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Question Number 38024 by Tinkutara last updated on 20/Jun/18

Answered by ajfour last updated on 20/Jun/18

f(x) < 0 and 2f(sin x)+(√2)f(−cos x)=−tan x Then for x=150° 2f((1/2))+(√2)f(((√3)/2))=(1/(√3)) ......(i) for x= 120° 2f(((√3)/2))+(√2)f((1/2))=(√3) ....(ii) (i)×(√2) −(ii) gives (√2)f((1/2)) = ((√2)/(√3))−(√3) ⇒ f((1/2)) = (1/(√3))−((√3)/(√2)) or f((1/2)) = (((√2)−3)/(√6)) (A) .

$${f}\left({x}\right)\:<\:\mathrm{0}\:\:{and} \\ $$$$\mathrm{2}{f}\left(\mathrm{sin}\:{x}\right)+\sqrt{\mathrm{2}}{f}\left(−\mathrm{cos}\:{x}\right)=−\mathrm{tan}\:{x} \\ $$$${Then}\:\:{for}\:{x}=\mathrm{150}° \\ $$$$\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\mathrm{2}}{f}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\:\:......\left({i}\right) \\ $$$${for}\:{x}=\:\mathrm{120}° \\ $$$$\mathrm{2}{f}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+\sqrt{\mathrm{2}}{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\mathrm{3}}\:\:\:\:....\left({ii}\right) \\ $$$$\left({i}\right)×\sqrt{\mathrm{2}}\:−\left({ii}\right)\:{gives} \\ $$$$\:\:\:\sqrt{\mathrm{2}}{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}}−\sqrt{\mathrm{3}} \\ $$$$\:\:\:\Rightarrow\:\:\:{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}−\frac{\sqrt{\mathrm{3}}}{\sqrt{\mathrm{2}}}\: \\ $$$$\:\:\:{or}\:\:\:{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\sqrt{\mathrm{2}}−\mathrm{3}}{\sqrt{\mathrm{6}}}\:\:\:\:\:\left({A}\right)\:. \\ $$

Commented by Tinkutara last updated on 20/Jun/18

Commented by Tinkutara last updated on 20/Jun/18

Sir is this wrong anywhere?

Commented by ajfour last updated on 21/Jun/18

at x=((2π)/3), f(cos x) = f(−(1/2))

$${at}\:\:\:{x}=\frac{\mathrm{2}\pi}{\mathrm{3}},\:\:\:\:{f}\left(\mathrm{cos}\:{x}\right)\:=\:{f}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

Commented by Tinkutara last updated on 24/Jun/18

But it is f(−cos x) there.

$${But}\:{it}\:{is}\:{f}\left(−\mathrm{cos}\:{x}\right)\:{there}. \\ $$

Commented by ajfour last updated on 24/Jun/18

No, it is f(cos x) there; your eq. (2) where you set x=((2π)/3) .

$${No},\:{it}\:{is}\:{f}\left(\mathrm{cos}\:{x}\right)\:{there};\:{your}\:\:{eq}. \\ $$$$\left(\mathrm{2}\right)\:{where}\:{you}\:{set}\:{x}=\frac{\mathrm{2}\pi}{\mathrm{3}}\:. \\ $$

Commented by Tinkutara last updated on 24/Jun/18

No Sir, I first changed x→−x f(−cos x)=f(−cos (−x)) I had put a little minus sign.

$${No}\:{Sir},\:{I}\:{first}\:{changed}\:{x}\rightarrow−{x} \\ $$$${f}\left(−\mathrm{cos}\:{x}\right)={f}\left(−\mathrm{cos}\:\left(−{x}\right)\right) \\ $$$${I}\:{had}\:{put}\:{a}\:{little}\:{minus}\:{sign}. \\ $$

Commented by ajfour last updated on 24/Jun/18

In quwstion it is said f(x) is always negative. Nothing wrong with your steps, just that this answer is discarded becoz of the condition in question f(x) < 0 .

$${In}\:{quwstion}\:{it}\:{is}\:{said}\:{f}\left({x}\right)\:{is} \\ $$$${always}\:{negative}.\:{Nothing} \\ $$$${wrong}\:{with}\:{your}\:{steps},\:{just}\:{that} \\ $$$${this}\:{answer}\:{is}\:{discarded}\: \\ $$$${becoz}\:{of}\:{the}\:{condition}\:{in} \\ $$$${question}\:{f}\left({x}\right)\:<\:\mathrm{0}\:. \\ $$

Commented by Tinkutara last updated on 25/Jun/18

Thanks Sir!

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