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QJEEM23Ma1062: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1062 Solution: Let be two non-zero real numbers. Then the number of elements…

Let a≠ba \neq b be two non-zero real numbers. Then the number of elements in the set X={z∈C:Re⁡(az2+bz)=aX=\left\{z \in \mathbb{C}: \operatorname{Re}\left(a z^{2}+b z\right)=a\right. and Re⁡(bz2+az)=b}\left.\operatorname{Re}\left(b z^{2}+a z\right)=b\right\} is equal to :

Multiple Choice Options

  • A 0
  • B 2
  • C 1
  • D Infinite

Step-by-Step Verified Solution

Let z=x+iyz=x+i y

amp;⇒z2=x2−y2+2ixyamp;∴az2+bzamp;=a(x2−y2+2ixy)+b(x+iy)amp;=a(x2−y2)+bx+2aixy+biy\begin{array}{ll}&\Rightarrow z^2=x^2-y^2+2 i x y \\\\&\therefore a z^2+b z \\\\& =a\left(x^2-y^2+2 i x y\right)+b(x+i y) \\\\& =a\left(x^2-y^2\right)+b x+2 a i x y+b i y\end{array}

Re⁡(az2+bz)=b⇒a(x2−y2)+bx=a⇒x2−y2+bax=1.........(i)\begin{array}{ll}\operatorname{Re}\left(a z^2+b z\right)=b \\\\ \Rightarrow a\left(x^2-y^2\right)+b x=a \\\\ \Rightarrow x^2-y^2+\frac{b}{a} x=1 .........(i)\end{array}

amp; and bz2+azamp;=b(x2−y2+2ixy)+a(x+iy)amp;=b(x2−y2)+ax+2bixy+aiyamp;Re⁡(bz2+az)=bamp;⇒b(x2−y2)+ax=bamp;⇒x2−y2+abx=1..........(ii)\begin{aligned} & \text { and } b z^2+a z \\\\ & =b\left(x^2-y^2+2 i x y\right)+a(x+i y) \\\\ & =b\left(x^2-y^2\right)+a x+2 b i x y+a i y \\\\ & \operatorname{Re}\left(b z^2+a z\right)=b \\\\ & \Rightarrow b\left(x^2-y^2\right)+a x=b \\\\ & \Rightarrow x^2-y^2+\frac{a}{b} x=1 ..........(ii)\end{aligned}

On subtracting Equation (ii) from Equation (i), we get

amp;bax−abx=0amp;⇒(ba−ab)x=0amp;⇒x=0 or ba−ab=0amp;⇒b2−a2=0amp;⇒a=±bamp;⇒a=−b( since a≠b)\begin{aligned}& \frac{b}{a} x-\frac{a}{b} x=0 \\\\& \Rightarrow \left(\frac{b}{a}-\frac{a}{b}\right) x=0 \\\\& \Rightarrow x=0 \text { or } \frac{b}{a}-\frac{a}{b}=0 \\\\& \Rightarrow b^2-a^2=0 \\\\& \Rightarrow a= \pm b \\\\& \Rightarrow a=-b(\text { since } a \neq b)\end{aligned}

From (i), when x=0x=0 , then

amp;0−y2=1⇒y2=−1amp;⇒y∈ϕ⇒z∈ϕ has no solution. \begin{aligned}& 0-y^2=1 \Rightarrow y^2=-1 \\\\& \Rightarrow y \in \phi \Rightarrow z \in \phi \text { has no solution. }\end{aligned}

When, a=−ba=-b , then x2−y2−x=1x^2-y^2-x=1 has infinitely many solutions.

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