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UPSC NDA I 2025 Mathematics Paper

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Question : 50 of 120

Marks: +1, -0

If

\frac{1}{k}

, then what is the value of k?

1
$1/2$
$1/3$
$1/4$

Solution:

Concept:

Use the inverse tangent addition formula:

\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)

, valid when

ab < 1

.

Explanation:

Given

\tan^{-1}k + \tan^{-1}\frac12 = \frac{\pi}{4}

.

Apply the formula:

\tan^{-1}\left(\frac{k+\frac12}{1-k\cdot\frac12}\right) = \frac{\pi}{4}

.

Take tangent on both sides:

\frac{k+\frac12}{1-\frac{k}{2}} = \tan\frac{\pi}{4} = 1

.

Simplify:

k + \frac12 = 1 - \frac{k}{2}

.

Multiply by 2:

2k + 1 = 2 - k

.

Bring like terms:

2k + k = 2 - 1

, so

3k = 1

.

Thus

k = \frac13

.

Answer:

k = \frac13

, which is Option C.

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