JEE 2023 — Mathematics PYQ

Solution

1. Newton’s Theorem: Agar $x^2-7x-1=0$ ke roots $a,b$ hain aur $S_n=a^n+b^n$, toh relation hoga: $S_n-7S_{n-1}-S_{n-2}=0⟹S_n-S_{n-2}=7S_{n-1}$.

2. Numerator simplify karein: Hame diya gaya hai $\frac{S_{21}+S_{17}}{S_{19}}$.

3. Equation use karein:

   • $n=21$ ke liye: $S_{21}-7S_{20}-S_{19}=0⟹S_{21}-S_{19}=7S_{20}$

   • $n=19$ ke liye: $S_{19}-7S_{18}-S_{17}=0⟹S_{19}+S_{17}=7S_{18}$... par yahan direct substitution behtar hai.

   • Roots property se: $a^2-7a-1=0⟹a^2+1=7a$ aur $b^2+1=7b$.

   • Numerator $=a^{19}(a^2+1)+b^{19}(b^2+1)+(a^{17}+b^{17})$... Actually, best tarika hai:

   • $a^2-1=7a⟹a^2+1\ne 7a$. Wait, equation is $x^2-7x-1=0$, so $a^2-1=7a$.

   • $a^{21}-a^{19}=7a^{20}$ (Nahi).

     Let's use $a^2-7a-1=0⟹a^{21}-7a^{20}-a^{19}=0$.

     Hame chahiye $\frac{(a^{21}+a^{17})+(b^{21}+b^{17})}{a^{19}+b^{19}}$.

     $a^2-1=7a⟹a^4-2a^2+1=49a^2⟹a^4+1=51a^2$.

     Isi tarah $b^4+1=51b^2$.

     Numerator $=a^{17}(a^4+1)+b^{17}(b^4+1)=a^{17}(51a^2)+b^{17}(51b^2)=51(a^{19}+b^{19})$.

4. Final Calculation:

   $$\frac{51(a^{19}+b^{19})}{a^{19}+b^{19}}=51$$

Answer: 51