The period of the function $f (x) = tan (4 x - 1)$ is

π

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\frac{π}{2}

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2 π

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\frac{π}{4}

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\frac{3 π}{4}

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Solution

The correct option is D $\frac{π}{4}$
We have $f (x) = tan (4 x - 1)$
time period of $tan x$ is $π$
$∴$ Time period of $tan (4 x - 1)$ will be equal to $\frac{π}{4}$

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Identify the function based on the description.

1.It is periodic with period $2 π$ .

2.Domain of the function is R and the range is [-1, 1]

3.F(x) decreases strictly from 1 to -1 as x increases from 0 to $π$ . [For eg. If $x_{2} > x_{1}, F (x_{1}) > f (x_{2}), x ϵ [0, π]$

4.F(x) increases strictly from -1 to 1 as x increases from $π t o 2 π$ . (foreg. If $x_{2} > x_{1}, f (x_{2}) > f (x_{1}), x ϵ [π, 2 π]$

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