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Question Number 173033 by cortano1 last updated on 05/Jul/22

	Answered by greougoury555 last updated on 05/Jul/22
	$f ′(x)= cos x−sin x −((4k cos 2x)/(sin^2 2x)) =0 (cos x−sin x)(1−((4k(cos x+sin x))/(sin^2 2x)))=0 (°) tan x=1⇒f =(√2) +2k (°°) 1 =((4k(cos x+sin x))/(sin^2 2x)) (cos x+sin x)^2 −sin 2x = ((4k(cos x+sin x))/(sin^2 2x)) { ((cos x+sin x=u)),((sin 2x = v)) :} u^2 −v = ((4ku)/v^2 ) ; (uv)^2 −v^3 = 4ku$
	$f^{'} (x) = \cos x - \sin x - \frac{4 k \cos 2 x}{\sin^{2} 2 x} = 0$ $(\cos x - \sin x) (1 - \frac{4 k (\cos x + \sin x)}{\sin^{2} 2 x}) = 0$ $(°) \tan x = 1 \Rightarrow f = \sqrt{2} + 2 k$ $(° °) 1 = \frac{4 k (\cos x + \sin x)}{\sin^{2} 2 x}$ ${(\cos x + \sin x)}^{2} - \sin 2 x = \frac{4 k (\cos x + \sin x)}{\sin^{2} 2 x}$ ${\begin{cases} \cos x + \sin x = u \\ \sin 2 x = v \end{cases}$ $u^{2} - v = \frac{4 k u}{v^{2}}; {(u v)}^{2} - v^{3} = 4 k u$
	Answered by a.lgnaoui last updated on 05/Jul/22

	$f (x) = \sin x + \cos x + \frac{2 k}{\sin 2 x}$ $l i m f {(x)}_{x \to 0} = 1 + \frac{1}{x} {l i m}_{x \to 0} \frac{2 x}{\sin 2 x} k = 1 + \frac{k}{x}$ $i f k > 0 f \to + \infty; i f k < 0 f \to - \infty$ ${l i m}_{x \to \frac{π}{2}} f (x) = 1 + {l i m}_{x \to \frac{π}{2}} \frac{2 k}{\sin 2 x} = + \infty i f k > 0 o r - \infty i f k < 0$ $d o n c f d e c r o i s s a n t e t c r i i s s a n t s i k > 0 e t c o n t r a i r e p o u r k > 0$ $f^{'} (x) = \cos x - \sin x - k \frac{\cos^{2} x - \sin^{2} x}{\sin^{2} x \cos^{2} x} = (\cos x - \sin x) [1 - k (\frac{1}{\cos x} + \frac{1}{\sin x})]$ $r e m a r q u e x = \frac{π}{4} \Rightarrow f^{'} (x) = 0 d o n c \frac{π}{4} i s m i n i m u m o f f (x) .$
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