JEE Main 2023 — Mathematics Algebra

Given,

$a,b$a,b be two real numbers such that $ab<0$ab<0, $|\frac{1+ai}{b+i}|=1$∣b+i1+ai​∣=1, $a+ib$a+ib lies on the circle $|z-1|=|2z|$∣z−1∣=∣2z∣ and $ab<0$ab<0

Now using $|\frac{1+ai}{b+i}|=1$∣b+i1+ai​∣=1 we get,

$$\Rightarrow |1+ai|=|b+i|$$⇒∣1+ai∣=∣b+i∣

Now squaring both side we get,

$$\Rightarrow {|1+ai|}^{2}={|b+i|}^{2}$$⇒∣1+ai∣2=∣b+i∣2

$$\Rightarrow (1+ai)(1-ai)=(b+i)(b-i){{|z|}^{2}=z\bar{z}}$$⇒(1+ai)(1−ai)=(b+i)(b−i)∣z∣2=zzˉ

$$\Rightarrow 1+{a}^{2}={b}^{2}+1$$⇒1+a2=b2+1

$\Rightarrow a=\pm b\Rightarrow b=-a$⇒a=±b⇒b=−a as $ab<0$ab<0

Now given,

$(a,b)$(a,b) lies on $|z-1|=|2z|$∣z−1∣=∣2z∣

So, $|a+ib-1|=2|a+ib|$∣a+ib−1∣=2∣a+ib∣

$$\Rightarrow {|a-1+ib|}^{2}=4{|a+ib|}^{2}$$⇒∣a−1+ib∣2=4∣a+ib∣2

$$\Rightarrow {(a-1)}^{2}+{b}^{2}=4({a}^{2}+{b}^{2})$$⇒(a−1)2+b2=4(a2+b2)

$$\Rightarrow {(a-1)}^{2}+{a}^{2}=4(2{a}^{2}){\text{as}b=-a}$$⇒(a−1)2+a2=4(2a2)asb=−a

$$\Rightarrow 6{a}^{2}+2a-1=0$$⇒6a2+2a−1=0

$$\Rightarrow a=\frac{-2\pm \sqrt{28}}{12}=\frac{-1\pm \sqrt{7}}{6}$$⇒a=12−2±28​​=6−1±7​​

Taking positive sign we get,

a=\frac{\sqrt{7}-1}{6}&b=\frac{1-\sqrt{7}}{6} ${\text{as}b=-a}$asb=−a

Now the value of $[a]=0$[a]=0 as $0<\frac{\sqrt{7}-1}{6}<1$0<67​−1​<1

$$\therefore \frac{1+[a]}{4b}=\frac{6}{4(1-\sqrt{7})}=-(\frac{1+\sqrt{7}}{4})$$∴4b1+[a]​=4(1−7​)6​=−(41+7​​)

Now taking negative sign we get, a=\frac{-1-\sqrt{7}}{6}&b=\frac{\sqrt{7}+1}{6}

Now $[a]=-1{\text{as}-1<\frac{-1-\sqrt{7}}{6}<0}$[a]=−1as−1<6−1−7​​<0

So, the value of $\frac{1+[a]}{4b}=\frac{1+(-1)}{4b}=\frac{0}{4b}=0$4b1+[a]​=4b1+(−1)​=4b0​=0

Note- Official answer key was drop by NTA, here we have modified the option.