The number of solutions of tan - 1 4 x + tan - 1 6 x = 6 , where - 1 2 6 x 1 2 6 , is equal to [2026]

Question

The number of solutions of $\tan^{- 1} 4 x + \tan^{- 1} 6 x = \frac{π}{6},$ where $- \frac{1}{2 \sqrt{6}} < x < \frac{1}{2 \sqrt{6}},$ is equal to [2026]

Smart Achievers · Accepted Answer

(2)

$\tan^{- 1} (4 x) + \tan^{- 1} (6 x) = \frac{π}{6}$

$\Rightarrow \tan^{- 1} (\frac{4 x + 6 x}{1 + 24 x^{2}}) = \frac{π}{6}$

$\Rightarrow \frac{10 x}{1 - 24 x^{2}} = \frac{1}{\sqrt{3}}$

$\Rightarrow 24 x^{2} + 10 \sqrt{3} x - 1 = 0$

$x = \frac{- 10 \sqrt{3} \pm \sqrt{300 + 96}}{48}$

$x = \frac{\sqrt{396} - 10 \sqrt{3}}{48}$

$Only 1 solution in (- \frac{1}{2 \sqrt{6}}, \frac{1}{2 \sqrt{6}})$

Solution

Q.
The number of solutions of $\tan^{- 1} 4 x + \tan^{- 1} 6 x = \frac{π}{6},$ where $- \frac{1}{2 \sqrt{6}} < x < \frac{1}{2 \sqrt{6}},$ is equal to [2026]

1 2

2 1

3 0

4 3

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Solution

Q. The number of solutions of tan-14x+tan-16x=π6, where -126<x<126, is equal to [2026]

1 2

2 1

3 0

4 3

Ans. (2) tan-1(4x)+tan-1(6x)=π6 ⇒tan-1(4x+6x1+24x2)=π6 ⇒10x1-24x2=13 ⇒24x2+103 x-1=0 x=-103±300+9648 x=396-10348 Only 1 solution in (-126, 126)

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Q.
The number of solutions of $\tan^{- 1} 4 x + \tan^{- 1} 6 x = \frac{π}{6},$ where $- \frac{1}{2 \sqrt{6}} < x < \frac{1}{2 \sqrt{6}},$ is equal to [2026]