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secondary 3 | A Maths
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need help with e(ii) only
Date Posted: 3 years ago
Views: 180
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done
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clear
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An idea
Date Posted:
3 years ago
im still confused
First y = x^3 - 2x + 1
Second y = 2x + 2
Basically to satisfy an inequality like that
x^3 - 2x + 1 <= 2x + 2
it means that the value of the first y must be lower than the value of the second y. This means that the curve is at a lower height than the line. I have marked this as shaded in red.
There are three intersections of the curve with the line if you observe carefully. One at x1 = -1.8, one at x2 = -0.25 and one at x3 = 2.1. At these points, the first y and the second y are at the same height level.
To the left of x1 = -1.8, the line is above the curve at every single point.
In between x1 = -1.8 and x2 = -0.25, the curve is above the line in the position.
In between x2 = -0.25 and x3 = 2.1, the line is above the curve.
Beyond x3 = 2.1, the curve is above the line all the way.
The question wishes us to find the portions of the graph (to the right of x = -1) where the line is above the curve. This is represented by my shaded region in red, where -0.25 <= x <= 2.1.
Hence, the solutions to that inequality are -0.25 <= x <= 2.1.
This technique (of spotting which curve is above which) is a must-know in Sec 4 A Maths under Integration: Area of curves.
Second y = 2x + 2
Basically to satisfy an inequality like that
x^3 - 2x + 1 <= 2x + 2
it means that the value of the first y must be lower than the value of the second y. This means that the curve is at a lower height than the line. I have marked this as shaded in red.
There are three intersections of the curve with the line if you observe carefully. One at x1 = -1.8, one at x2 = -0.25 and one at x3 = 2.1. At these points, the first y and the second y are at the same height level.
To the left of x1 = -1.8, the line is above the curve at every single point.
In between x1 = -1.8 and x2 = -0.25, the curve is above the line in the position.
In between x2 = -0.25 and x3 = 2.1, the line is above the curve.
Beyond x3 = 2.1, the curve is above the line all the way.
The question wishes us to find the portions of the graph (to the right of x = -1) where the line is above the curve. This is represented by my shaded region in red, where -0.25 <= x <= 2.1.
Hence, the solutions to that inequality are -0.25 <= x <= 2.1.
This technique (of spotting which curve is above which) is a must-know in Sec 4 A Maths under Integration: Area of curves.
So far do you understand my idea?
This concept is difficult to explain unless I am explaining that concept to my students in a tuition centre as I can voice out my thoughts on this.
This concept is difficult to explain unless I am explaining that concept to my students in a tuition centre as I can voice out my thoughts on this.
but why isnt x= - 1.8 counted tho
in the question they stated 2 inequalities, are they trying to tell us to find the 2 inequalities or do they mean something else?
"but why isnt x= - 1.8 counted tho"
"in the question they stated 2 inequalities, are they trying to tell us to find the 2 inequalities or do they mean something else?"
Basically here there are two regions in which the line is above the curve or equal to the curve, that is,
1. x <= -1.8
2. -0.25 <= x <= 2.1
These two inequality ranges of x satisfy the first inequality criteria of the question.
Only the second case, -0.25 <= x <= 2.1, satisfy the second given condition that x >= -1.
The connector word "and" in e(ii) means that the two conditions must be satisfied at the same time.
"in the question they stated 2 inequalities, are they trying to tell us to find the 2 inequalities or do they mean something else?"
Basically here there are two regions in which the line is above the curve or equal to the curve, that is,
1. x <= -1.8
2. -0.25 <= x <= 2.1
These two inequality ranges of x satisfy the first inequality criteria of the question.
Only the second case, -0.25 <= x <= 2.1, satisfy the second given condition that x >= -1.
The connector word "and" in e(ii) means that the two conditions must be satisfied at the same time.
thx :)
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