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UPSC CDS 2 2022 Math Paper

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Question Numbers: 41-42

Consider the following for the next two (02) items that follow:

Consider the equation

6x^{2}-25x+\frac{6}{x^{2}}+\frac{25}{x}+12=0

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Question : 42 of 100

Marks: +1, -0

What is one of the possible values of

x^{2} + \frac{1}{x^{2}}

?

6
$\frac{62}{9}$
8
$\frac{82}{9}$

Solution:

Concept:

Use the identity: if

x - \frac{1}{x} = a

, then

x^{2} + \frac{1}{x^{2}} = a^{2} + 2

.

Explanation:

Given equation:

6x^{2} - 25x + \frac{6}{x^{2}} + \frac{25}{x} + 12 = 0

.

Rewrite as

6\left(x^{2} + \frac{1}{x^{2}}\right) - 25\left(x - \frac{1}{x}\right) + 12 = 0

.

Let

a = x - \frac{1}{x}

. Then

x^{2} + \frac{1}{x^{2}} = a^{2} + 2

.

Substitute:

6(a^{2} + 2) - 25a + 12 = 0

→

6a^{2} + 12 - 25a + 12 = 0

→

6a^{2} - 25a + 24 = 0

.

Factor:

(2a - 3)(3a - 8) = 0

, so

a = \frac{3}{2}

or

a = \frac{8}{3}

.

For

a = \frac{3}{2}

:

x^{2} + \frac{1}{x^{2}} = \left(\frac{3}{2}\right)^{2} + 2 = \frac{9}{4} + 2 = \frac{17}{4}

(not in options).

For

a = \frac{8}{3}

:

x^{2} + \frac{1}{x^{2}} = \left(\frac{8}{3}\right)^{2} + 2 = \frac{64}{9} + 2 = \frac{82}{9}

.

Thus one possible value is

\frac{82}{9}

.

Answer:

\frac{82}{9}

(Option D).

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