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UPSC NDA I 2023 Mathematics Solved Paper

Question Numbers: 79-80

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

Let

f(x) = \sin(\pi^{2})x + \cos(-\pi^{2})x

where

[.]

is a greatest ‘integer function

Question : 79 of 120

Marks: +1, -0

What is

f(\frac{\pi }{2})

equal to ?

Solution:

Concept:

The greatest integer function

[x]

gives the largest integer

\le x

When

x

is negative,

[x]

moves to the next lower integer.

Explanation:

Given

f(x) = \sin([\pi^2]x) + \cos([-\pi^2]x)

Compute

\pi^2 \approx 9.869

, so

[\pi^2] = 9

For

-\pi^2 \approx -9.869

, the greatest integer

\le -9.869

-10

, so

[-\pi^2] = -10

Thus

f(x) = \sin(9x) + \cos(-10x)

Since

\cos(-\theta) = \cos\theta

, we have

f(x) = \sin 9x + \cos 10x

Now evaluate at

x = \frac{\pi}{2}

f\!\left(\frac{\pi}{2}\right) = \sin\left(9 \cdot \frac{\pi}{2}\right) + \cos\left(10 \cdot \frac{\pi}{2}\right) = \sin\frac{9\pi}{2} + \cos 5\pi

\sin\frac{9\pi}{2} = \sin\left(4\pi + \frac{\pi}{2}\right) = \sin\frac{\pi}{2} = 1

\cos 5\pi = \cos(4\pi + \pi) = \cos\pi = -1

Sum:

1 + (-1) = 0

Answer:

0

(Option B).

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