For every pair of numbers a and b, the function f satisfies $b^{2} f (a) = a^{2} f (b)$ $b^2 f(a) = a^2 f(b)$ . If $f (2)$ $f(2)$ does not equal 0, find the value of $\frac{f (5) - f (1)}{f (2)}$ $[f(5) - f(1)] / f(2)$ ?

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For every pair of numbers a and b, the function f satisfies $b^{2} f (a) = a^{2} f (b)$ $b^2 f(a) = a^2 f(b)$ . If $f (2)$ $f(2)$ does not equal 0, find the value of $\frac{f (5) - f (1)}{f (2)}$ $[f(5) - f(1)] / f(2)$ ?

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Answer 1 · 2017-06-02T15:25:51

$6$ $6$

Explanation:

Given:

$b^{2} f (a) = a^{2} f (b)$ $b^2f(a) = a^2f(b)$

Let $k = \frac{f (2)}{4} \neq 0$ $k=f(2)/4 != 0$

Then, putting $b = 2$ $b=2$ we find that for any $a$ $a$

$4 f (a) = a^{2} \cdot f (2) = 4 k a^{2}$ $4f(a) = a^2*f(2) = 4ka^2$

Dividing both ends by $4$ $4$ , we find:

$f (a) = k a^{2}$ $f(a) = ka^2$

So:

$\frac{f (5) - f (1)}{f (2)} = \frac{k \cdot 5^{2} - k \cdot 1^{2}}{k \cdot 2^{2}} = \frac{25 - 1}{4} = 6$ $(f(5)-f(1))/f(2) = (k*5^2-k*1^2)/(k*2^2) = (25-1)/4 = 6$

Answer 2 · 2017-06-02T16:30:07

$6$ $6$

Explanation:

Making $b = λ a$ $b = lambda a$

$b^{2} f (a) = a^{2} f (b) \Rightarrow f (λ a) = λ^{2} f (a)$ $b^2f(a)=a^2f(b) rArr f(lambda a) = lambda^2f(a)$ now making $a = 1$ $a=1$ we have

$f (λ) = λ^{2} f (1)$ $f(lambda)=lambda^2f(1)$ and then

$\frac{f (5) - f (1)}{f (2)} = \frac{5^{2} f (1) - f (1)}{2^{2} f (1)} = 6$ $(f(5)-f(1))/f(2)=(5^2f(1)-f(1))/(2^2 f(1))=6$

NOTE: $f (1) \neq 0$ $f(1) ne 0$ because otherwise $f (2) = 0$ $f(2)=0$

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For every pair of numbers a and b, the function f satisfies $b^{2} f (a) = a^{2} f (b)$ $b^2 f(a) = a^2 f(b)$ . If $f (2)$ $f(2)$ does not equal 0, find the value of $\frac{f (5) - f (1)}{f (2)}$ $[f(5) - f(1)] / f(2)$ ?

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For every pair of numbers a and b, the function f satisfies b2f(a)=a2f(b)b^2 f(a) = a^2 f(b). If f(2)f(2) does not equal 0, find the value of f(5)−f(1)f(2)[f(5) - f(1)] / f(2)?

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Explanation:

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For every pair of numbers a and b, the function f satisfies $b^{2} f (a) = a^{2} f (b)$ $b^2 f(a) = a^2 f(b)$ . If $f (2)$ $f(2)$ does not equal 0, find the value of $\frac{f (5) - f (1)}{f (2)}$ $[f(5) - f(1)] / f(2)$ ?