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UPSC NDA I 2022 Mathematics Solved Paper

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Question : 36 of 120

Marks: +1, -0

If

an^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{x}{3}\right)=\frac{\pi}{4},

where 0 < x < 6, then what is x equal to?

1
2
3
5

Solution:

Concept:

The sum formula for inverse tangents: if

xy < 1

, then

\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)

.

Explanation:

Given

\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{x}{3}\right) = \frac{\pi}{4}

, with

0 < x < 6

.

Here,

x = \frac{1}{2}

and

y = \frac{x}{3}

, so

xy = \frac{1}{2} \cdot \frac{x}{3} = \frac{x}{6}

.

Since

0 < x < 6

, we have

\frac{x}{6} < 1

, so

xy < 1

. Hence we use the first case.

Apply the formula:

\tan^{-1}\left(\frac{\frac{1}{2} + \frac{x}{3}}{1 - \frac{1}{2} \cdot \frac{x}{3}}\right) = \frac{\pi}{4}

.

Simplify the fraction inside:

Numerator:

\frac{1}{2} + \frac{x}{3} = \frac{3 + 2x}{6}

.

Denominator:

1 - \frac{x}{6} = \frac{6 - x}{6}

.

Thus

\frac{\frac{3+2x}{6}}{\frac{6-x}{6}} = \frac{3+2x}{6-x}

.

Now

\tan^{-1}\left(\frac{3+2x}{6-x}\right) = \frac{\pi}{4}

implies

\frac{3+2x}{6-x} = \tan\frac{\pi}{4} = 1

.

So

3+2x = 6-x

.

Rearranging:

2x + x = 6 - 3

, i.e.

3x = 3

.

Hence

x = 1

.

Answer:

x = 1

(Option A).

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