If f(x) is an even function and satisfies the relation x2f(x)−2f(1x)=g(x), where g(x) is an odd function, then f(5) equals to

g(−x)=−g(x)&f(−x)=f(x)⇒x2f(−x)−2f(−1x)=g(−x)⇒g(x)=0∴x2f(x)−2f(1x)=0⇒1x2f(1x)−2f(x)=0⇒f(x)=0∀x∈R−{0}

If f(x) is an even function and satisfies the relation x2f(x)−2f(1x)=g(x), where g(x) is an odd function, then f(5) equals to

g(−x)=−g(x)&f(−x)=f(x)⇒x2f(−x)−2f(−1x)=g(−x)⇒g(x)=0∴x2f(x)−2f(1x)=0⇒1x2f(1x)−2f(x)=0⇒f(x)=0∀x∈R−{0}

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If $f (x)$ is an even function and satisfies the relation $x^{2} f (x) - 2 f (\frac{1}{x}) = g (x),$ where $g (x)$ is an odd function, then $f (5)$ equals to

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$\frac{49}{75}$

$\frac{52}{37}$

$\frac{50}{73}$

answer is C.

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$g (- x) = - g (x) & f (- x) = f (x) \Rightarrow x^{2} f (- x) - 2 f (\frac{- 1}{x}) = g (- x) \Rightarrow g (x) = 0$

$∴ x^{2} f (x) - 2 f (\frac{1}{x}) = 0 \Rightarrow \frac{1}{x^{2}} f (\frac{1}{x}) - 2 f (x) = 0 \Rightarrow f (x) = 0 \forall x \in R - {0}$

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If f(x) is an even function and satisfies the relation x2f(x)−2f(1x)=g(x), where g(x) is an odd function, then f(5) equals to

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If $f (x)$ is an even function and satisfies the relation $x^{2} f (x) - 2 f (\frac{1}{x}) = g (x),$ where $g (x)$ is an odd function, then $f (5)$ equals to