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Domain and Range of Basic Inverse Trigonometric Functions

Considering o...

Question

Considering only the principal values of inverse functions, the set
$A = {x \geq 0 : {tan}^{- 1} (2 x) + {tan}^{- 1} (3 x) = \frac{π}{4}}$

is an empty set

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Contains more than two elements

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Contains two elements

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is a singleton

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Solution

The correct option is C is a singleton
${tan}^{- 1} (2 x) + {tan}^{- 1} (3 x) = π / 4$
$\Rightarrow \frac{5 x}{1 - 6 x^{2}} = 1$
$\Rightarrow 6 x^{2} + 5 x - 1 = 0$
$x = - 1$ or $x = \frac{1}{6}$
$x = \frac{1}{6} ∵ x > 0$

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Similar questions

A = {x \geq 0 : {tan}^{- 1} (2 x) + {tan}^{- 1} (3 x) = \frac{π}{4}}

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

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{t a n}^{- 1} 2 x + {t a n}^{- 1} 3 x = n π + (π / 4)

{t a n}^{- 1} x + {c o s}^{- 1} (\frac{y}{\sqrt{1 + y^{2}}}) = {s i n}^{- 1} (\frac{3}{\sqrt{10}})

tan [{cos}^{- 1} \frac{1}{\sqrt{2}} - {sin}^{- 1} \frac{4}{\sqrt{17}}]

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{t a n}^{- 1} \frac{1}{4} + 2 {t a n}^{- 1} \frac{1}{5} + {t a n}^{- 1} \frac{1}{6} + {t a n}^{- 1} \frac{1}{x} = \frac{π}{4}

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

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