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Question Number 120049 by bramlexs22 last updated on 29/Oct/20

$$f(x+2)+f(x-2)=f\left(x\right)$$$$f\left(1\right)=1,f\left(2\right)=2,f\left(3\right)=3,f\left(4\right)=4$$$$thenf\left(100\right)=?$$

Answered by bemath last updated on 29/Oct/20

$$\Rightarrow f(x-2)-f\left(x\right)+f(x+2)=0$$$$replacexbyx+2\Rightarrow f\left(x\right)-f(x+2)+f(x+4)=0\left(i\right)$$$$replaceagainxbyx+2\Rightarrow f(x+2)-f(x+4)+f(x+6)=0\left(ii\right)$$$$adding\left(i\right)and\left(ii\right)$$$$\Rightarrow f\left(x\right)+f(x+6)=0\left(iii\right)$$$$replacexbyx+6\Rightarrow f(x+6)+f(x+12)=0\left(iv\right)$$$$substract\left(iii\right)by\left(iv\right)\Rightarrow f(x+12)=f\left(x\right)$$$$itdoesmeantf\left(x\right)isperiodicfunction$$$$withperiod12.$$$$\Rightarrow f\left(100\right)=f(8.12+4)=f\left(4\right)=4$$$$$$$$$$

Answered by Ar Brandon last updated on 29/Oct/20

$$f(\mathrm{x}+2)+f(\mathrm{x}-2)=f\left(\mathrm{x}\right)...\mathrm{eqn}\left(1\right)$$$$\mathrm{Replacing}\mathrm{x}\mathrm{with}\mathrm{x}-2\mathrm{we}\mathrm{have}$$$$f\left(\mathrm{x}\right)+f(\mathrm{x}-4)=f(\mathrm{x}-2)...\mathrm{eqn}\left(2\right)$$$$\mathrm{eqn}\left(1\right)+\mathrm{eqn}\left(2\right)$$$$\Rightarrow f(\mathrm{x}+2)=-f(\mathrm{x}-4)$$$$\Rightarrow f\left(\mathrm{x}\right)=-f(\mathrm{x}-6)\{\mathrm{replacing}\mathrm{x}=\mathrm{x}-2\}$$$$\Rightarrow f(\mathrm{x}+6)=-f\left(\mathrm{x}\right)=f(\mathrm{x}-6)$$$$\Rightarrow f\left(\mathrm{x}\right)=f(\mathrm{x}-12)$$$$\mathrm{Period}\mathrm{is}12\Rightarrow f\left(100\right)=f[4+12\left(8\right)]$$$$\Rightarrow f\left(100\right)=f\left(4\right)=4$$

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