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MHTCET 18 Apr 2026 Shift 2 Paper

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Question : 117 of 150

Marks: +1, -0

If

\tan ^{-1} a x+\tan ^{-1} 3 x=\frac{\pi}{4}

3 a x^{2} < 1

, then value of

a

x=\frac{1}{6}

is …

2
3
4
9

Solution:

Concept:

Use the inverse tangent addition formula:

\tan^{-1}p + \tan^{-1}q = \tan^{-1}\left(\frac{p+q}{1-pq}\right)

, valid when

pq < 1

.

Explanation:

Here,

p = ax

and

q = 3x

.

So,

pq = 3ax^2 < 1

, which is given, so the formula is directly applicable.

Given:

\tan^{-1} ax + \tan^{-1} 3x = \frac{\pi}{4}

Taking tangent on both sides:

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

Since

\tan\frac{\pi}{4} = 1

, we get:

\frac{x(a+3)}{1 - 3ax^2} = 1

Substitute

x = \frac{1}{6}

:

\frac{\frac{1}{6}(a+3)}{1 - 3a\left(\frac{1}{6}\right)^2} = 1

\frac{\frac{a+3}{6}}{1 - \frac{a}{12}} = 1

So:

\frac{a+3}{6} = 1 - \frac{a}{12}

Multiplying by 12:

2(a+3) = 12 - a

2a + 6 = 12 - a

3a = 6

a = 2

Answer:

a = 2

, which is Option A.

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