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QJEEM25Ma9fe2: JEE Main 2025 Mathematics - Class 11 JEEM25Ma9fe2 Solution: Let the curve , divide the region into two parts of areas and . …

Let the curve z(1+i)+zˉ(1−i)=4,z∈Cz(1+i)+\bar{z}(1-i)=4, z \in C , divide the region ∣z−3∣≤1|z-3| \leq 1 into two parts of areas α\alpha and β\beta . Then ∣α−β∣|\alpha-\beta| equals :

Multiple Choice Options

  • A 1+π31+\frac{\pi}{3}
  • B 1+π61+\frac{\pi}{6}
  • C 1+π21+\frac{\pi}{2}
  • D 1+π41+\frac{\pi}{4}

Step-by-Step Verified Solution

JEE Main 2025 (Online) 22nd January Evening Shift Mathematics - Complex Numbers Question 6 English Explanation

amp; Let z=x+iyamp;(x+iy)(1+i)+(x−iy)(1−i)=4amp;x+ix+iy−y+x−ix−iy−y=4amp;2x−2y=4amp;x−y=2amp;∣z−3∣≤1amp;(x−3)2+y2≤1\begin{aligned}& \text { Let } z=x+i y \\& (x+i y)(1+i)+(x-i y)(1-i)=4 \\& x+i x+i y-y+x-i x-i y-y=4 \\& 2 x-2 y=4 \\& x-y=2 \\& |z-3| \leq 1 \\& (x-3)^2+y^2 \leq 1\end{aligned}

amp; Area of shaded region =π⋅124−12⋅1⋅1=π4−12amp; Area of unshaded region inside the circle amp;amp;=34π⋅12+12⋅1⋅1=3π4+12amp;∴ difference of area =(3π4+12)−(π4−12)amp;=π2+1\begin{aligned}&\text { Area of shaded region }=\frac{\pi \cdot 1^2}{4}-\frac{1}{2} \cdot 1 \cdot 1=\frac{\pi}{4}-\frac{1}{2}\\&\text { Area of unshaded region inside the circle }\\&\begin{aligned}& =\frac{3}{4} \pi \cdot 1^2+\frac{1}{2} \cdot 1 \cdot 1=\frac{3 \pi}{4}+\frac{1}{2} \\& \therefore \text { difference of area }=\left(\frac{3 \pi}{4}+\frac{1}{2}\right)-\left(\frac{\pi}{4}-\frac{1}{2}\right)\end{aligned}\\&=\frac{\pi}{2}+1\end{aligned}

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