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Question Number 190894 by mnjuly1970 last updated on 13/Apr/23
f(x)= mx + cos(x) , m>1 , f^( −1) (3)= 2f^( −1) (2 ) find : f ( (1/m) )= ?
$$ \\ $$$$\:\:\:\:\:{f}\left({x}\right)=\:{mx}\:+\:{cos}\left({x}\right)\:\:,\:\:{m}>\mathrm{1} \\ $$$$\:\:\:\:,\:\:\:{f}^{\:−\mathrm{1}} \left(\mathrm{3}\right)=\:\mathrm{2}{f}^{\:−\mathrm{1}} \left(\mathrm{2}\:\right) \\ $$$$\:\:\:\:\:{find}\::\:\:\:\:\:\:{f}\:\left(\:\frac{\mathrm{1}}{{m}}\:\right)=\:? \\ $$
Answered by mehdee42 last updated on 13/Apr/23
f^(−1) (3)=a⇒f(a)=3 & f^(−1) (2)=b⇒f(b)=2⇒a=2b 3=ma+cosa=2mb+cos2b (1) & 2=mb+cosb (2) (1),(2)⇒cos^2 b−cosb=0 if cosb=0⇒b=(π/2)⇒m=(4/π) ✓⇒f((1/m))=1+((√2)/2) if cosb=1⇒b=2π⇒m=(1/(2π)) ×
$${f}^{−\mathrm{1}} \left(\mathrm{3}\right)={a}\Rightarrow{f}\left({a}\right)=\mathrm{3}\:\:\:\&\:\:{f}^{−\mathrm{1}} \left(\mathrm{2}\right)={b}\Rightarrow{f}\left({b}\right)=\mathrm{2}\Rightarrow{a}=\mathrm{2}{b} \\ $$$$\mathrm{3}={ma}+{cosa}=\mathrm{2}{mb}+{cos}\mathrm{2}{b}\:\:\:\left(\mathrm{1}\right)\:\:\:\&\:\:\mathrm{2}={mb}+{cosb}\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right),\left(\mathrm{2}\right)\Rightarrow{cos}^{\mathrm{2}} {b}−{cosb}=\mathrm{0} \\ $$$${if}\:\:{cosb}=\mathrm{0}\Rightarrow{b}=\frac{\pi}{\mathrm{2}}\Rightarrow{m}=\frac{\mathrm{4}}{\pi}\:\checkmark\Rightarrow{f}\left(\frac{\mathrm{1}}{{m}}\right)=\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${if}\:\:{cosb}=\mathrm{1}\Rightarrow{b}=\mathrm{2}\pi\Rightarrow{m}=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\:\:\:× \\ $$$$ \\ $$
Answered by mr W last updated on 14/Apr/23
pm+cos p=3 ⇒f^(−1) (3)=p qm+cos q=2 ⇒f^(−1) (2)=q f^(−1) (3)=2f^(−1) (2) ⇒p=2q 2qm+cos 2q=3 2qm+2cos q=4 2cos q−2cos^2 q+1=1 cos q(1−cos q)=0 cos q=0 ⇒q=(((2k+1)π)/2) ⇒m=(4/((2k+1)π)) cos q=1 ⇒q=2kπ ⇒m=(1/(2kπ)) f((1/m))=1+cos ((((2k+1)π)/4))=1±(1/( (√2))) ✓ or f((1/m))=1+cos (2kπ)=1 ✓
$${pm}+\mathrm{cos}\:{p}=\mathrm{3}\:\Rightarrow{f}^{−\mathrm{1}} \left(\mathrm{3}\right)={p} \\ $$$${qm}+\mathrm{cos}\:{q}=\mathrm{2}\:\Rightarrow{f}^{−\mathrm{1}} \left(\mathrm{2}\right)={q} \\ $$$${f}^{−\mathrm{1}} \left(\mathrm{3}\right)=\mathrm{2}{f}^{−\mathrm{1}} \left(\mathrm{2}\right)\:\Rightarrow{p}=\mathrm{2}{q} \\ $$$$\mathrm{2}{qm}+\mathrm{cos}\:\mathrm{2}{q}=\mathrm{3} \\ $$$$\mathrm{2}{qm}+\mathrm{2cos}\:{q}=\mathrm{4} \\ $$$$\mathrm{2cos}\:{q}−\mathrm{2cos}^{\mathrm{2}} \:{q}+\mathrm{1}=\mathrm{1} \\ $$$$\mathrm{cos}\:{q}\left(\mathrm{1}−\mathrm{cos}\:{q}\right)=\mathrm{0} \\ $$$$\mathrm{cos}\:{q}=\mathrm{0}\:\Rightarrow{q}=\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}}\:\Rightarrow{m}=\frac{\mathrm{4}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \\ $$$$\mathrm{cos}\:{q}=\mathrm{1}\:\Rightarrow{q}=\mathrm{2}{k}\pi\:\Rightarrow{m}=\frac{\mathrm{1}}{\mathrm{2}{k}\pi} \\ $$$${f}\left(\frac{\mathrm{1}}{{m}}\right)=\mathrm{1}+\mathrm{cos}\:\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{4}}\right)=\mathrm{1}\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\checkmark \\ $$$${or} \\ $$$${f}\left(\frac{\mathrm{1}}{{m}}\right)=\mathrm{1}+\mathrm{cos}\:\left(\mathrm{2}{k}\pi\right)=\mathrm{1}\:\checkmark \\ $$
Commented by mehdee42 last updated on 14/Apr/23
sir due to the question (m>1) only m=(4/π) it is true.>
$${sir} \\ $$$${due}\:{to}\:{the}\:{question}\:\left({m}>\mathrm{1}\right)\:{only}\:\:{m}=\frac{\mathrm{4}}{\pi}\:{it}\:{is}\:{true}.> \\ $$
Commented by mr W last updated on 14/Apr/23
you are right sir.
$${you}\:{are}\:{right}\:{sir}.\: \\ $$

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