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Monotonically Increasing Functions

Let f x =sin ...

Question

Let $f (x) = {sin}^{4} x + {cos}^{4} x$ . Then $f$ is an increasing function in the interval :

A

[\frac{5 π}{8}, \frac{3 π}{4}]

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B

[\frac{π}{2}, \frac{5 π}{8}]

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C

[0, \frac{π}{4}]

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D

[\frac{π}{4}, \frac{π}{2}]

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Solution

The correct option is D $[\frac{π}{4}, \frac{π}{2}]$
$f (x) = {sin}^{4} x + {cos}^{4} x f^{'} (x) = 4 {sin}^{3} x cos x - 4 {cos}^{3} x sin x = 4 sin x cos x ({sin}^{2} x - {cos}^{2} x) = - 2 sin 2 x . cos 2 x = - sin 4 x f^{'} (x) \geq 0 for increasing function \Rightarrow sin 4 x \leq 0 \Rightarrow π \leq 4 x \leq 2 π$
$x \in [\frac{π}{4}, \frac{π}{2}]$

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Similar questions

g_{i} : [\frac{π}{8}, \frac{3 π}{8}] \to R, i = 1, 2

f : [\frac{π}{8}, \frac{3 π}{8}] \to R

be functions such that

g_{1} (x) = 1, g_{2} (x) = | 4 x - π |

f (x) = {sin}^{2} x,

x \in [\frac{π}{8}, \frac{3 π}{8}] .

S_{i} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} f (x) \cdot g_{i} (x) d x, i = 1, 2

\frac{16 S_{1}}{π}

g_{i} : [\frac{π}{8}, \frac{3 π}{8}] \to R, i = 1, 2

f : [\frac{π}{8}, \frac{3 π}{8}] \to R

be functions such that

g_{1} (x) = 1, g_{2} (x) = | 4 x - π |

f (x) = {sin}^{2} x,

x \in [\frac{π}{8}, \frac{3 π}{8}] .

S_{i} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} f (x) \cdot g_{i} (x) d x, i = 1, 2

\frac{48 S_{2}}{π^{2}}

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

f (x)

is continuous in

[0, \frac{π}{2}]

f (\frac{π}{4})

f (x) = \frac{1 - tan x}{4 x - π}

x \neq π / 4

x \in [0, \frac{π}{2}]

f (x)

is continuous in

[0, \frac{π}{2}]

f (\frac{π}{4})

Q. The function

f (x) = sin 2 x + cos 2 x \forall x \in [0, \frac{π}{2}]

is strictly decreasing in the interval

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Monotonically Increasing Functions

Standard XII Mathematics

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