Short Questions (Question 7 & 8)


Question 7:
Diagram below shows a circle with centre O and radius 12 cm.

Given that A, B and C are points such that OA = AB and  ∠OAC = 90°, find
(a)   ∠BOC, in radians,
(b)  the area, in cm2, of the shaded region.   

Solution:
(a) For triangle OAC,
  cos  ∠AOC = 6/12 
  ÐAOC = 1.047 rad (change calculator to Rad mode)
  ÐBOC = 1.047 rad

(b) 
Area of the shaded region
= Area of BOC – Area of triangle AOC
½ (12)2 (1.047) – ½ (6) (12) sin 1.047 (change calculator to Rad mode)
= 75.38 – 31.17
= 44.21 cm2



Question 8:
Diagram below shows a sector QOR of a circle with centre O.

It is given that PS = 8 cm and QP = PO= OS = SR = 5 cm.
Find
(a) the length, in cm, of the arc QR,
(b) the area, in cm2, of the shaded region.

Solution:
(a) Length of arc QR = θ = 10 (1.75) = 17.5 cm

(b)
Area of the shaded region
= Area of sector QOR – Area of triangle POS
½ (10)2 (1.75) – ½ (5) (5) sin 1.75 (change calculator to Rad mode)
= 87.5 – 12.30
= 75.2 cm2

9.7 Second-Order Differentiation, Turning Points, Maximum and Minimum Points (Examples)


Example 1 (Maximum Value of Quadratic Function)
Given that y = 3x (4 – x), calculate
(a) the value of x when y is a maximum,
(b) the maximum value of y.

Solution:
(a)
y = 3 x ( 4 x ) y = 12 x 3 x 2 d y d x = 12 6 x When  y  is maximum,  d y d x = 0 0 = 12 6 x x = 2

(b) 
y = 12 x 3 x 2 When  x = 2 , y = 12 ( 2 ) 3 ( 2 ) 2 y = 12  

Example 2 (Determine the Turning Points and Second Derivative Test)
Find the coordinates of the turning points on the curve y = 2x3 + 3x2 – 12x + 7 and determine the nature of these turning points.

Solution:
y = 2 x 3 + 3 x 2 12 x + 7 d y d x = 6 x 2 + 6 x 12 At turning point,  d y d x = 0
6x2 + 6x – 12 = 0
x2 + x – 2 = 0
(x – 1) (x + 2) = 0
x = 1 or x = –2

When x = 1
y = 2(1)3 + 3(1)2 – 12(11) + 7
y = 0
(1, 0) is a turning point.

When x = –2
y = 2(–2)3 + 3(–2)2 – 12(–2) + 7
y = 27
(–2, 27) is a turning point.

d 2 y d x 2 = 12 x + 6 When  x = 1 , d 2 y d x 2 = 12 ( 1 ) + 6 = 18 > 0  (positive)

Hence, the turning point (1, 0) is a minimum point.

When  x = 2 , d 2 y d x 2 = 12 ( 2 ) + 6 = 18 < 0  (negative)

Hence, the turning point (–2, 27) is a maximum point.

SPM Practice Question 2


Question 2:
The third term and the sixth term of a geometric progression are 24 and 7 1 9 respectively. Find
(a) the first term and the common ratio,
(b) the sum of the first five terms,
(c) the sum of the first n terms with n is very big approaching rn ≈ 0.

Solution:
(a)
Given  T 3 =24  a r 2 =24 ...........( 1 ) Given  T 6 =7 1 9  a r 5 = 64 9  ...........( 2 ) ( 2 ) ( 1 ) : a r 5 a r 2 = 64 9 24    r 3 = 8 27    r= 2 3

Substitute r= 2 3  into ( 1 )    a ( 2 3 ) 2 =24 a( 4 9 )=24    a=24× 9 4  =54  the first term 54 and the common ratio is  2 3 .

(b)
S 5 = 54[ 1 ( 2 3 ) 5 ] 1 2 3    =54× 211 243 × 3 1    =140 2 3  sum of the first five term is 140 2 3 .

(c)
When 1<r<1 and n becomes  very big approaching  r n 0,   S n = a 1r    = 54  1   2 3      =162
Therefore, sum of the first n terms with n is very big approaching rn ≈ 0 is 162.

Short Questions (Question 7 & 8)


Question 7:
The masses of mangoes in a stall have a normal distribution with a mean of 200 g and a standard deviation of 30 g.
(a) Find the mass, in g, of a mango whose z-score is 0.5.
(b) If a mango is chosen at random, find the probability that the mango has a mass of at least 194 g.

Solution:
µ = 200 g
σ = 30 g
Let X be the mass of a mango.

(a)
X 200 30 = 0.5 X = 0.5 ( 30 ) + 200 X = 215 g

(b)
P ( X 194 ) = P ( Z 194 200 30 ) = P ( Z 0.2 ) = 1 P ( Z > 0.2 ) = 1 0.4207 = 0.5793



Question 8:
Diagram below shows a standard normal distribution graph.


The probability represented by the area of the shaded region is 0.3238.
(a) Find the value of k.
(b) X is a continuous random variable which is normally distributed with a mean of 80 and variance of 9.
Find the value of X when the z-score is k.

Solution:
(a)
P(Z > k) = 0.5 – 0.3238 
= 0.1762
k = 0.93

(b)
µ = 80,
σ2 = 9, σ = 3
X 80 3 = 0.93 X = 3 ( 0.93 ) + 80 X = 82.79

Short Questions (Question 5 & 6)


Question 5:
Diagram below shows the graph of a binomial distribution of X.

