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UPSC NDA I 2025 Mathematics Paper

Question Numbers: 31-33

Consider the following for the three (03) items that follow: Let

p=sin35^{∘},q=sin25^{∘}

and

r=sin(−95^{∘})

Question : 33 of 120

Marks: +1, -0

What is

(p^{2}+q^{2}+r^{2})

equal to?

Solution:

Concept:

We use the algebraic identity

(p+q+r)^2 = p^2+q^2+r^2 + 2(pq+qr+rp)

and the given condition

p+q+r=0

to find the sum of squares.

Explanation:

Given

p+q+r=0

, substitute into the identity:

0 = p^2+q^2+r^2 + 2(pq+qr+rp)

From earlier calculations,

pq+qr+rp = -\frac{3}{4}

Replace this value:

0 = p^2+q^2+r^2 + 2\left(-\frac{3}{4}\right)

Simplify:

0 = p^2+q^2+r^2 - \frac{3}{2}

Rearrange:

p^2+q^2+r^2 = \frac{3}{2}

Answer:

p^2+q^2+r^2 = \frac{3}{2}

, which corresponds to option C.

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