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Yahoo!奇摩知識+ - 分類問答 - 科學常識 - 已解決
Yahoo!奇摩知識+ - 分類問答 - 科學常識 - 已解決 
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Show that, if x is real, the v
Apr 30th 2014, 00:11

(Q.a) Let y = (2x + 7)(2x - 1)/(x - 1), so
y(x - 1) = (2x + 7)(2x - 1)
==> yx - y = 4x^2 + 12x - 7
==> 4x^2 + (12 - y)x + (y - 7) = 0
As x is real, so
(12 - y)^2 >= 4(4)(y - 7)
==> 144 - 24y + y^2 >= 16y - 112
==> y^2 - 40y + 256 >= 0
==> (y - 8)(y - 32) >= 0
==> y = 32
That means y cannot lie between 8 and 32.

(Q.b) y = ax^2 + bx + c
So, y' = 2ax + b
As the gradient on x=1 and x=2 are 4 and 5 respectively, therefore,
4 = 2a(1) + b
==> 2a + b = 4 ⋯⋯ (i)
5 = 2a(2) + b
==> 4a + b = 5 ⋯⋯ (ii)
Solving (i), (ii), we get,
a = 0.5, b = 3
As the curve also passes through origin, so c = 0.

Therefore, a = 0.5, b = 3, c = 0


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