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If-f-x-is-an-even-function-and-satisfies-the-relation-x-2-f-x-2f-x-g-x-where-g-x-is-an-odd-function-then-the-value-of-f-5-is-a-0-b-37-75-c-4-d-51-77-




Question Number 128424 by bramlexs22 last updated on 07/Jan/21
If f(x) is an even function and  satisfies the relation x^2 f(x)−2f(x)=g(x)  where g(x) is an odd function   then the value of f(5) is   (a) 0  (b) ((37)/(75))   (c) 4    (d) ((51)/(77))
$$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{function}\:\mathrm{and} \\ $$$$\mathrm{satisfies}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{x}^{\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)−\mathrm{2f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\mathrm{where}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function}\: \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{5}\right)\:\mathrm{is}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{0}\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{37}}{\mathrm{75}}\:\:\:\left(\mathrm{c}\right)\:\mathrm{4}\:\:\:\:\left(\mathrm{d}\right)\:\frac{\mathrm{51}}{\mathrm{77}} \\ $$
Answered by mr W last updated on 07/Jan/21
x^2 f(x)−2f(x)=g(x)   ...(i)  (−x)^2 f(−x)−2f(−x)=g(−x)  x^2 f(x)−2f(x)=−g(x)   ...(ii)  (i)+(ii):  2(x^2 f(x)−2f(x))=0  (x^2 −2)f(x)=0  (5^2 −2)f(5)=0  ⇒f(5)=0
$${x}^{\mathrm{2}} {f}\left({x}\right)−\mathrm{2}{f}\left({x}\right)={g}\left({x}\right)\:\:\:…\left({i}\right) \\ $$$$\left(−{x}\right)^{\mathrm{2}} {f}\left(−{x}\right)−\mathrm{2}{f}\left(−{x}\right)={g}\left(−{x}\right) \\ $$$${x}^{\mathrm{2}} {f}\left({x}\right)−\mathrm{2}{f}\left({x}\right)=−{g}\left({x}\right)\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} {f}\left({x}\right)−\mathrm{2}{f}\left({x}\right)\right)=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{2}\right){f}\left({x}\right)=\mathrm{0} \\ $$$$\left(\mathrm{5}^{\mathrm{2}} −\mathrm{2}\right){f}\left(\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}\left(\mathrm{5}\right)=\mathrm{0} \\ $$

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