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UPSC NDA II 2025 Mathematics Paper

Question Numbers: 29-30

For the following two (02) items:

Let

cos(2x+3y)=\frac{1}{2}

and

cos(3x+2y)=\frac{\sqrt{3}}{2}

where

-\pi <(2x+3y) < \pi

and

-\pi <(3x+2y) < \pi .

Question : 30 of 120

Marks: +1, -0

How many values does (y-x) have?

Solution:

Concept:

Use the range of inverse cosine to find all possible angle values for the given linear combinations.

Explanation:

Given

\cos(2x+3y)=\frac12

, and

-\pi<2x+3y < \pi

Thus

2x+3y = \frac{\pi}{3}

-\frac{\pi}{3}

Given

\cos(3x+2y)=\frac{\sqrt3}{2}

, and

-\pi<3x+2y < \pi

Thus

3x+2y = \frac{\pi}{6}

-\frac{\pi}{6}

Now subtract the equations:

(3x+2y)-(2x+3y)=x-y

x-y

takes the values

\frac{\pi}{6}-\frac{\pi}{3}

\frac{\pi}{6}+\frac{\pi}{3}

-\frac{\pi}{6}-\frac{\pi}{3}

-\frac{\pi}{6}+\frac{\pi}{3}

Each gives a distinct

x-y

, hence

y-x

also has four distinct values.

Answer:

Option C (Four)

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