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NIMCET 2025 Paper

Section: Mathematics

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Question : 14 of 120

Marks: +1, -0

Given the equations:

x+y=1, \quad x^2+y^2=2, \quad x^5+y^5=A.

N

be the number of ordered solution pairs

(x, y)

Find the value of

A N

9
$\frac{7}{2}$
3
$\frac{19}{2}$

Solution:

Given:

x+y=1, \quad x^2+y^2=2

Use:

\;(x+y)^2=x^2+y^2+2xy

\;1=2+2xy \Rightarrow xy=-\frac{1}{2}

So

x, y

are roots of:

t^2-t-\frac{1}{2}=0

This gives two distinct real roots, so the ordered pairs

(x, y)

are:

(x, y) \quad \text{ and } \quad (y, x),

hence

N=2

Find

A=x^5+y^5

Let

S_n=x^n+y^n

.

Recurrence:

S_n=(x+y)S_{n-1}-xyS_{n-2}.

Given:

S_0=2, \quad S_1=1, \quad xy=-\frac{1}{2}.

Compute:

S_2=2,

S_3=S_2+\frac{1}{2}S_1=2+0.5=2.5,

S_4=S_3+\frac{1}{2}S_2=2.5+1=3.5,

S_5=S_4+\frac{1}{2}S_3=3.5+1.25=4.75=\frac{19}{4}.

Thus:

A=\frac{19}{4}.

Finally:

A N=\frac{19}{4} \cdot 2=\frac{19}{2}.

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