What kind of negative number does not show -ve sign? And, why is it important for you to recognise them when you see them - particularly in inequalities?
Now, you know that in an inequality, say,
2 < 3
when you x or ÷ both sides of the inequality by a -ve number, say, by -1,
it becomes: -2 (sign) -3, and
the inequality sign is the reverse of the earlier sign because:
-2 > -3
This is basic O-Maths:
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Thus, for SPM 2013 Modern Maths (1449/1) Paper 1 Q 24:
Q 24: Find the solution for (x - 6)/(-3) < 5
You can easily solve Q24 by multiplying both sides of the inequality by -3
and, you reverse the inequality sign to finally get the answer x > -9 (Answer C)
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Now, in Add. Maths:
You may come across inequality like this (in solving GP question):
n log 0.25 > log 0.0005
for you to find the greatest integer value of n.
To solve this Q, you must know that: log (0 < x < 1) = negative numbers, although they do not carry the -ve signs with them. Let me explain:
Thus, all Add-Maths students must know this:
log (0 < x < 1) = negative numbers
(even though the -ve sign is not shown)
Thus, log 0.00001, log 0.25, log (1/5), log 0.99999 (and the likes, where 0 < x < 1)
are all -ve numbers!
So, in the inequality: n log 0.25 > log 0.0005,
when you divide both sides of the inequality by log 0.25, to isolate and find n;
you have to reverse the > sign since log 0.25 is a negative number (-0.602059991),
Thus, n < log 0.0005/log 0.25
n < 5.482892142
Therefore, the greatest integer value of n = 5
(This skill is tested in Q9(f) of Edexcel 4PMO 2013 Jan Paper 2 on "Series" (pl see below). So, all A-Maths students - SPM and IGCSE Cambridge alike - beware! You may face a logarithmic inequality question of the above nature in time to come if you have not already been tested this way! By the way, there are also other negative numbers without the negative signs - can you name some of them? :))
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Edexcel PMO 2013 Jan Paper 2 Q9
9) The third and fifth terms of a geometric series S are 48 and 768 respectively. Find
(a) the two possible values of the common ratio of S, (3)
(b) the first term of S. (1)
Given that the sum of the first 5 terms of S is 615
(c) find the sum of the first 9 terms of S. (4)
Another geometric series T has the same first term as S. The common ratio of T is 1
r where r is one of the values obtained in part (a). The nth term of T is tn.
Given that t2 > t3
(d) find the common ratio of T. (1)
The sum of the first n terms of T is Tn
(e) Writing down all the numbers on your calculator display, find T9. (2)
The sum to infinity of T is T
Given that T – Tn > 0.002
(f) find the greatest value of n. (5)
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are all -ve numbers!
So, in the inequality: n log 0.25 > log 0.0005,
when you divide both sides of the inequality by log 0.25, to isolate and find n;
you have to reverse the > sign since log 0.25 is a negative number (-0.602059991),
Thus, n < log 0.0005/log 0.25
n < 5.482892142
Therefore, the greatest integer value of n = 5
(This skill is tested in Q9(f) of Edexcel 4PMO 2013 Jan Paper 2 on "Series" (pl see below). So, all A-Maths students - SPM and IGCSE Cambridge alike - beware! You may face a logarithmic inequality question of the above nature in time to come if you have not already been tested this way! By the way, there are also other negative numbers without the negative signs - can you name some of them? :))
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Edexcel PMO 2013 Jan Paper 2 Q9
9) The third and fifth terms of a geometric series S are 48 and 768 respectively. Find
(a) the two possible values of the common ratio of S, (3)
(b) the first term of S. (1)
Given that the sum of the first 5 terms of S is 615
(c) find the sum of the first 9 terms of S. (4)
Another geometric series T has the same first term as S. The common ratio of T is 1
r where r is one of the values obtained in part (a). The nth term of T is tn.
Given that t2 > t3
(d) find the common ratio of T. (1)
The sum of the first n terms of T is Tn
(e) Writing down all the numbers on your calculator display, find T9. (2)
The sum to infinity of T is T
Given that T – Tn > 0.002
(f) find the greatest value of n. (5)
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