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Question Number 80199 by peter frank last updated on 31/Jan/20

Commented by mr W last updated on 08/Feb/20

$a l l a r e c l e a r i f y o u h a v e a l o o k a t$ $t h e g r a p h o f t h e f u n c t i o n s .$
	Answered by mr W last updated on 08/Feb/20

	$(a)$ $\cosh x = \frac{e^{x} + e^{- x}}{2}$ $\sinh x = \frac{e^{x} - e^{- x}}{2}$ $a \cosh x + b \sinh x = 0$ $\frac{a}{2} (e^{x} + e^{- x}) + \frac{b}{2} (e^{x} - e^{- x}) = 0$ $(a + b) e^{x} + (a - b) e^{- x} = 0$ $e^{2 x} = \frac{b - a}{b + a} = \frac{\frac{b}{a} - 1}{\frac{b}{a} + 1}$ $w i t h ∣ b ∣>∣ a ∣:$ $\frac{b}{a} > 1 o r \frac{b}{a} < - 1$ $w i t h \frac{b}{a} > 1 :$ $\frac{b}{a} - 1 > 0 a n d \frac{b}{a} + 1 > 0$ $\Rightarrow \frac{\frac{b}{a} - 1}{\frac{b}{a} + 1} > 0$ $w i t h \frac{b}{a} < - 1 :$ $\frac{b}{a} - 1 < 0 a n d \frac{b}{a} + 1 < 0$ $\Rightarrow \frac{\frac{b}{a} - 1}{\frac{b}{a} + 1} > 0$ $t h a t m e a n s i f ∣ b ∣>∣ a ∣,$ $e^{2 x} = \frac{b - a}{b + a} = \frac{\frac{b}{a} - 1}{\frac{b}{a} + 1} > 0$ $i t h a s o n e a n d o n l y o n e r o o t :$ $x = \frac{1}{2} \ln (\frac{b - a}{b + a})$
	Answered by mr W last updated on 08/Feb/20

	$(b)$ $(i) :$ $\cosh x = \frac{e^{x} + e^{- x}}{2} ⩾ \frac{2 \sqrt{e^{x} e^{- x}}}{2} = \frac{2}{2} = 1$ $(i i) :$ $i f x ⩾ 0 :$ $\sinh x = \frac{e^{x} - e^{- x}}{2} ⩽ \frac{e^{x} + e^{- x}}{2} = \cosh x$ $\Rightarrow∣ \sinh x ∣⩽ \cosh x$ $i f x ⩽ 0 :$ $\sinh x = \frac{e^{x} - e^{- x}}{2} ⩾ \frac{- e^{x} - e^{- x}}{2} = - \cosh x$ $\Rightarrow∣ \sinh x ∣⩽ \cosh x$
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