)2 < 0 M1

{ x : x < 1 or x > 5 } M1 A1

(1, 3) is min point A1

(5, 1

3) is a maximum point A1

Asymptotes are x = – 1, x = 2 and A1

y = 0 A1

x = – 1 y x = 2

centre shape D1

4.5 sides D1

(1, 3) (5, 1/3) asymptotes & points D1

All correct D1

0 2 3 x

Y x = 3

Asymptotes D1

(5, 3) shapes D1

All D1

(1, 1/3)

2/9

– 1 0 2 x

Y Y= 𝑒𝑥

3 1 intersection point, ∴ 1 real root B1

8.

D1D1

Y = 3 – 2x

0

1 2

1

x

f(0) =1-3+0 = -2

f(1) = e-3+2 = 1.718 f(x) changes sign, ∴ there is a root between 0 and 1. M1A1

substitute X0 = 0.6,

Using NR method xn+1 = x0 - 𝑓(𝑥)

𝑓′(𝑥) , f(x)= 𝑒𝑥 − 3 + 2𝑥, f’(x) = 𝑒𝑥 + 2, M1

X1 = 0.6 – 0,022119

3.82212 = 0.59421 M1A1

X2 = 0.59421 – 0,00002

3.81160 = 0.5942 M1

∴ the approximate root is 0.5942 ( 4dp) A1

(a) 0.5942 B1

b) 0.5942 B1

Y x = 3

2

ln3

D1D1

0 1 x

the intersection point is (0.5942, 0.5942) B1