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Fantezi Cebir => Analiz-Cebir => Konuyu başlatan: stuart clark - Temmuz 30, 2011, 06:09:42 öö

Başlık: $\sin^2x +acosx +a^2> 1+cosx$ holds for any $x$ belongs to $\mathbb{R}$
Gönderen: stuart clark - Temmuz 30, 2011, 06:09:42 öö

if the inequality $\sin^2x +acosx +a^2> 1+cosx$ holds for any $x$ belongs to $\mathbb{R}$ then the largest negative integral

value of a is

options:

(a)-4     

(b)-3

(c)-2

(d)-1

Başlık: Ynt: Trigonometric equation
Gönderen: senior - Ağustos 01, 2011, 01:09:43 ös

Replace sin²x with 1 - cos²x, then we get the inequality
cos²x + (1-a)cosx < a², Let f(x) = cos²x + (1-a)cosx and examine the extreme values.
f'(x) = -2cosxsinx + (a-1)sinx = 0 --> sinx = 0 or cosx = (a-1)/2
sinx = 0 means cosx = -1 or 1(also note that these are the boundary values of cosx, so we need not to check them,too; otherwise we would have to check them). Let's try these:
Replace cosx = (a-1)/2 --> f(x) = -(a-1)²/4 and f(x) < a² clearly for all x
Replace cosx = 1 --> f(x) = 1 + (1-a) --> 2-a < a² and it is true for all a < -2
Replace cosx = -1 --> f(x) = 1 + (a-1) < a², and it is true for all integer a not equal 1 or 0
Then we have an upper bound a < -2, and the greatest negative integer that a can take is -3

Başlık: Ynt: Trigonometric equation
Gönderen: proble_m - Ağustos 02, 2011, 12:44:21 öö

Very nice solution..

Başlık: Ynt: Trigonometric equation
Gönderen: stuart clark - Ağustos 02, 2011, 07:02:40 öö

Thanks moderator for Nice solution