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Question Number 190894 by mnjuly1970 last updated on 13/Apr/23

$$$$ $$f\left(x\right)=mx+cos\left(x\right),m>1$$ $$,f^{-1}\left(3\right)=2f^{-1}\left(2\right)$$ $$find:f\left(\frac{1}{m}\right)=?$$

Answered by mehdee42 last updated on 13/Apr/23

$$f^{-1}\left(3\right)=a\Rightarrow f\left(a\right)=3\mathrm{\&}{f}^{-1}\left(2\right)=b\Rightarrow f\left(b\right)=2\Rightarrow a=2b$$ $$3=ma+cosa=2mb+cos2b\left(1\right)\mathrm{\&}2=mb+cosb\left(2\right)$$ $$\left(1\right),\left(2\right)\Rightarrow {cos}^{2}b-cosb=0$$ $$ifcosb=0\Rightarrow b=\frac{\pi }{2}\Rightarrow m=\frac{4}{\pi }✓\Rightarrow f\left(\frac{1}{m}\right)=1+\frac{\sqrt{2}}{2}$$ $$ifcosb=1\Rightarrow b=2\pi \Rightarrow m=\frac{1}{2\pi }\times$$ $$$$

Answered by mr W last updated on 14/Apr/23

$$pm+\cos p=3\Rightarrow f^{-1}\left(3\right)=p$$ $$qm+\cos q=2\Rightarrow f^{-1}\left(2\right)=q$$ $$f^{-1}\left(3\right)=2f^{-1}\left(2\right)\Rightarrow p=2q$$ $$2qm+\cos 2q=3$$ $$2qm+2\cos q=4$$ $$2\cos q-{2\cos }^{2}q+1=1$$ $$\cos q(1-\cos q)=0$$ $$\cos q=0\Rightarrow q=\frac{(2k+1)\pi }{2}\Rightarrow m=\frac{4}{(2k+1)\pi }$$ $$\cos q=1\Rightarrow q=2k\pi \Rightarrow m=\frac{1}{2k\pi }$$ $$f\left(\frac{1}{m}\right)=1+\cos \left(\frac{(2k+1)\pi }{4}\right)=1\pm \frac{1}{\sqrt{2}}✓$$ $$or$$ $$f\left(\frac{1}{m}\right)=1+\cos \left(2k\pi \right)=1✓$$

Commented bymehdee42 last updated on 14/Apr/23

$$sir$$ $$duetothequestion(m>1)onlym=\frac{4}{\pi }itistrue.>$$

Commented bymr W last updated on 14/Apr/23

$$youarerightsir.$$

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