Let f : R - { 0 } R be a function satisfying f ( x y ) = f ( x ) f ( y ) for all x , y , ? f ( y ) ? 0 . If f ' ( 1 ) = 2024 , then [2024]

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Q.
Let $f : R - {0} \to R$ be a function satisfying $f (\frac{x}{y}) = \frac{f (x)}{f (y)} for all x, y, f (y) \neq 0 .$ If $f' (1) = 2024,$ then [2024]

1 $x f' (x) + f (x) = 2024$

2 $x f' (x) - 2024 f (x) = 0$

3 $x f' (x) - 2023 f (x) = 0$

4 $x f' (x) + 2024 f (x) = 0$

Ans.
(2)

$We have, f (\frac{x}{y}) = \frac{f (x)}{f (y)} and f' (1) = 2024$

$On putting x = y = 1, we get$

$f (1) = 1$ ...(i)

$Also, on putting x = 1, we get$

$f (\frac{1}{y}) = \frac{f (1)}{f (y)} = \frac{1}{f (y)} (Using (i))$

$\Rightarrow f (y) = \pm y^{n} (∵ I f f (x) f (\frac{1}{x}) = 1, then f (x) = \pm x^{n})$

$\Rightarrow f (y) = y^{n} (∵ f (1) = 1)$

$\Rightarrow f' (y) = n y^{n - 1} \Rightarrow f' (1) = n = 2024$

$Now, f (x) = x^{2024}$

$\Rightarrow f' (x) = 2024 x^{2023} \Rightarrow x f' (x) = 2024 x^{2024}$

$∴ x f' (x) - 2024 x^{2024} = 0$

Solution

Q.
Let $f : R - {0} \to R$ be a function satisfying $f (\frac{x}{y}) = \frac{f (x)}{f (y)} for all x, y, f (y) \neq 0 .$ If $f' (1) = 2024,$ then [2024]

1 $x f' (x) + f (x) = 2024$

2 $x f' (x) - 2024 f (x) = 0$

3 $x f' (x) - 2023 f (x) = 0$

4 $x f' (x) + 2024 f (x) = 0$

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Solution

Q. Let f:R-{0}→R be a function satisfying f(xy)=f(x)f(y) for all x,y, f(y)≠0. If f'(1)=2024, then [2024]

1 xf'(x)+f(x)=2024

2 xf'(x)-2024f(x)=0

3 xf'(x)-2023f(x)=0

4 xf'(x)+2024f(x)=0

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Q.
Let $f : R - {0} \to R$ be a function satisfying $f (\frac{x}{y}) = \frac{f (x)}{f (y)} for all x, y, f (y) \neq 0 .$ If $f' (1) = 2024,$ then [2024]

1 $x f' (x) + f (x) = 2024$

2 $x f' (x) - 2024 f (x) = 0$

3 $x f' (x) - 2023 f (x) = 0$

4 $x f' (x) + 2024 f (x) = 0$