Another JEE Mains problem

For $f(x)=\dfrac 1{\sqrt{\{x+1\}-x^2+2x}}$ to be real and defined,

$\begin{aligned} {\color{#3D99F6}\{x+1\}}-x^2+2x & > 0 & \small \color{#3D99F6} \text{Note that }\{x+1\} = \{x\} \\ {\color{#3D99F6}\{x\}}-x^2+2x & > 0 \\ x^2 - 2x - \{x\} & < 0 & \small \color{#3D99F6} \text{Note that } x = \lfloor x \rfloor + \{x\} \\ (\lfloor x \rfloor + \{x\})^2 - 2(\lfloor x \rfloor + \{x\}) - \{x\} & < 0 \\ \lfloor x \rfloor(\lfloor x \rfloor - 2) + (2\lfloor x \rfloor + \{x\} - 3)\{x\} & < 0 \end{aligned}$

We note that for $x \ge 2$ , the $\text{LHS } \ge 0$ , then $f(x)$ is not real. Also that $x=0$ , the $\text{LHS }=0$ , $f(x)$ is not defined. $f(x)$ is real and defined for $x \in (0, 2)$ .

Now consider for $x < 0$ . From

$\begin{aligned} x^2 - 2x - \{x\} < 0 \\ (x-1)^2 & < 1 + \{x\} < 2 \\ \implies x & > -1 \end{aligned}$

Again, consider

$\begin{aligned} x^2 - 2x - \{x\} & < 0 & \small \color{#3D99F6} \text{For } - 1< x < 0, \ \{x\} = 1+x \\ x^2 - 3x - 1 & < 0 \\ x & > \frac {3-\sqrt {13}}2 & \small \color{#3D99F6} \text{Note that } x < 0 \end{aligned}$

Therefore, $f(x)$ is real and defined for $x \in \left(\dfrac {3-\sqrt{13}}2, 2\right) \backslash \{0\}$ .

$\implies a+b+c+d = 3+13+2+0 = \boxed{18}$

We note that $f(x)$ is real and defined when $(x-1)^2 < 1+\{x\}$ . The following graph clearly shows the solution.