QJEEM20Ma0a89: JEE Main 2020 Mathematics - Class 11 JEEM20Ma0a89 Solution: Let f : R → R be a function which satisfies f(x + y) = f(x) +…

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) = $$\sum _{k=1}^{\left(n-1\right)}f\left(k\right)$$

= f(1) + f(2) +.........+ f(n - 1)

= 2 + 4 + 6 + ......+ 2(n - 1)

= 2 $$\times$$ $$\frac{(n-1)\left(n\right)}{2}$$

= n² - n

Given g(n) = 20

$$\Rightarrow$$ n²– n = 20

$$\Rightarrow$$ n² – n – 20 = 0

$$\Rightarrow$$ n = 5