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junior college 1 | H2 Maths
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Date Posted: 4 years ago
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a)
f(x) = (x + k)/(x + 1)
f(x) = (x + 1 + (k - 1))/(x + 1)
f(x) = 1 + (k -1)/(x + 1)
We can see the function has a horizontal asymptote at y = 1
k > 1 so k - 1 > 0
k > 1 so -k < -1.
Since x ≤ -k, x ≤ -k < -1
So x < - 1 , which means x + 1 < 0
Since x + 1 < 0 and k - 1 > 0, (k -1)/(x+1) < 0 as a positive numerator over a negative denominator would give a negative value.
So 1 + (k -1)/(x + 1) < 1 , meaning f(x) < 1
Since x ≤ -k, x + 1 ≤ 1 - k
1 ≥ (1 - k)/(x + 1) (flip signs as we know x + 1 is negative since x + 1< 0 , and dividing by a known negative value requires a flip of sign)
-1 ≤ (k - 1)/(x + 1)
0 ≤ 1 + (k - 1)/(x + 1). So f(x) ≥ 0.
We can now say 0 ≤ f(x) < 1 based on the two results we got.
So Rf = [0,1)
Dg = [0,∞)
Since Rf ⊂ Dg, gf exists
b)
gf(x) = 1 + cos[3π((x + k)/(x +1))/2]
= 1 + cos[3π(x+k)/(2(x +1))]
-1 ≤ cos[3π(x+k)/(2(x +1))] ≤ 1 for real values of x
So 0 ≤ 1 + cos[3π(x+k)/(2(x +1))] ≤ 2 for real values of x
Rgf = [0,2]
f(x) = (x + k)/(x + 1)
f(x) = (x + 1 + (k - 1))/(x + 1)
f(x) = 1 + (k -1)/(x + 1)
We can see the function has a horizontal asymptote at y = 1
k > 1 so k - 1 > 0
k > 1 so -k < -1.
Since x ≤ -k, x ≤ -k < -1
So x < - 1 , which means x + 1 < 0
Since x + 1 < 0 and k - 1 > 0, (k -1)/(x+1) < 0 as a positive numerator over a negative denominator would give a negative value.
So 1 + (k -1)/(x + 1) < 1 , meaning f(x) < 1
Since x ≤ -k, x + 1 ≤ 1 - k
1 ≥ (1 - k)/(x + 1) (flip signs as we know x + 1 is negative since x + 1< 0 , and dividing by a known negative value requires a flip of sign)
-1 ≤ (k - 1)/(x + 1)
0 ≤ 1 + (k - 1)/(x + 1). So f(x) ≥ 0.
We can now say 0 ≤ f(x) < 1 based on the two results we got.
So Rf = [0,1)
Dg = [0,∞)
Since Rf ⊂ Dg, gf exists
b)
gf(x) = 1 + cos[3π((x + k)/(x +1))/2]
= 1 + cos[3π(x+k)/(2(x +1))]
-1 ≤ cos[3π(x+k)/(2(x +1))] ≤ 1 for real values of x
So 0 ≤ 1 + cos[3π(x+k)/(2(x +1))] ≤ 2 for real values of x
Rgf = [0,2]
Thank you!
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Edit to part b) :
0 ≤ gf(x) ≤ 2 for real values of x , x ≤ -k, k > 1 instead of all real values of x, because it's undefined when x = -1, and because the function has a restricted domain.
0 ≤ gf(x) ≤ 2 for real values of x , x ≤ -k, k > 1 instead of all real values of x, because it's undefined when x = -1, and because the function has a restricted domain.
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