NIMCET 2025 Mathematics PYQ — Given the equation , , and . Let be the number of solution pairs … | Mathem Solvex | Mathem Solvex
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NIMCET 2025 — Mathematics PYQ
NIMCET | Mathematics | 2025
Given the equation x+y=1, x2+y2=2, and x5+y5=A. Let N be the number of solution pairs (x,y) to this system of equations. Then the value of AN is:
Choose the correct answer:
A.
9
B.
3
C.
7/2
D.
19/2
(Correct Answer)
Correct Answer:
19/2
Explanation
1. Finding the number of solutions (N):
Given the first two equations:
x+y=1
x2+y2=2
From (1), we have y=1−x. Substituting this into (2):
x2+(1−x)2=2x2+1−2x+x2=22x2−2x−1=0Using the quadratic formula x=2a−b±b2−4ac:
x=42±4−4(2)(−1)=42±12=21±3This gives two distinct values for x, each with a corresponding unique value for y. Thus, there are two solution pairs (x,y).
N=2**2. Finding the value of A (x5+y5):**
We use the properties of symmetric polynomials. Let sn=xn+yn.
From (1), we have y=1−x. Substituting this into (2):
x2+(1−x)2=2x2+1−2x+x2=22x2−2x−1=0Using the quadratic formula x=2a−b±b2−4ac:
x=42±4−4(2)(−1)=42±12=21±3This gives two distinct values for x, each with a corresponding unique value for y. Thus, there are two solution pairs (x,y).
N=2**2. Finding the value of A (x5+y5):**
We use the properties of symmetric polynomials. Let sn=xn+yn.
