JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 14)
If f(x + y) = f(x)f(y) and $$\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$$ , x, y $$ \in $$ N, where N is the set of all natural number, then the
value of
$${{f\left( 4 \right)} \over {f\left( 2 \right)}}$$ is :
$${2 \over 3}$$
$${1 \over 9}$$
$${1 \over 3}$$
$${4 \over 9}$$
Explanation
f(x + y) = f(x)f(y)
$$\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$$
$$ \Rightarrow $$ f(1) + f(2) + f(3) + ........$$\infty $$ = 2 ....(1)
On f(x + y) = f(x) f(y)
* Put x = 1, y = 1
f(2) = (f(1))2
* Put x = 2, y = 1
f(3) = f(2). f(1) = f((1))3
* Put x = 2, y = 2
f(4) = f((2))2 = f((1))4
Now put these values in equation (1)
f(1) + f((1))2 + f((1))3 + ....... = 2
$$ \Rightarrow $$ $${{f\left( 1 \right)} \over {1 - f\left( 1 \right)}}$$ = 2
$$ \Rightarrow $$ f(1) = $${2 \over 3}$$
Now f(2) = $${\left( {{2 \over 3}} \right)^2}$$
and f(4) = $${\left( {{2 \over 3}} \right)^4}$$
$$ \therefore $$ $${{f\left( 4 \right)} \over {f\left( 2 \right)}}$$
= $${{{{\left( {{2 \over 3}} \right)}^4}} \over {{{\left( {{2 \over 3}} \right)}^2}}}$$ = $${4 \over 9}$$
$$\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$$
$$ \Rightarrow $$ f(1) + f(2) + f(3) + ........$$\infty $$ = 2 ....(1)
On f(x + y) = f(x) f(y)
* Put x = 1, y = 1
f(2) = (f(1))2
* Put x = 2, y = 1
f(3) = f(2). f(1) = f((1))3
* Put x = 2, y = 2
f(4) = f((2))2 = f((1))4
Now put these values in equation (1)
f(1) + f((1))2 + f((1))3 + ....... = 2
$$ \Rightarrow $$ $${{f\left( 1 \right)} \over {1 - f\left( 1 \right)}}$$ = 2
$$ \Rightarrow $$ f(1) = $${2 \over 3}$$
Now f(2) = $${\left( {{2 \over 3}} \right)^2}$$
and f(4) = $${\left( {{2 \over 3}} \right)^4}$$
$$ \therefore $$ $${{f\left( 4 \right)} \over {f\left( 2 \right)}}$$
= $${{{{\left( {{2 \over 3}} \right)}^4}} \over {{{\left( {{2 \over 3}} \right)}^2}}}$$ = $${4 \over 9}$$
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