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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