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Author: Tinku Tara

f-R-R-g-R-R-f-x-y-f-x-g-y-g-x-f-y-g-x-y-f-x-f-y-g-x-g-y-f-x-g-x-

Question Number 1979 by prakash jain last updated on 27/Oct/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right){g}\left({y}\right)+{g}\left({x}\right){f}\left({y}\right) \\ $$$${g}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)+{g}\left({x}\right){g}\left({y}\right) \\ $$$${f}\left({x}\right)=? \\ $$$${g}\left({x}\right)=? \\ $$ Answered by…

f-R-R-g-R-R-f-x-y-f-x-g-x-f-y-g-y-g-x-y-f-x-f-y-g-x-g-y-f-x-2-g-x-2-f-x-g-y-g-x-f-y-f-x-g-x-

Question Number 1976 by 123456 last updated on 27/Oct/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right){g}\left({x}\right)+{f}\left({y}\right){g}\left({y}\right) \\ $$$${g}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)+{g}\left({x}\right){g}\left({y}\right) \\ $$$$\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\left[{g}\left({x}\right)\right]^{\mathrm{2}} =? \\ $$$$\left[{f}\left({x}\right)+{g}\left({y}\right)\right]\left[{g}\left({x}\right)+{f}\left({y}\right)\right]=?? \\ $$$${f}\left({x}\right)=??? \\…

f-0-R-xf-x-f-f-x-f-x-f-x-

Question Number 1970 by 123456 last updated on 27/Oct/15 $${f}:\left[\mathrm{0},+\infty\right)\rightarrow\mathbb{R} \\ $$$${xf}\left({x}\right)={f}\left[{f}\left({x}\right)\right]{f}\left({x}\right) \\ $$$${f}\left({x}\right)=? \\ $$ Answered by prakash jain last updated on 27/Oct/15 $${f}\left({x}\right)\neq\mathrm{0}\:\mathrm{then}\:{f}\left({f}\left({x}\right)={x}\right.…

Show-that-1n-3-2n-3n-2-is-divisible-by-2-and-3-for-all-positive-integers-n-

Question Number 67501 by TawaTawa last updated on 28/Aug/19 $$\mathrm{Show}\:\mathrm{that}\:\:\mathrm{1n}^{\mathrm{3}} \:+\:\mathrm{2n}\:+\:\mathrm{3n}^{\mathrm{2}} \:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}\:\mathrm{and}\:\mathrm{3}\:\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{n}. \\ $$ Commented by Prithwish sen last updated on 28/Aug/19 $$\mathrm{Another}\:\mathrm{approch} \\ $$$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}…