Let f:R→R be a function which satisfies fx+y=fx+fy, ∀x, y∈R . If f1=2 and gn=∑k=1n-1fk, n∈N then the value of n, for which gn=20, is

Correct answer: (A) 5. ∵ fx+y=fx+fy ∴ fx=kx, ∀x∈R Now f1=2 ⇒ k=2 ∴ fx=2x, ∀x∈R gn=f1+f2+……..+fn-1 =21+2+3+……+n-1 =2·(n-1)/2n=20 ⇒n^2-n-20=0 ⇒n=5.

Q28 / 75 · JEE Main 2020 (02 Sep Shift 2) · Mathematics

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 $g (n) = \sum_{k = 1}^{(n - 1)} f (k),$ $n \in N$ then the value of $n$ , for which $g (n) = 20$ , is

$∵ f (x + y) = f (x) + f (y)$

$∴ f (x) = k x, \forall x \in R$

Now $f (1) = 2 \Rightarrow k = 2$

$∴ f (x) = 2 x, \forall x \in R$

$g (n) = f (1) + f (2) + \dots \dots . . + f (n - 1)$

$= 2 [1 + 2 + 3 + \dots \dots + (n - 1)]$

$= 2 \cdot \frac{(n - 1)}{2} n = 20$

$\Rightarrow n^{2} - n - 20 = 0$

$\Rightarrow n = 5$ .

Hence, the correct answer is option: 'A'.