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UPSC NDA II 2024 Mathematics Paper

Question Numbers: 83-84

Direction : Consider the following for the items that follow :

Let

f(x) = \lfloor x \rfloor^{2} - \lfloor x^{2} \rfloor

Question : 83 of 120

Marks: +1, -0

What is f(0.999) + f(1.001) equal to

Solution:

Concept:

The function uses the greatest integer function

[x]

. Compute its value at two points near 1.

Explanation:

Given

f(x) = [x]^2 - [x^2]

For

x = 0.999

[0.999] = 0

, so

[0.999]^2 = 0

0.999^2 = 0.998001

, so

[0.999^2] = 0

Thus

f(0.999) = 0 - 0 = 0

For

x = 1.001

[1.001] = 1

, so

[1.001]^2 = 1

(1.001)^2 = 1.002001

, so

[(1.001)^2] = 1

Thus

f(1.001) = 1 - 1 = 0

Now

f(0.999) + f(1.001) = 0 + 0 = 0

Answer:

0 (Option B).

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