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Higher Order Derivatives

The fundament...

Question

The fundamental period of the function $f (x) = | sin 2 x | + | cos 2 x |$ is

\frac{π}{2}

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π

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2 π

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\frac{π}{4}

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Solution

The correct option is A $\frac{π}{2}$
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Similar questions

f (x) = 2 c o s \frac{1}{3} (x - π)

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

f (x) = c^{⎡ ⎢ ⎣ {sin}^{2} x + {sin}^{2} (x + \frac{π}{3}) + cos x cos (x + \frac{π}{3}) ⎤ ⎥ ⎦}

c

\frac{\sin x}{e^{x}}

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f (x) = \frac{x}{2} - \sin \frac{π x}{6} on [- 1, 0]

f (x) = \frac{6 x}{π} - 4 \sin^{2} x on [0, π / 6]

[0, \frac{π}{2}]

f (x) = lim n \to \infty {\frac{x}{1 + {3 {tan}^{2} x}^{n}}}

f (x)

π

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} 0, & - \frac{π}{2} < x < - \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = - \frac{π}{6} \end{matrix} \begin{matrix} x, & - \frac{π}{6} < x < \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = \frac{π}{6} \end{matrix} \begin{matrix} 0, & \frac{π}{6} < x < \frac{π}{2} \end{matrix} \end{matrix}

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