Let $f (x)= \{x\} [x]$; $g(x)=ax^2$ the sum of all real solutions of equation satisfying $f(x) = g(x)$ is 420 (where a is (+ve) rational number), then a is equal to (where [.] and {.} represents greatest integer function and fractional part function respectively
A) $\frac{28}{900}$
B)$\frac{29}{900}$
C)$\frac{31}{900}$
D)$\frac{37}{900}$
Let $n = \lfloor x \rfloor$ and $r = \{x\} = x - n$ be the integer and fractional parts of $x$.
The equation $f(x) = g(x)$ then becomes $nr = a(n + r)^2$. Or equivalently,
$$ar^2 + (2an - n)r + an^2 = 0$$
We can solve for $r$ with the quadratic formula.
$$r = n \frac{1 - 2a \pm \sqrt{1 - 4a}}{2a}$$
(The requirement that all numbers involved be real gives us the constraint $a \le \frac{1}{4}$, but all of the answer choices satisfy this anyway.)
But, by definition of the fractional part, we know $r \in [0, 1)$. Thus,
$$0 \le n \frac{1 - 2a \pm \sqrt{1 - 4a}}{2a} < 1$$ $$0 \le n (1 - 2a \pm \sqrt{1 - 4a}) < 2a$$ $$0 \le n < \frac{2a}{1 - 2a \pm \sqrt{1 - 4a}}$$
This gives us a constraint on the possible values for $n$ (and thus on $x$). For brevity, let $m = \frac{2a}{1 - 2a \pm \sqrt{1 - 4a}}$. Let $x_n$ be the solution to $f(x) = g(x)$ corresponding to a choice of $n$ within this range. Then we are given:
$$\sum_{n=0}^{m-1} x_i = 420$$ $$\sum_{n=0}^{m-1} (n + n \frac{1 - 2a \pm \sqrt{1 - 4a}}{2a}) = 420$$ $$(1 + \frac{1 - 2a \pm \sqrt{1 - 4a}}{2a}) \sum_{n=0}^{m-1} n = 420$$ $$(1 + \frac{1}{m}) \frac{m(m-1)}{2} = 420$$ $$(m+1)(m-1) = 840$$ $$m^2 - 841 = 0$$ $$m = 29$$
Now that we have our upper bound $m$, let's solve for $a$.
$$\frac{2a}{1 - 2a \pm \sqrt{1 - 4a}} = 29$$ $$2a = 29(1 - 2a \pm \sqrt{1 - 4a})$$ $$2a = 29 - 58a \pm 29\sqrt{1 - 4a})$$ $$60a - 29 = \pm 29\sqrt{1 - 4a})$$ $$(60a - 29)^2 = 29^2(1-4a)$$ $$3600a^2 - 3480a + 841 = 841 - 3364a$$ $$3600a^2 - 116a = 0$$ $$a(3600a - 116) = 0$$ $$a = 0 \lor a = \frac{116}{3600} = \frac{29}{900}$$
Since $a = 0$ is not a valid solution, we must have $\boxed{a = \frac{29}{900}}$.