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QJEEM23Mafb24: JEE Main 2023 Mathematics - Class 11 JEEM23Mafb24 Solution: Let be two real numbers such that . IF the complex number is of …

Let a,ba,b be two real numbers such that ab < 0 . IF the complex number 1+aib+i\frac{1+ai}{b+i} is of unit modulus and a+iba+ib lies on the circle ∣z−1∣=∣2z∣|z-1|=|2z| , then a possible value of 1+[a]4b\frac{1+[a]}{4b} , where [t][t] is greatest integer function, is :

Multiple Choice Options

  • A (1+74)\left(\frac{1+\sqrt{7}}{4}\right)
  • B 12\frac{1}{2}
  • C 0
  • D −- 1

Step-by-Step Verified Solution

amp;∣1+aib+i∣=1amp;∣1+ia∣=∣b+i∣amp;a2+1=b2+1⇒a=±b⇒b=−a as ablt;0amp;(a+ib) lies on ∣z−1∣=∣2z∣amp;∣a+ib−1∣=2∣a+ib∣amp;(a−1)2+b2=4(a2+b2)amp;(a−1)2=a2=4(2a2)amp;1−2a=6a2⇒6a2+2a−1=0amp;a=−2±2812=−1±76amp;a=7−16 and b=1−76\begin{aligned} & \left|\frac{1+a i}{b+i}\right|=1 \\\\ & |1+i a|=|b+i| \\\\ & a^2+1=b^2+1 \Rightarrow \mathrm{a}=\pm \mathrm{b} \Rightarrow \mathrm{b}=-\mathrm{a} \quad \text { as } \mathrm{ab}<0 \\\\ & (\mathrm{a} +\mathrm{ib}) \text { lies on }|z-1|=|2 z| \\\\ & |a+i b-1|=2|a+i b| \\\\ & (a-1)^2+b^2=4\left(a^2+b^2\right) \\\\ & (a-1)^2=a^2=4\left(2 a^2\right) \\\\ & 1-2 a=6 a^2 \Rightarrow 6 a^2+2 a-1=0 \\\\ & a=\frac{-2 \pm \sqrt{28}}{12}=\frac{-1 \pm \sqrt{7}}{6} \\\\ & a=\frac{\sqrt{7}-1}{6} \text { and } b=\frac{1-\sqrt{7}}{6}\end{aligned}

amp;[a]=0amp;∴1+[a]4b=64(1−7)=−(1+74)\begin{aligned}& {[a]=0} \\\\& \therefore \frac{1+[a]}{4 b}=\frac{6}{4(1-\sqrt{7})}=-\left(\frac{1+\sqrt{7}}{4}\right)\end{aligned}

Similarly when a=−1−76  and  b=1+76a = {{ - 1 - \sqrt 7 } \over 6}\,\text { and }\,b = {{1 + \sqrt 7 } \over 6}

then [ aa ] = -1

∴\therefore 1+[a]4b=1−14×1+76{{1 + \left[ a \right]} \over {4b}} = {{1 - 1} \over {4 \times {{1 + \sqrt 7 } \over 6}}} = 0

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