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1.Relation and Function
hard
ધારો કે $f= R \rightarrow(0, \infty)$ વિકલનીય વિધેય છે,જ્યાં $5 f(x+y)=f(x) . f(y), \forall x, y \in R$. જો $f(3)=320$ હોય,તો $\sum \limits_{ n =0}^5 f( n )=.......$
A
$6875$
B
$6575$
C
$6825$
D
$6528$
(JEE MAIN-2023)
Solution
$5 f ( x + y )= f ( x ) \cdot f ( y )$
$5 f (0)= f (0)^2 \Rightarrow f (0)=5$
$5 f ( x +1)= f ( x ) \cdot f (1)$
$\frac{ f ( x +1)}{ f ( x )}=\frac{ f (1)}{5}$
$\frac{ f (1)}{ f (0)} \cdot \frac{ f (2)}{ f (1)} \cdot \frac{ f (3)}{ f (2)}=\left(\frac{ f (1)}{5}\right)^3$
$\frac{320}{5}=\frac{( f (1))^3}{5^3} \Rightarrow f (1)=20$
$5 f ( x +1)=20 \cdot f ( x ) \Rightarrow f ( x +1)=4 f ( x )$
$\sum \limits_{ n =0}^5 f ( n )=5+5.4+5.4^2+5.4^3+5.4^4+5.4^5$
$=\frac{5\left[4^6-1\right]}{3}=6825$
Standard 12
Mathematics