(a) the value of h,
(b) P (X ≥ 3)

Solution:
(a)
P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1
1 16 + 1 4 + h + 1 4 + 1 16 = 1 h = 1 5 8 h = 3 8

(b)
P (X ≥ 3) = P (X = 3) + P (X = 4)
P ( X 3 ) = 1 4 + 1 16 = 5 16



Question 6:
The random variable X represents a binomial distribution with 10 trails and the probability of success is ¼.
(a) the standard deviation of the distribution,
(b) the probability that at least one trial is success.

Solution:
(a)
n = 10, p = ¼
Standard deviation = n p q = 10 × 1 4 × 3 4 = 1.875

(b)
P ( X = r ) = C 10 r ( 1 4 ) r ( 3 4 ) 10 r P ( X 1 ) = 1 P ( X < 1 ) = 1 P ( X = 0 ) = 1 C 10 0 ( 1 4 ) 0 ( 3 4 ) 10 = 0.9437

SPM Practice 2 (Question 11 & 12)


Question 11 (3 marks):
Diagram 6 shows the graph of a straight line x 2 y  against  1 x .  

Diagram 11

Based on Diagram 6, express y in terms of x.


Solution:

m= 4( 5 ) 60 = 3 2 c=5 Y= x 2 y X= 1 x Y=mX+c x 2 y = 3 2 ( 1 x )+( 5 ) x 2 y = 3 2x 5 x 2 y = 310x 2x y x 2 = 2x 310x y= 2 x 3 310x



Question 12 (3 marks):
The variables x and y are related by the equation y=x+ r x 2 , where r is a constant. Diagram 8 shows a straight line graph obtained by plotting ( yx ) against  1 x 2 .

Diagram 12

Express h in terms of p and r.


Solution:

y=x+ r x 2 yx=r( 1 x 2 )+0 Y=mX+c m=r, c=0 m= y 2 y 1 x 2 x 1 r= 5p0 h 2 0 hr 2 =5p hr=10p h= 10p r

SPM Practice 2 (Linear Law) – Question 1


Question 1 (10 marks):
Use a graph to answer this question.
Table 1 shows the values of two variables, x and y, obtained from an experiment. A straight line will be obtained when a graph of y 2 x  against  1 x is plotted.


(a) Based on Table 1, construct a table for the values of 1 x  and  y 2 x .  

( b ) Plot  y 2 x  against  1 x , using a scale of 2 cm to 0.1 unit on the  1 x -axis   and 2cm to 2 units on the  y 2 x -axis.   Hence, draw the line of best fit.

(c) Using the graph in 1(b)
(i) find the value of y when x = 2.7,
(ii) express y in terms of x.


Solution:
(a)


(b)



(c)(i)
When x=2.7,  1 x =0.37 From graph, y 2 x =5.2 y 2 2.7 =5.2 y=3.75



(c)(ii)

Form graph, y-intercept, c = –4 gradient, m= 16( 4 ) 0.80 =25 Y=mX+c y 2 x =25( 1 x )4 y= 254x


SPM Practice 3 (Linear Law) – Question 6

Question 6
The table below shows the values of two variables, x and y, obtained from an experiment. The variables x and y are related by the equation a y = b x + 1 , where k and p are constants.


(a) Based on the table above, construct a table for the values of 1 x and 1 y . Plot 1 y against 1 x , using a scale of  2 cm to 0.1 unit on the 1 x - axis and  2 cm to 0.2 unit on the 1 y - axis. Hence, draw the line of best fit.
(b) Use the graph from  (b)  to find the value of
(i)  a,
(ii)  b.


Solution

Step 1 : Construct a table consisting X and Y.




Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit

Steps to draw line of best fit - Click here




Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph




Step 4 : Rewrite the original equation given and reduce it to linear form

Step 5 : Compare with the values of m and c obtained, find the values of the unknown required

SPM Practice 3 (Linear Law) – Question 5

Question 5
The following table shows the corresponding values of two variables, x and y, that are related by the equation y = p k x , where p and k are constants.


(a) Plot log 10 y against x  .  Hence, draw the line of best fit

(b) Use your graph in (a) to find the values of p and k.


Solution
Step 1 : Construct a table consisting X and Y.



Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit

For steps to draw line of best fit - Click here



Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph


Step 4 : Rewrite the original equation given and reduce it to linear form


Step 5
:
Compare with the values of m and c obtained, find the values of the unknown required

SPM Practice 2 (Linear Law) – Question 4

Question 4
The table below shows the corresponding values of two variables, x and y, that are related by the equation y = q x + p q x , where p and q are constants.


One of the values of y is incorrectly recorded.
(a) Using scale of 2 cm to 5 units on the both axis, plot the graph of xy against x 2  .  Hence, draw the line of best fit

(b) Use your graph in (a) to answer the following questions:
(i) State the values of y which is incorrectly recorded and determine its actual value.
(ii) Find the value of p and of q.

Solution
Step 1 : Construct a table consisting X and Y.


Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit


Steps to draw line of best fit - Click here

(b) (i) State the values of y which is incorrectly recorded and determine its actual value.


Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph

Step 4 : Rewrite the original equation given and reduce it to linear form

Step 5 : Compare with the values of m and c obtained, find the values of the unknown required