JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 3)
Let f : R $$ \to $$ R be a function which satisfies
f(x + y) = f(x) + f(y) $$\forall $$ x, y $$ \in $$ R. If f(1) = 2 and
g(n) = $$\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)} $$, n $$ \in $$ N then the value of n, for which g(n) = 20, is :
f(x + y) = f(x) + f(y) $$\forall $$ x, y $$ \in $$ R. If f(1) = 2 and
g(n) = $$\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)} $$, n $$ \in $$ N then the value of n, for which g(n) = 20, is :
20
9
5
4
Explanation
Given f(1) = 2 ;
f(x + y) = f(x) + f(y)
When x = y = 1 $$ \Rightarrow $$ f(2) = 2 + 2 = 4
When x = 2, y = 1 $$ \Rightarrow $$ f(3) = 4 + 2 = 6
g(n) = $$\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)} $$
= f(1) + f(2) +.........+ f(n - 1)
= 2 + 4 + 6 + ......+ 2(n - 1)
= 2 $$ \times $$ $$\frac{\left( n-1\right) \left( n\right) }{2} $$
= n2 - n
Given g(n) = 20
$$ \Rightarrow $$ n2ā n = 20
$$ \Rightarrow $$ n2 ā n ā 20 = 0
$$ \Rightarrow $$ n = 5
f(x + y) = f(x) + f(y)
When x = y = 1 $$ \Rightarrow $$ f(2) = 2 + 2 = 4
When x = 2, y = 1 $$ \Rightarrow $$ f(3) = 4 + 2 = 6
g(n) = $$\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)} $$
= f(1) + f(2) +.........+ f(n - 1)
= 2 + 4 + 6 + ......+ 2(n - 1)
= 2 $$ \times $$ $$\frac{\left( n-1\right) \left( n\right) }{2} $$
= n2 - n
Given g(n) = 20
$$ \Rightarrow $$ n2ā n = 20
$$ \Rightarrow $$ n2 ā n ā 20 = 0
$$ \Rightarrow $$ n = 5
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