Considering only the principal values of inverse trigonometric functions, the number of positive real values of satisfying is

Question

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan^{- 1} (x) + \tan^{- 1} (2 x) = \frac{π}{4}$ is: [2024]

Smart Achievers · Accepted Answer

(1)

Given, $\tan^{- 1} (x) + \tan^{- 1} (2 x) = \frac{π}{4}$

$\Rightarrow \tan^{- 1} (\frac{3 x}{1 - 2 x^{2}}) = \frac{π}{4}$ $(∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y}))$

$\Rightarrow \frac{3 x}{1 - 2 x^{2}} = \tan \frac{π}{4} \Rightarrow \frac{3 x}{1 - 2 x^{2}} = 1 \Rightarrow 2 x^{2} + 3 x - 1 = 0$

$\Rightarrow x = \frac{- 3 \pm \sqrt{17}}{4}$ $∴$ $Possible value of x = \frac{- 3 + \sqrt{17}}{4}$

Hence, only 1 positive real value of $x$ satisfies the equation.

Solution

Q.
Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan^{- 1} (x) + \tan^{- 1} (2 x) = \frac{π}{4}$ is: [2024]

1 1

2 more than 2

3 2

4 0

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Solution

Q. Considering only the principal values of inverse trigonometric functions, the number of positive real values of x satisfying tan-1(x)+tan-1(2x)=π4 is: [2024]

1 1

2 more than 2

3 2

4 0

Ans. (1) Given, tan-1(x)+tan-1(2x)=π4 ⇒tan-1(3x1-2x2)=π4 (∵ tan-1x+tan-1y=tan-1(x+y1-xy)) ⇒3x1-2x2=tanπ4⇒3x1-2x2=1⇒2x2+3x-1=0 ⇒x=-3±174 ∴ Possible value of x=-3+174 Hence, only 1 positive real value of x satisfies the equation.

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Q.
Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan^{- 1} (x) + \tan^{- 1} (2 x) = \frac{π}{4}$ is: [2024]