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

Answered by ajfour last updated on 20/Jun/18

$$f\left(x\right)<0and$$$$2f\left(\sin x\right)+\sqrt{2}f(-\cos x)=-\tan x$$$$Thenforx=150°$$$$2f\left(\frac{1}{2}\right)+\sqrt{2}f\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{\sqrt{3}}......\left(i\right)$$$$forx=120°$$$$2f\left(\frac{\sqrt{3}}{2}\right)+\sqrt{2}f\left(\frac{1}{2}\right)=\sqrt{3}....\left(ii\right)$$$$\left(i\right)\times \sqrt{2}-\left(ii\right)gives$$$$\sqrt{2}f\left(\frac{1}{2}\right)=\frac{\sqrt{2}}{\sqrt{3}}-\sqrt{3}$$$$\Rightarrow f\left(\frac{1}{2}\right)=\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{\sqrt{2}}$$$$orf\left(\frac{1}{2}\right)=\frac{\sqrt{2}-3}{\sqrt{6}}\left(A\right).$$

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

$$atx=\frac{2\pi }{3},f\left(\cos x\right)=f(-\frac{1}{2})$$

Commented by Tinkutara last updated on 24/Jun/18

$$Butitisf(-\cos x)there.$$

Commented by ajfour last updated on 24/Jun/18

$$No,itisf\left(\cos x\right)there;youreq.$$$$\left(2\right)whereyousetx=\frac{2\pi }{3}.$$

Commented by Tinkutara last updated on 24/Jun/18

$$NoSir,Ifirstchangedx\to -x$$$$f(-\cos x)=f(-\cos (-x))$$$$Ihadputalittleminussign.$$

Commented by ajfour last updated on 24/Jun/18

$$Inquwstionitissaidf\left(x\right)is$$$$alwaysnegative.Nothing$$$$wrongwithyoursteps,justthat$$$$thisanswerisdiscarded$$$$becozoftheconditionin$$$$questionf\left(x\right)<0.$$

Commented by Tinkutara last updated on 25/Jun/18

Thanks Sir!

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