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Question Number 25023 by Tinkutara last updated on 01/Dec/17

$$\mathrm{Consider}\mathrm{the}\mathrm{function}f\left(x\right)\mathrm{which}$$$$\mathrm{satisfying}\mathrm{the}\mathrm{functional}\mathrm{equation}$$$$2f\left(x\right)+f(1-x)=x^2+1,\mathrm{\forall }x\in R$$$$\mathrm{and}g\left(x\right)=3f\left(x\right)+1.\mathrm{The}\mathrm{range}\mathrm{of}$$$$\varphi \left(x\right)=g\left(x\right)+\frac{1}{g\left(x\right)+1}\mathrm{is}$$

Answered by prakash jain last updated on 01/Dec/17

$$2f\left(x\right)+f(1-x)=x^2+1\left(1\right)$$$$\mathrm{substitue}xby1-x$$$$2f(1-x)+f\left(x\right)={(1-x)}^2+1$$$$4f\left(x\right)+2f(1-x)=2x^2+2\left(multiply\left(1\right)by2\right)$$$$\mathrm{subtract}$$$$3f\left(x\right)=2x^2-{(1-x)}^2+1$$$$3f\left(x\right)=2x^2-1+2x-x^2+1$$$$=x^2+2x$$$$g\left(x\right)=3f\left(x\right)+1={(x+1)}^2$$$$\mathrm{\varnothing }\left(x\right)={(x+1)}^2+\frac{1}{{(x+1)}^2+1}$$$$\mathrm{I}\mathrm{think}\mathrm{you}\mathrm{can}\mathrm{calculate}\mathrm{range}\mathrm{now}.$$

Commented by Tinkutara last updated on 02/Dec/17

$$\mathrm{I}\mathrm{already}\mathrm{have}\mathrm{solved}\mathrm{upto}\mathrm{this}.\mathrm{I}$$$$\mathrm{have}\mathrm{doubt}\mathrm{in}\mathrm{finding}\mathrm{range}\mathrm{only}.$$

Commented by prakash jain last updated on 02/Dec/17

$$\mathrm{for}x\in \mathbb{R}$$$$\mathrm{Range}\mathrm{of}\varphi \left(x\right)=[1,\mathrm{\infty })$$

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