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)=Σ????f(k), n∈N, then the value of n for which g(n)=20 is

$3-2^x-2^{1-x}\ge 0put{2}^x=t$

$3-t-\frac{2}{t}\ge 0\Rightarrow t^2-3t+2\le 0$

$\left(t-1\right)\left(t-2\right)\le 0\Rightarrow t\in \left[1,2\right]$

? P? = 6!/2! = 360

19.00
Only '2' in range → 1 function
one element out of 1, 3, 4 is in range with '2'
number of ways = ³C? × (3!/(2!1!)) × 2! = 18
(Select one from 1, 3, 4 and distribute among a, b, c)
Total function = 1 + 18 = 19

$f\left(x\right)=\frac{1-x}{1+x},0<x<1$ (on simplification)

$f\text{'}\left(x\right)=\frac{-2}{{\left(1+x\right)}^2}$

${\left(1-x\right)}^{2}f\text{'}\left(x\right)=-2{\left(f\left(x\right)\right)}^2$

Use [x + n] = n + [x], where $n\in z$  f (x) = [x] – 10 will be minimum for $x\in [0,10]$  break the limits as G.I.F. is discontinuous at integral points.

${\displaystyle \underset{0}{\overset{10}{\int }}f\left(x\right)dx=-10-9-....-1}$    

$=\frac{-10.11}{2}$

${\displaystyle \underset{0}{\overset{10}{\int }}\left|f\left(x\right)\right|dx=10+9+....+1}$