Solve $|x\,-\,2| + |x\,-\,1| = x\,-\,3$

If a function $g(x)$ is defined in $[-1, 1]$ and two vertices of an equilateral triangle are $(0, 0)$ and $(x, g(x))$ and its area is $\frac{\sqrt 3}{4}$ , then $g(x)$ equals :-

Suppose $\quad f : R \rightarrow(0, \infty)$ be a differentiable function such that $5 f ( x + y )= f ( x ) \cdot f ( y ), \forall x , y \in R$. If $f(3)=320$, then $\sum \limits_{n=0}^5 f(n)$ is equal to :

Let a function $f : R \rightarrow  R$ is defined such that $3f(2x^2 -3x + 5) + 2f(3x^2 -2x + 4) = x^2 -7x + 9\ \ \  \forall  x \in R$, then the value of $f(5)$ is-

If $f(x)$ is a polynomial function satisfying the condition $f(x) . f(1/x) = f(x) + f(1/x)$ and $f(2) = 9$ then :

Set of all values of $x$ satisfying

$\frac{{{x^4} - 4{x^3} + 3{x^2}}}{{({x^2} - 4)({x^2} - 7x + 10)}} \ge 0$