JEE 2023 — Mathematics PYQ

Solution

Since, $|1+ai|=|b+i|$

$\Rightarrow a^2+1=b^2+1\Rightarrow a^2=b^2$

And $|a+ib-1|=|2a+2bi|$

$\Rightarrow a^2+1-2a+b^2=4a^2+4b^2$

$\Rightarrow 3a^2+3b^2+2a-1=0$

$\Rightarrow 6a^2+2a-1=0(\because a^2=b^2)$

$\Rightarrow a=\frac{-2\pm \sqrt{4+24}}{12}$

$=\frac{-2\pm 2\sqrt{7}}{12}=\frac{-1\pm \sqrt{7}}{6}$

$\therefore (a,b)\equiv \left(\frac{-1+\sqrt{7}}{6},\frac{1-\sqrt{7}}{6}\right)$ or $\left(\frac{-1-\sqrt{7}}{6},\frac{1+\sqrt{7}}{6}\right)$

$\Rightarrow [a]=0$ (for the first case)

$\frac{1+[a]}{4b}=0$ or $\frac{-(1+\sqrt{7})}{4}$

Correct Option: (4)