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UPSC NDA I 2016 Mathematics Solved Paper

Question Numbers: 16-17

Direction: For the next two (2) items that follow:

Consider the functions

f(x)=xg(x) and g(x)=[\frac{1}{x}]

Where [⋅] is the greatest integer function.

Question : 16 of 120

Marks: +1, -0

What is

\overset{\frac{1}{2}}{\underset{\frac{1}{3}}{∫}}g(x)dx

equal to?

Solution:

Concept:

The greatest integer function

[x]

gives the largest integer less than or equal to

x

\frac{1}{x}

lies between two consecutive integers, then

[1/x]

is constant over that interval.

Explanation:

We need

I = \int_{1/3}^{1/2} g(x)\,dx

where

g(x)=[1/x]

For

x

(\frac{1}{3},\frac{1}{2})

, we have

\frac{1}{3} < x < \frac{1}{2}

Taking reciprocals reverses the inequalities:

3 > \frac{1}{x} > 2

, i.e.

2 < \frac{1}{x} < 3

Therefore

[1/x] = 2

for every

x

in this interval.

Hence

I = \int_{1/3}^{1/2} 2 \, dx = 2\left[x\right]_{1/3}^{1/2} = 2\left( \frac{1}{2} - \frac{1}{3} \right)

Compute:

2 \times \frac{1}{6} = \frac{1}{3}

Answer:

\frac{1}{3}

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