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Question Number 33571 by Joel578 last updated on 19/Apr/18

f(x) = x^(20) + a_1 x^(19) + a_2 x^(18) + ... + a_(20) If f(1) = f(2) = f(3) = ... = f(20) What is the value of a_1 ?

$${f}\left({x}\right)\:=\:{x}^{\mathrm{20}} \:+\:{a}_{\mathrm{1}} {x}^{\mathrm{19}} \:+\:{a}_{\mathrm{2}} {x}^{\mathrm{18}} \:+\:...\:+\:{a}_{\mathrm{20}} \\ $$$$\mathrm{If}\:{f}\left(\mathrm{1}\right)\:=\:{f}\left(\mathrm{2}\right)\:=\:{f}\left(\mathrm{3}\right)\:=\:...\:=\:{f}\left(\mathrm{20}\right) \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}_{\mathrm{1}} \:? \\ $$

Answered by MJS last updated on 19/Apr/18

let me downsize this f(x)=x^2 +a_1 x+a_2 f(1)=1+a_1 +a_2 f(2)=4+2a_1 +a_2 f(2)−f(1)=0 3+a_1 =0 ⇒ a_1 =−3 f(x)=x^3 +a_1 x^2 +a_2 x+a_3 f(1)=1+a_1 +a_2 +a_3 f(2)=8+4a_1 +2a_2 +a_3 f(3)=27+9a_1 +3a_2 +a_3 f(3)−f(2)=0 19+5a_1 +a_2 =0 (i) f(2)−f(1)=0 7+3a_1 +a_2 =0 (ii) (i)−(ii) 12+2a_1 =0 a_1 =−6 f(x)=x^4 +... ⇒ ⇒ a_1 =−10 f(x)=x^5 +... ⇒ ⇒ a_1 =−15 f(x)=x^n +... ⇒ ⇒ a_1 =−(n/2)(n+1) ⇒ f(x)=x^(20) +... ⇒ a_1 =−210

$$\mathrm{let}\:\mathrm{me}\:\mathrm{downsize}\:\mathrm{this} \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{4}+\mathrm{2}{a}_{\mathrm{1}} +{a}_{\mathrm{2}} \\ $$$${f}\left(\mathrm{2}\right)−{f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{3}+{a}_{\mathrm{1}} =\mathrm{0}\:\Rightarrow\:{a}_{\mathrm{1}} =−\mathrm{3} \\ $$$$ \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} +{a}_{\mathrm{1}} {x}^{\mathrm{2}} +{a}_{\mathrm{2}} {x}+{a}_{\mathrm{3}} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{8}+\mathrm{4}{a}_{\mathrm{1}} +\mathrm{2}{a}_{\mathrm{2}} +{a}_{\mathrm{3}} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{27}+\mathrm{9}{a}_{\mathrm{1}} +\mathrm{3}{a}_{\mathrm{2}} +{a}_{\mathrm{3}} \\ $$$${f}\left(\mathrm{3}\right)−{f}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{19}+\mathrm{5}{a}_{\mathrm{1}} +{a}_{\mathrm{2}} =\mathrm{0}\:\left({i}\right) \\ $$$${f}\left(\mathrm{2}\right)−{f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{7}+\mathrm{3}{a}_{\mathrm{1}} +{a}_{\mathrm{2}} =\mathrm{0}\:\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right) \\ $$$$\mathrm{12}+\mathrm{2}{a}_{\mathrm{1}} =\mathrm{0} \\ $$$${a}_{\mathrm{1}} =−\mathrm{6} \\ $$$$ \\ $$$${f}\left({x}\right)={x}^{\mathrm{4}} +...\:\Rightarrow \\ $$$$\Rightarrow\:{a}_{\mathrm{1}} =−\mathrm{10} \\ $$$$ \\ $$$${f}\left({x}\right)={x}^{\mathrm{5}} +...\:\Rightarrow \\ $$$$\Rightarrow\:{a}_{\mathrm{1}} =−\mathrm{15} \\ $$$$ \\ $$$${f}\left({x}\right)={x}^{{n}} +...\:\Rightarrow \\ $$$$\Rightarrow\:{a}_{\mathrm{1}} =−\frac{{n}}{\mathrm{2}}\left({n}+\mathrm{1}\right) \\ $$$$ \\ $$$$\Rightarrow\:{f}\left({x}\right)={x}^{\mathrm{20}} +...\:\Rightarrow\:{a}_{\mathrm{1}} =−\mathrm{210} \\ $$

Commented by Joel578 last updated on 20/Apr/18

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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