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junior college 1 | H2 Maths
2 Answers Below
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please help with part 2:) to solve *without* using part i graph
Date Posted: 3 years ago
Views: 219
|(x + 1)/(1 - x)| < x + 2
-(x + 2) < (x + 1)/(1 - x) < x + 2
-(x + 2) < (x + 1)/(1 - x) and (x + 1)/(1 - x) < x + 2
0 < x + 2 + (x + 1)/(1 - x) and (x + 1)/(1 - x) - (x + 2) < 0
0 < ((x + 2)(1 - x) + (x + 1)) / (1 - x) and (x + 1 - (1 - x)(x + 2) ) / (1 - x) < 0
0 < (3 - x²)/(1 - x) and (x² + 2x - 1)/(1 - x) < 0
0 < (√3 - x)(√3 + x)/(1 - x) and (x - (√2 - 1))(x + (√2 + 1)/(1 - x) < 0
Checking the signs,
x > √3 or -√3 < x < 1
and
x > 1 or -√2 - 1 < x < √2 - 1
Combining the two,
x > √3 or -√3 < x < √2 - 1
-(x + 2) < (x + 1)/(1 - x) < x + 2
-(x + 2) < (x + 1)/(1 - x) and (x + 1)/(1 - x) < x + 2
0 < x + 2 + (x + 1)/(1 - x) and (x + 1)/(1 - x) - (x + 2) < 0
0 < ((x + 2)(1 - x) + (x + 1)) / (1 - x) and (x + 1 - (1 - x)(x + 2) ) / (1 - x) < 0
0 < (3 - x²)/(1 - x) and (x² + 2x - 1)/(1 - x) < 0
0 < (√3 - x)(√3 + x)/(1 - x) and (x - (√2 - 1))(x + (√2 + 1)/(1 - x) < 0
Checking the signs,
x > √3 or -√3 < x < 1
and
x > 1 or -√2 - 1 < x < √2 - 1
Combining the two,
x > √3 or -√3 < x < √2 - 1
See 2 Answers
done
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clear
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Hang on for the second part. I misjudged my paper space available.
Date Posted:
3 years ago
done
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clear
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The remaining
Date Posted:
3 years ago
Denominator is 1 - x. This implies that x ≠ 1.
So for your signs table, the overall should be undefined for the row where x = 1.
Also, for the rows where you put all 0s, some of the factors do not equal 0 as well.
So for your signs table, the overall should be undefined for the row where x = 1.
Also, for the rows where you put all 0s, some of the factors do not equal 0 as well.
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