The period of function $f (x) = cos 4 x + tan 3 x$ is

\frac{π}{12}

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\frac{π}{6}

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\frac{π}{2}

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π

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2 π

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Solution

The correct option is C $π$
We have, $f (x) = cos 4 x + tan 3 x$

Clearly, period of $cos 4 x$ is $\frac{2 π}{4} = \frac{π}{2}$ and that of $tan 3 x$ is $\frac{π}{3}$ .

Therefore, period of $f (x) = \frac{LCM of (π and π)}{HCF of (2 and 3)}$ $= \frac{π}{1} = π$

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {(1 + | s i n x |)}^{\frac{a}{| s i n x |}} & ; - \frac{π}{6} < x < 0 b & ; x = 0 e^{(\frac{t a n 2 x}{t a n 3 x})} & ; 0 < x < \frac{π}{6} \end{matrix}

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Let $f (x) = {\begin{matrix} - π, if & - π < x \leq 0 π, if & 0 < x \leq π \end{matrix}$
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The period of function f(x)=cos4x+tan3x is

The correct option is C πWe have, f(x)=cos4x+tan3xClearly, period of cos4x is 2π4=π2 and that of tan3x is π3.Therefore, period of f(x)=LCM of (π and π)HCF of( 2and 3) =π1=π

The period of function $f (x) = cos 4 x + tan 3 x$ is

The correct option is C $π$
We have, $f (x) = cos 4 x + tan 3 x$

Clearly, period of $cos 4 x$ is $\frac{2 π}{4} = \frac{π}{2}$ and that of $tan 3 x$ is $\frac{π}{3}$ .

Therefore, period of $f (x) = \frac{LCM of (π and π)}{HCF of (2 and 3)}$ $= \frac{π}{1} = π$