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UPSC NDA I 2024 Mathematics Paper

Question Numbers: 89-90

Consider the following for the items that follow :

Let

\varphi(a)=\int_a^{a+100 \pi} | \sin x | d x

Question : 90 of 120

Marks: +1, -0

What is φ'(a) equal to ?

Solution:

Concept:

Leibniz rule for differentiating an integral with variable limits is used here.

Explanation:

Given

\varphi(a) = \int_{a}^{a+100\pi} |\sin x| \, dx

We differentiate with respect to

a

Here

f(x,a) = |\sin x|

does not depend on

a

, so its partial derivative is zero.

Lower limit

u(a)=a

, upper limit

v(a)=a+100\pi

Their derivatives:

u'(a)=1

v'(a)=1

By Leibniz rule:

\varphi'(a) = f(v(a),a) \cdot v'(a) - f(u(a),a) \cdot u'(a)

Thus

\varphi'(a) = |\sin(a+100\pi)| \cdot 1 - |\sin a| \cdot 1

Since

\sin(x)

has period

2\pi

\sin(a+100\pi) = \sin(a+50 \cdot 2\pi) = \sin a

Therefore

\varphi'(a) = |\sin a| - |\sin a| = 0

Answer:

0

(Option A).

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