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Question Number 182721 by mnjuly1970 last updated on 13/Dec/22

	Answered by Acem last updated on 13/Dec/22
	$It period: T= (1/a) f(x)=1 ⇒ b=1 lim_(x→((15^( −) )/2)) f(x)= −∞ ⇒ cos πax= cos (π^( +) /2) ⇒ a<0 π a x= π/2 ± 2πk when x= ((15)/2) : a= (1/(15)) ± (4/(15)) k^(am at problem of k) this term should be negative cause a< 0 so for k=1 and with minus mark_( But why not 2, 3, ..?) a= (1/(15)) −(4/(15)) ⇒ a= ((−1)/5) f(x)= tan ((−π)/5) x +1 _(continue...) f(x)= 0 ⇒ tan (π/(15)) x = −1 = tan (((3 π)/4) +πk) According to the diagram k=0 , k=1 x_A = ((45)/4) , x_B = ((105)/4) ⇒ Σ= ((75)/2) ≠ {a, b, c, d}$
	$I t p e r i o d : T = \frac{1}{a}$ $f (x) = 1 \Rightarrow b = 1$ $\lim_{x \to \frac{15^{-}}{2}} f (x) = - \infty \Rightarrow \cos π a x = \cos \frac{π^{+}}{2} \Rightarrow a < 0$ $π a x = π / 2 \pm 2 π k$ $w h e n x = \frac{15}{2} : a = \frac{1}{15} \underset{⏟}{\pm \frac{4}{15} k}^{a m a t p r o b l e m o f k}$ $t h i s t e r m s h o u l d b e n e g a t i v e c a u s e a < 0$ $s o f o r k = 1 a n d w i t h m i n u s {m a r k}_{B u t w h y n o t 2, 3, . . ?}$ $a = \frac{1}{15} - \frac{4}{15} \Rightarrow a = \frac{- 1}{5}$ $f (x) = \tan \frac{- π}{5} x + 1_{c o n t i n u e . . .}$ $f (x) = 0 \Rightarrow \tan \frac{π}{15} x = - 1 = \tan (\frac{3 π}{4} + π k)$ $A c c o r d i n g t o t h e d i a g r a m k = 0, k = 1$ $x_{A} = \frac{45}{4}, x_{B} = \frac{105}{4} \Rightarrow Σ = \frac{75}{2} \neq {a, b, c, d}$
	Answered by mahdipoor last updated on 13/Dec/22

	$f (0) = b = 1 \Rightarrow$ $x^{-} \to \frac{15}{2} \Rightarrow f \to - \infty \Rightarrow a < 0$ $l i m f (x) = \pm \infty \Leftrightarrow π a x \to \frac{π}{2} + π k k \in Z$ $\Rightarrow x \to \frac{1}{2 a} + \frac{k}{a} \Rightarrow x \to \frac{15}{2} = \frac{1}{2 a} + \frac{- 2}{a} = \frac{- 3}{2 a}$ $\Rightarrow a = - \frac{1}{5}$ $f (x) = t a n (- \frac{π}{5} x) + 1 = 0 \Rightarrow x = \frac{5}{4}, 5 + \frac{5}{4}, . . .$ $x_{A} + x_{B} = (\frac{5}{4}) + (\frac{5}{4} + 5) = 7.5$
	Commented by Acem last updated on 13/Dec/22

	$H e l l o S i r! w h y k = - 2 w h e n x \to \frac{15}{2}$
	Commented by mahdipoor last updated on 13/Dec/22

	$H i, w h e n a < 0 a n d x > 0 \Rightarrow π a x < 0$ $i n f i r s t v e r t i c a l l i n e : π a x = - \frac{π}{2}$ $a n d i n s e c e n d l i n e : π a x = \frac{- π}{2} - π = \frac{- 3 π}{2}$ $. . .$ $x = \frac{15}{2} i s s e c e n d l i n e s o$ $π a x = \frac{π}{2} + k π = \frac{- 3 π}{2} \Rightarrow k = - 2$
	Commented by Acem last updated on 13/Dec/22

	$H i S i r! t h a n k y o u v e r y m u c h$ $I n f a c t t h e r e a r e s t i l l t o w p r o b l e m s :$ $a ∙ h o w d o i k n o w t h a t {i t}^{'} s t h e f i r s t a s s y m p t o t$ $b ∙ y o u s a i d t h a t π a x = \frac{- π}{2} t h a t m e a n k = - 1$ $h o w d o i k n o w t h a t k h e r e = - 1$ $T h o u g h t t h a t w e m a y c a n s a y t h a t t h e 1 s t$ $a s s y m p t o t i s w h e n k = 0 n o t - 1$
	Commented by mahdipoor last updated on 14/Dec/22

	$[k] i s n o t i m p o r t a n t, [π {a x}_{2} = π a \frac{15}{2} = \frac{- 3 π}{2}]$ $i s i m p o r t a n t, i m s a y [π a x = \frac{π}{2} + k π] s o$ $k = - 2 w h e n [π {a x}_{2} = \frac{- 3 π}{2} = \frac{π}{2} + (- 2) π]$
	Commented by mnjuly1970 last updated on 14/Dec/22

	$m a m n o o n j e n a b e m a h d i p o o r$ $a l i b o o d$
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