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UPSC NDA I 2023 Mathematics Solved Paper

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Question Numbers: 84-85

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

Let

l=∫_{a}^{b}\frac{ | x | }{x}dx,

a < b

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Question : 84 of 120

Marks: +1, -0

What is l equal to when

a < 0

< b ?

a + b
a - b
b - a
$\frac{(a+b)}{2}$

Solution:

Concept:

The integral

\int_{a}^{b} \frac{|x|}{x} \, dx

is split at

x=0

because the sign of

\frac{|x|}{x}

changes at that point.

Explanation:

Given

a < 0 < b

, we evaluate

l = \int_{a}^{b} \frac{|x|}{x} \, dx

.

For

a < x < 0

,

|x| = -x

, so

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

.

For

0 < x < b

,

|x| = x

, so

\frac{|x|}{x} = \frac{x}{x} = 1

.

Thus

l = \int_{a}^{0} (-1) \, dx + \int_{0}^{b} 1 \, dx

.

Compute:

\int_{a}^{0} (-1) \, dx = [ -x ]_{a}^{0} = -(0 - a) = a

.

\int_{0}^{b} 1 \, dx = [ x ]_{0}^{b} = b - 0 = b

.

Therefore

l = a + b

.

Answer:

a + b

(Option A).

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