Let f- - be a function which satisfies fx-y-fx-fy - x- y- If f1-2 and gn-n-1k-1 fk- n- then the value of n- for which gn-20- is-

F(1)=2; f(x+y)=f(x)+f(y) x=y=1⇒ f(2)=2+2=4 x=2, y=1⇒ f(3)=4+2=6 g(n)=f(1)+f(2)+........+f(n 1) =2+4+6+.........+2(n 1) =2 ∑ (n 1) =2(n 1).n2 =n^2 n Given g(n) ...

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Standard XII

Mathematics

Condition for Monotonically Decreasing Function

Let f: ℝ→ℝ be...

Question

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 $g (n) = (n - 1) \sum k = 1 f (k), n \in N$ then the value of $n$ , for which $g (n) = 20$ , is:

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Solution

The correct option is B $5$
$f (1) = 2; f (x + y) = f (x) + f (y)$
$x = y = 1 \Rightarrow f (2) = 2 + 2 = 4$
$x = 2, y = 1 \Rightarrow f (3) = 4 + 2 = 6$

$g (n) = f (1) + f (2) + . . . . . . . . + f (n - 1)$
$= 2 + 4 + 6 + . . . . . . . . . + 2 (n - 1)$
$= 2 \sum (n - 1)$
$= 2 \frac{(n - 1) . n}{2}$
$= n^{2} - n$

Given $g (n) = 20 \Rightarrow n^{2} - n = 20$
$n^{2} - n - 20 = 0$
$n = 5 (∵ n \in N)$

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Let f:R→R be a function which satisfies f(x+y)=f(x)+f(y) ∀x,y∈R. If f(1)=2 and g(n)=(n−1)∑k=1 f(k),n∈N then the value of n, for which g(n)=20, is:

The correct option is B 5f(1)=2;f(x+y)=f(x)+f(y) x=y=1⇒f(2)=2+2=4 x=2,y=1⇒f(3)=4+2=6 g(n)=f(1)+f(2)+........+f(n−1) =2+4+6+.........+2(n−1) =2 ∑(n−1) =2(n−1).n2 =n2−n Given g(n)=20 ⇒n2−n=20 n2−n−20=0 n=5(∵n∈N)

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 $g (n) = (n - 1) \sum k = 1 f (k), n \in N$ then the value of $n$ , for which $g (n) = 20$ , is: