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UPSC NDA I 2025 Mathematics Paper

Question Numbers: 34-35

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

p=∣sinα−sin(α−90∘)∣

Question : 35 of 120

Marks: +1, -0

p

Solution:

Concept:

The maximum value of an expression like

|\sin\alpha \pm \cos\alpha|

\sqrt{2}

because it can be rewritten as a single sine function.

Explanation:

Given

p = |\sin\alpha - \sin(\alpha-90^\circ)|

Use the identity

\sin(\alpha-90^\circ) = -\cos\alpha

. Then

p = |\sin\alpha - (-\cos\alpha)| = |\sin\alpha + \cos\alpha|

Rewrite

\sin\alpha + \cos\alpha = \sqrt{2} \sin(\alpha+45^\circ)

Since

|\sin(\theta)| \leq 1

, the maximum of

\sqrt{2}\sin(\alpha+45^\circ)

\sqrt{2}

Hence, the maximum value of

p

\sqrt{2}

Answer:

Option B:

\sqrt{2}

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