Period of function f (x) = frac{{|sin x| + |cos x|}}{{|sin x - cos x|}} is

English

Gujarati

Hindi

1.Relation and Function

normal

The period of the function $f (x) =$$\frac{{|\sin x| + |\cos x|}}{{|\sin x - \cos x|}}$ is

A

$\pi /2$

B

$\pi /4$

C

$\pi$

D

$2\pi$

Solution

$f(x)=\frac{|\sin x|-|\cos x|}{|\sin x \cos x|}$

$=\frac{|\sin x|-|\cos x|}{\left|\frac{1}{2} 2 \sin x \cos x\right|}=\frac{|\sin x|-|\cos x|}{\left|\frac{1}{2} \sin 2 x\right|}$

As period of $|\sin x|,|\cos x|$ is $\pi,$ and period of $|\sin 2 x|$ is $\frac{\pi}{2}$

Therefore period of $f(x)$ is $\pi$

Standard 12

Mathematics

Share

Similar Questions

The graph of the function $y = f(x)$ is symmetrical about the line $x = 2$, then

hard

(AIEEE-2004)

If $f(x) = \frac{1}{{\sqrt {x + 2\sqrt {2x – 4} } }} + \frac{1}{{\sqrt {x – 2\sqrt {2x – 4} } }}$ for $x > 2$, then $f(11) = $

easy

If $y = f(x) = \frac{{ax + b}}{{cx – a}}$, then $x$ is equal to

easy

If non-zero real numbers $b$ and $c$ are such that $min \,f\left( x \right) > \max \,g\left( x \right)$, where $f\left( x \right) = {x^2} + 2bx + 2{c^2}$ and $g\left( x \right) = {-x^2} – 2cx + {b^2}$$\left( {x \in R} \right)$; then $\left| {\frac{c}{b}} \right|$ lies in the interval

hard

(JEE MAIN-2014)

Let a function $f : R \rightarrow R$ is defined such that $3f(2x^2 -3x + 5) + 2f(3x^2 -2x + 4) = x^2 -7x + 9\ \ \ \forall x \in R$, then the value of $f(5)$ is-

normal