JEE 2023 — Mathematics PYQ

Solution (Samadhan)

Ise solve karne ke liye hum complex numbers ka polar form istemal karenge:

1. $p$ aur $q$ ki value nikaalna:

   Hume pata hai ki $1-\sqrt{3}i$ ko hum aise likh sakte hain:

   $$2\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)=2\left(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}\right)=2e^{-i\pi /3}$$

   Ab power $200$ lene par:

   $$(1-\sqrt{3}i{)}^{200}=2^{200}{\left(e^{-i\pi /3}\right)}^{200}=2^{200}{e}^{-i200\pi /3}$$

   Angle ko simplify karne par: $\frac{200\pi }{3}=66\pi +\frac{2\pi }{3}$. Kyonki $66\pi$ ek complete rotation hai, hum ise $e^{-i2\pi /3}$ likh sakte hain:

   $$2^{200}\left(\cos \frac{2\pi }{3}-i\sin \frac{2\pi }{3}\right)=2^{200}\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)=2^{199}(-1-\sqrt{3}i)$$

   Equation se compare karne par:

   $2^{199}(p+iq)=2^{199}(-1-\sqrt{3}i)$

   Isse humein milta hai: $p=-1$ aur $q=-\sqrt{3}$.

2. Roots nikaalna:

   Pehla root $(\alpha )=p+q+q^2=-1-\sqrt{3}+(-\sqrt{3}{)}^2=-1-\sqrt{3}+3=2-\sqrt{3}$

   Doosra root $(\beta )=p-q+q^2=-1-(-\sqrt{3})+(-\sqrt{3}{)}^2=-1+\sqrt{3}+3=2+\sqrt{3}$

3. Quadratic Equation banana:

   • Roots ka sum $(\alpha +\beta )=(2-\sqrt{3})+(2+\sqrt{3})=4$

   • Roots ka product $(\alpha \beta )=(2-\sqrt{3})(2+\sqrt{3})=2^2-(\sqrt{3}{)}^2=4-3=1$

   Equation hoti hai: $x^2-(\text{sum})x+(\text{product})=0$

   $$x^2-4x+1=0$$