UPSC NDA I 2025 Mathematics Paper

We are given that k is a root of x2−4x+1=0
Solving the quadratic equation:
‌x=‌

−(−4)±√(−4)2−4(1)(1)

2(1)

=‌

4±√16−4

2

=‌

4±√12

2

=‌

4±2√3

2

‌⇒x=2±√‌3
Thus, the two possible values of k are
⇒k=2+√‌3‌‌‌ or ‌‌‌k=2−√‌3
We need to find tan−1(k)+tan−1(‌

1

k

)
Using the identity for the sum of inverse tangents
‌tan−1(a)+tan−1(b)=tan−1(‌

a+b

1−ab

)
‌⇒tan−1(k)+tan−1(‌

1

k

)=tan−1(‌

k+ 1 k

1−k⋅‌ 1 k

)
‌=tan−1(‌

k+ 1 k

1−1

)=tan−1(‌

k+ 1 k

0

)
This expression results in an undefined value, but we know from properties of the inverse tangent that ⇒tan−1(k)+tan−1(‌

1

k

)=‌

π

2