NIMCET 2025 Paper

Section: Mathematics

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Question : 14 of 120

Marks: +1, -0

Given the equations:

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

Let

N

be the number of ordered solution pairs

(x, y)

.
Find the value of

A N

.
Options:

9
$\;{7}/{2}$
3
$\;{19}/{2}$

Solution:

Given:

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

Use:

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

\;1=2+2 x y ⇒ x y=-\;{1}/{2}

So

x, y

are roots of:

t^2-t-\;{1}/{2}=0

This gives two distinct real roots, so the ordered pairs

(x, y)

are:

(x, y) \;\; \;\text " and "\; \;\; (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}-x y S_{n-2} .

Given:

S_0=2, \;\; S_1=1, \;\; x y=-\;{1}/{2} .

Compute:

S_2=2,

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

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

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

Thus:

A=\;{19}/{4} .

Finally:

A N=\;{19}/{4} ⋅ 2=\;{19}/{2} .

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