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junior college 2 | H1 Maths
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Date Posted: 1 year ago
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ln[(e^y)/(x²–1)] = ln(x+1)–ln(x–1) = ln[(x+1)/(x–1)] = ln{[(x+1)/(x–1)]×[(x+1)/(x+1)]} = ln[(x+1)²/(x+1)(x–1)] = ln[(x+1)²/(x²–1)]
Therefore e^y = (x+1)² (shown)
e^y+e^(y/2) = 6x → (1)
Sub e^y = (x+1)² into (1)
(x+1)²+[(x+1)²]^½ = 6x
x²+2x+1+x+1 = 6x
x²–3x+2 = 0
(x–2)(x–1) = 0
x–2 = 0 or x–1 = 0
x = 2 or x = 1 (rej since x > 1)
ln[(e^y)/(x²–1)] = ln(x+1)–ln(x–1) = ln[(x+1)²/(x²–1)] → (2)
Sub x = 2 into (2)
ln[(e^y)/(2²–1)] = ln[(2+1)²/(2²–1)]
ln[(e^y)/3] = ln[9/3]
e^y = 9
ln(e^y) = ln9
y = ln9
Therefore e^y = (x+1)² (shown)
e^y+e^(y/2) = 6x → (1)
Sub e^y = (x+1)² into (1)
(x+1)²+[(x+1)²]^½ = 6x
x²+2x+1+x+1 = 6x
x²–3x+2 = 0
(x–2)(x–1) = 0
x–2 = 0 or x–1 = 0
x = 2 or x = 1 (rej since x > 1)
ln[(e^y)/(x²–1)] = ln(x+1)–ln(x–1) = ln[(x+1)²/(x²–1)] → (2)
Sub x = 2 into (2)
ln[(e^y)/(2²–1)] = ln[(2+1)²/(2²–1)]
ln[(e^y)/3] = ln[9/3]
e^y = 9
ln(e^y) = ln9
y = ln9
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