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SPM Practice Question 13 – 15

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Question 13:
An arithmetic progression consists of 10 terms. The sum of the last 5 terms is 5 and the fourth term is 9. Find the sum of this progression.

Solution:




Question 14:
The sum of the first 6 terms of an arithmetic progression is 39 and the sum of the next 6 terms is –69. Find
(a) The first term and the common difference.
(b) The sum of all the terms from the 15th term to the 25th term.

Solution:






Question 15:
An arithmetic progression has 9 terms. The sum of the first four terms is 24 and the sum of all the odd number terms is 55. Find
(a) The first term and common difference,
(b) The seventh term.

Solution:


SPM Practice Question 4 – 6

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Question 4:
It is given that -7, h, k, 20,… are the first four terms of an arithmetic progression. Find the value of h and of k.

Solution:





Question 5:
51, 58, 65,... 191 are the first n terms of an arithmetic progression. Find the value of n.

Solution:





Question 6:
Find the number of the multiples of 8 between 100 and 300.

Solution:


The nth Term of Arithmetic Progression (Example 3 & 4)

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Example 3:
The volume of water in a tank is 75 litres on the first day. Subsequently, 15 litres of water is added to the tank everyday.
Calculate the volume, in litres, of water in the tank at the end of the 12th day.

Solution:
Volume of water on the first day = 75l
Volume of water on the second day = 75 + 15 = 90l
Volume of water on the third day = 90 + 15 = 105l
75, 90, 105, …..
AP, a = 75, d = 90 – 75 = 15
Volume of water on the 12th day,
T12 = a + 11d
T12 = 75 + 11 (15)
T12 = 240l



Example 4:
The first three terms of an arithmetic progression are 72, 65 and 58.
The nth term of this progression is negative.
Find the least value of n.


Solution:
72, 65, 58
AP, a = 72, d = 65 – 72 = –7
The nth term is negative,
Tn < 0
a + (n – 1) d < 0
72 + (n – 1) (–7) < 0
(n – 1) (–7) < –72
n – 1 > –72/ –7
n – 1 > 10.28
n > 11.28
n must be integer, n = 12, 13, 14, ….

Therefore, the least value of n = 12.

9.2.1 First Derivative for Polynomial Function (Examples)

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Example:
Find d y d x for each of the following functions:
(a) y = 12
(b) y = x4
(c) y = 3x
(d) y = 5x3
(e)  y = 1 x (f)  y = 2 x 4 (g)  y = 2 5 x 2 (h)  y = 3 x (i)  y = 4 x 3

Solution:
(a) 
y = 12
  d y d x = 0
   
(b) 
y = x4
  d y d x = 4x3

(c) 
y = 3x
  d y d x = 3

(d) 
y = 5x3
  d y d x = 3(5x2) = 15x2


(e)
y = 1 x = x − 1 d y d x = − x − 1 − 1 = − 1 x 2

(f)
y = 2 x 4 = 2 x − 4 d y d x = − 4 ( 2 x − 4 − 1 ) = − 8 x − 5 = − 8 x 5

(g)
y = 2 5 x 2 = 2 x − 2 5 d y d x = − 2 ( 2 x − 2 − 1 5 ) = − 4 x − 3 5 = − 4 5 x 3

(h)
y = 3 x = 3 ( x ) 1 2 d y d x = 1 2 ( 3 x 1 2 − 1 ) = 3 2 x − 1 2 = 3 2 x

(i)
y = 4 x 3 = 4 ( x 3 ) 1 2 = 4 x 3 2 d y d x = 3 2 ( 4 x 3 2 − 1 ) = 6 x 1 2 = 6 x

SPM Practice (Paper 1)

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Question 14:
Show that 6x−6−2k x 2 = x 2  has no real roots if k> 1 4 .

Solution:




Question 15:
The quadratic equation x 2 +px+q=0 has roots –2 and 6. Find
(a) the value of p and of q,
(b) the range of values of r for which the equation x 2 +px+q=r has no real roots.

Solution:




SPM Practice (Paper 1)

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Question 9:
The roots of the equation 6 x 2 +hx+1=0 are α and β, whereas 3α and 3β are the roots of the equation 2 x 2 −x+k=0 . Find the value of h and k.

Solution:
6 x 2 +hx+1=0 a=6, b=h, c=1 Roots=α,β sor: α+β=− b a α+β=− h 6 .........( 1 ) por: αβ= c a αβ= 1 6 .........( 2 ) 2 x 2 −x+k=0 a=2, b=−1, c=k Roots=3α, 3β sor: 3α+3β=− b a 3( α+β )=− ( −1 ) 2 α+β= 1 6 .........( 3 ) por: 3α( 3β )= c a 9αβ= k 2 k=18αβ.........( 4 ) Substitute ( 3 ) into ( 1 ) α+β=− h 6 1 6 =− h 6 h=−1 Substitute ( 2 ) into ( 4 ) k=18αβ k=18( 1 6 ) k=3




Question 10:
Find the range of values of p for which the equation 2 x 2 +5x+3−p=0 has two real distinct roots.

Solution:

SPM Practice (Paper 1)

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Question 3:
Solve the following quadratic equations by using quadratic formula. Give your answer in four significant figures.
(a) (x+1)(x−5)=15 (b)  x 2 +3x−2 x 2 −x−1 =3

Solution:








Question 4:
If the roots of 2 x 2 +4x−1=0 are α and β, find the equations whose roots are
(a)  α 2 , β 2 (b) α−β, β−α

Solution:







Question 5:
Write and simplify the equation whose roots are double the roots of 3 x 2 −5x−1=0 , without solving the given equation.

Solution:


5.5.1a Proving Trigonometric Identities Using Addition Formula And Double Angle Formulae (Part 2)

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Example 3:
(a) Given that sinP= 3 5  and sinQ= 5 13 ,  such that P is an acute angle and Q is an obtuse angle, without using tables or a calculator, find the value of cos (P + Q).

(b) Given that sinA=− 3 5  and sinB= 12 13 ,  such that A and B are angles in the third and fourth quadrants respectively, without using tables or a calculator, find the value of sin (A – B).

Solution:
(a)
sin P = 3 5 , cos P = 4 5 sin Q = 5 13 , cos Q = − 12 13 cos ( P + Q ) = cos A cos B − sin A sin B = ( 4 5 ) ( − 12 13 ) − ( 3 5 ) ( 5 13 ) = − 48 65 − 15 65 = − 63 65


(b)


sin A = − 3 5 , cos A = − 4 5 sin B = − 5 13 , cos B = 12 13 s i n ( A − B ) = s i n A cos B − c o s A sin B = ( − 3 5 ) ( 12 13 ) − ( − 4 5 ) ( − 5 13 ) = − 36 65 − 20 65 = − 56 65