If 0 < x < 1, the number of solutions of the equation $t a n^{- 1} (x - 1) + t a n^{- 1} x + t a n^{- 1} (x + 1) = t a n^{- 1} 3 x$ is

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 1
The given equation can be written as
$t a n^{- 1} (x - 1) + t a n^{- 1} (x + 1) = t a n^{- 1} 3 x - t a n^{- 1} x \Rightarrow \frac{2 x}{2 - x^{2}} = \frac{2 x}{1 + 3 x^{2}} \Rightarrow x + 3 x^{3} = 2 x - x^{3} \Rightarrow 4 x^{3} - x = 0 \Rightarrow (4 x^{2} - 1) = 0 \Rightarrow x = 0, x = \pm \frac{1}{2} . T h u s x = \frac{1}{2}$

Suggest Corrections

Similar questions

t a n^{- 1} (x - 1) + t a n^{- 1} x + t a n^{- 1} (x + 1) = t a n^{- 1} 3 x

\sqrt{2}

t a n^{- 1} (x - 1) + t a n^{- 1} x + t a n^{- 1} (x + 1) = t a n^{- 1} 3 x

t a n^{- 1} (x - 1) + t a n^{- 1} (x) + t a n^{- 1} (x + 1) = t a n^{- 1} (3 x)

Solution of the equation tan ^-1 (x - 1) + tan ^-1 x + tan ^-1 (x+1)=tan ^-1 3x is

(1 + x^{2}) \frac{d y}{d x} + 1 + y^{2} = 0

Join BYJU'S Learning Program