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Monotonically Increasing Functions

The function ...

Question

The function $f (x) = {tan}^{- 1} (sin x + cos x)$ is an increasing function in

(- \frac{π}{2}, \frac{π}{4})

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(0, \frac{π}{2})

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(- \frac{π}{2}, \frac{π}{2})

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(\frac{π}{4}, \frac{π}{2})

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Solution

The correct option is A $(- \frac{π}{2}, \frac{π}{4})$
$f (x) = {tan}^{- 1} (sin x + cos x)$

$f^{1} (x) = \frac{cos x - sin x}{1 + {(sin x + cos x)}^{2}} > 0$

$cos x > sin x$

$tan x < 1$

Since $(- \frac{π}{2}, \frac{π}{4})$ falls under the given interval $(- \frac{π}{2}, \frac{π}{4})$

Hence the correct answer is option $A$

Suggest Corrections

Similar questions

(0, \frac{π}{4})

(\frac{π}{4}, \frac{π}{2})

f (x) = \frac{1 - \tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{2}] .

[0, \frac{π}{2}]

f (\frac{π}{4}) =

f (x) = \frac{sin 2 x}{sin (x + \frac{π}{4})}

[0, \frac{π}{2}]

f (x) = {sec}^{- 1} x

α

β

f (x)

f (x) = - 2 sin x,

- π \leq x \leq - \frac{π}{2}

= α sin x + β,

- \frac{π}{2} < x < \frac{π}{2}

= cos x,

\frac{π}{2} \leq x \leq π

[- π, π]

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