1.Relation and Function
normal

Let $f$ be a function satisfying $f(xy) = \frac{f(x)}{y}$ for all positive real numbers $x$ and $y.$ If $ f(30) = 20,$ then the value of $f(40)$ is-

A

$15$

B

$20$

C

$40$

D

$60$

Solution

$f(x y)=\frac{f(x)}{y}$

Put $y=3 $ and $ x=10$

$f(30)=\frac{f(10)}{3} $

$\Rightarrow f(10)=3 \times f(30)=3 \times 20=60$

Now put $y=4 $ and $ x=10$

$f(40)=\frac{f(10)}{3}$

$ \Rightarrow f(40)=\frac{60}{4}=15$

Standard 12
Mathematics

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