Test Index

UPSC CDS 2 2022 Math Paper

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

Question : 41 of 100

Marks: +1, -0

\frac{1}{x}

Solution:

Concept:

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

, then

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

Explanation:

Start with the given equation:

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

Group the terms with

x^2

and

\frac{1}{x^2}

, and the terms with

x

and

\frac{1}{x}

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 into the equation:

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

Simplify:

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

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

Factor the quadratic:

6a^{2} - 16a - 9a + 24 = 0

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

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

So,

a = \frac{3}{2}

a = \frac{8}{3}

Among the given options, only

\frac{3}{2}

is present.

Answer:

x - \frac{1}{x} = \frac{3}{2}

Thus, the correct option is B.

Go to Question:

Prev Question

Next Question