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Question-80199

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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
all are clear if you have a look at the graph of the functions.
$${all}\:{are}\:{clear}\:{if}\:{you}\:{have}\:{a}\:{look}\:{at} \\ $$$${the}\:{graph}\:{of}\:{the}\:{functions}. \\ $$
Answered by mr W last updated on 08/Feb/20
(a) cosh x=((e^x +e^(−x) )/2) sinh x=((e^x −e^(−x) )/2) a cosh x+b sinh x=0 (a/2)(e^x +e^(−x) )+(b/2)(e^x −e^(−x) )=0 (a+b)e^x +(a−b)e^(−x) =0 e^(2x) =((b−a)/(b+a))=(((b/a)−1)/((b/a)+1)) with ∣b∣>∣a∣: (b/a)>1 or (b/a)<−1 with (b/a)>1: (b/a)−1>0 and (b/a)+1>0 ⇒(((b/a)−1)/((b/a)+1))>0 with (b/a)<−1: (b/a)−1<0 and (b/a)+1<0 ⇒ (((b/a)−1)/((b/a)+1))>0 that means if ∣b∣>∣a∣, e^(2x) =((b−a)/(b+a))=(((b/a)−1)/((b/a)+1))>0 it has one and only one root: x=(1/2)ln (((b−a)/(b+a)))
$$\left({a}\right) \\ $$$$\mathrm{cosh}\:{x}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$$\mathrm{sinh}\:{x}=\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}} \\ $$$${a}\:\mathrm{cosh}\:{x}+{b}\:\mathrm{sinh}\:{x}=\mathrm{0} \\ $$$$\frac{{a}}{\mathrm{2}}\left({e}^{{x}} +{e}^{−{x}} \right)+\frac{{b}}{\mathrm{2}}\left({e}^{{x}} −{e}^{−{x}} \right)=\mathrm{0} \\ $$$$\left({a}+{b}\right){e}^{{x}} +\left({a}−{b}\right){e}^{−{x}} =\mathrm{0} \\ $$$${e}^{\mathrm{2}{x}} =\frac{{b}−{a}}{{b}+{a}}=\frac{\frac{{b}}{{a}}−\mathrm{1}}{\frac{{b}}{{a}}+\mathrm{1}} \\ $$$${with}\:\mid{b}\mid>\mid{a}\mid: \\ $$$$\frac{{b}}{{a}}>\mathrm{1}\:{or}\:\frac{{b}}{{a}}<−\mathrm{1} \\ $$$$ \\ $$$${with}\:\frac{{b}}{{a}}>\mathrm{1}:\:\: \\ $$$$\frac{{b}}{{a}}−\mathrm{1}>\mathrm{0}\:{and}\:\frac{{b}}{{a}}+\mathrm{1}>\mathrm{0} \\ $$$$\Rightarrow\frac{\frac{{b}}{{a}}−\mathrm{1}}{\frac{{b}}{{a}}+\mathrm{1}}>\mathrm{0} \\ $$$${with}\:\frac{{b}}{{a}}<−\mathrm{1}:\: \\ $$$$\frac{{b}}{{a}}−\mathrm{1}<\mathrm{0}\:{and}\:\frac{{b}}{{a}}+\mathrm{1}<\mathrm{0} \\ $$$$\Rightarrow\:\frac{\frac{{b}}{{a}}−\mathrm{1}}{\frac{{b}}{{a}}+\mathrm{1}}>\mathrm{0} \\ $$$${that}\:{means}\:{if}\:\mid{b}\mid>\mid{a}\mid, \\ $$$${e}^{\mathrm{2}{x}} =\frac{{b}−{a}}{{b}+{a}}=\frac{\frac{{b}}{{a}}−\mathrm{1}}{\frac{{b}}{{a}}+\mathrm{1}}>\mathrm{0} \\ $$$${it}\:{has}\:{one}\:{and}\:{only}\:{one}\:{root}: \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\frac{{b}−{a}}{{b}+{a}}\right) \\ $$
Answered by mr W last updated on 08/Feb/20
(b) (i): cosh x=((e^x +e^(−x) )/2)≥((2(√(e^x e^(−x) )))/2)=(2/2)=1 (ii): if x≥0: sinh x=((e^x −e^(−x) )/2)≤((e^x +e^(−x) )/2)=cosh x ⇒∣sinh x∣≤cosh x if x≤0: sinh x=((e^x −e^(−x) )/2)≥((−e^x −e^(−x) )/2)=−cosh x ⇒∣sinh x∣≤cosh x
$$\left({b}\right) \\ $$$$\left({i}\right): \\ $$$$\mathrm{cosh}\:{x}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}\geqslant\frac{\mathrm{2}\sqrt{{e}^{{x}} {e}^{−{x}} }}{\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{2}}=\mathrm{1} \\ $$$$\left({ii}\right): \\ $$$${if}\:{x}\geqslant\mathrm{0}: \\ $$$$\mathrm{sinh}\:{x}=\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}\leqslant\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}=\mathrm{cosh}\:{x} \\ $$$$\Rightarrow\mid\mathrm{sinh}\:{x}\mid\leqslant\mathrm{cosh}\:{x} \\ $$$${if}\:{x}\leqslant\mathrm{0}: \\ $$$$\mathrm{sinh}\:{x}=\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}\geqslant\frac{−{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}=−\mathrm{cosh}\:{x} \\ $$$$\Rightarrow\mid\mathrm{sinh}\:{x}\mid\leqslant\mathrm{cosh}\:{x} \\ $$$$ \\ $$

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