Thursday, February 22, 2024
Homefsc 1st year math solutionsFind the periodic function Ex:11.1

Find the periodic function Ex:11.1

Here is the periodic function of all questions given here. Find the periods of the following functions:

Periodic Function Solved Questions

periods of trigonometric functions
Question 01: \sin 3 x

Given function is \sin 3 x
Since period of \sin is 2 \pi
Therefore f(x)=\f(x+P)
\sin 3 x=\sin (3 x+2 \pi)
\sin 3 x=\sin 3\left(x+\frac{2 \pi}{3}\right)
Thus period of \sin 3 x is \frac{2 \pi}{3}

Question 02: \cos 2 x

Given function is \cos 2 x
Since period of \cos is 2 \pi
Therefore f(x)=\f(x+P)
\cos 2 x=\cos (2 x+2 \pi)
\cos 2 x=\cos 2(x+\pi)
Thus period of \cos 2 x is \pi.

Question 03: \tan 4 x

Given function is \tan 4 x
Since period of \tan x is \pi
Therefore f(x)=\f(x+P)
\tan 4 x=\tan (4 x+\pi)
\tan 4 x=\tan 4\left(x+\frac{\pi}{4}\right)
Thus period of \tan 4 \times is \frac{\pi}{4}.

Periodic Function Solved Questions

Question 04: \cot \frac{x}{2}

Since period of \cot is \pi
Therefore f(x)=\f(x+P)
\cot \frac{x}{2}=\cot \left(\frac{x}{2}+\pi\right)
\cot \frac{x}{2}=\cot \frac{1}{2}(x+2 \pi)
Thus period of \cot \frac{x}{2} is 2 \pi

Question 05: \sin \frac{x}{3}

Given function is \sin \frac{x}{3}
Since period of \sin is 2 \pi
Therefore f(x)=\f(x+P)
\sin \frac{x}{3}=\sin \left(\frac{x}{3}+2 \pi\right)
\sin \frac{x}{3}=\sin \frac{1}{3}(x+6 \pi)
Thus period of \sin \frac{x}{3} is 6 \pi

Question 06: \cosec \frac{x}{4}

Given function is \operatorname{cosec} \frac{x}{4}
Since period of cosec is 2 \pi
Therefore f(x)=\f(x+P)
\operatorname{cosec} \frac{x}{4}=\operatorname{cosec}\left(\frac{x}{4}+2 \pi\right)
Thus period of \operatorname{cosec} \frac{x}{4} is 8 \pi

Periodic Function Solved Questions

Question 07: \sin \frac{x}{5}

Given function is \sin \frac{x}{5}
Since period of \sin x is 2 \pi
Therefore f(x)=f(x+P)
\sin \frac{x}{5}=\sin \left(\frac{x}{5}+2 \pi\right)
\sin \frac{x}{5}=\sin \frac{1}{5}(x+10 \pi)
Thus period of \sin \frac{x}{5} is 10 \pi

Question 08: \cos \frac{x}{6}

Given function is \cos \frac{x}{6}
Since period of \cos x is 2 \pi
f(x)=f(x+P)
Therefore \cos \frac{x}{6}=\cos \left(\frac{x}{6}+2 \pi\right)=\cos \frac{1}{6}(x+12 \pi)
Thus period of \cos \frac{x}{6} is 12 \pi

Question 09: \tan \frac{x}{7}

Given function is \tan \frac{x}{7}
Since period of \tan x is \pi
f(x)=f(x+P)
Therefore \tan \frac{x}{7}=\tan \left(\frac{x}{7}+\pi\right)
\tan \frac{x}{7}=\tan \frac{1}{7}(x+7 \pi)
Thus period of \tan \frac{x}{7} is 7 \pi

Periodic Function Solved Questions

Question 10: \cot 8 x

Given function is \cot 8 x
Since period of \cot x is \pi
f(x)=f(x+P)
Therefore \cot 8 x=\cot (8 x+\pi)
\cot 8 x=\cot 8\left(x+\frac{\pi}{8}\right)
Thus period of \cot 8 x is \frac{\pi}{8}

Question 11: \sec 9 \mathrm{x}

Given function is \sec 9 \mathrm{x}
Since period of \sec (x) is 2 \pi
f(x)=f(x+P)
Therefore \sec 9 x=\sec (9 x+2 \pi)
\sec 9 x=\sec 9\left(x+\frac{2 \pi}{9}\right)
Thus period of \sec 9 x is \frac{2 \pi}{9}

Question 12: \operatorname{cosec} 10 \mathrm{x}

Given function is \operatorname{cosec} 10 \mathrm{x}
Since period of \operatorname{cosec} x is 2 \pi
f(x)=f(x+P)
\operatorname{cosec} 10 x=\operatorname{cosec}(10 x+2 \pi)
\operatorname{cosec} 10 x=\operatorname{cosec} 10\left(x+\frac{2 \pi}{10}\right)
Thus period of \operatorname{cosec} 10 \mathrm{x} is \frac{\pi}{5}

Periodic Function Solved Questions

Question 13: 3 \sin x

Given function is 3 \sin x
Since period of \sin x is 2 \pi
f(x)=f(x+P)
Therefore 3 \sin \mathrm{x}=3 \sin (\mathrm{x}+2 \pi)
Thus period of 3 \sin x is 2 \pi.

Question 14: 2 \cos x

Given function is 2 \cos x
Since period of \cos x is 2 \pi
f(x)=f(x+P)
Therefore 2 \cos x=2 \cos (x+2 \pi)
Thus period of 2 \cos x is 2 \pi

Question 15: \operatorname{cosec} 10 \mathrm{x}

Given function is 3 \cos \frac{x}{5}
Since period of \cos x is 2 \pi
f(x)=f(x+P)
Therefore
3 \cos \frac{x}{5}=3 \cos \left(\frac{x}{5}+2 \pi\right)
3 \cos \frac{x}{5}=3 \cos \frac{1}{5}(x+10 \pi)
Thus
period of 3 \cos \frac{x}{5} is 10 \pi.

Periodic Function Solved Questions

Example: \operatorname{cosec} 10 \mathrm{x}

Given function is \operatorname{cosec} 10 \mathrm{x}
Since period of \operatorname{cosec} x is 2 \pi
f(x)=f(x+P)
\operatorname{cosec} 10 x=\operatorname{cosec}(10 x+2 \pi)
\operatorname{cosec} 10 x=\operatorname{cosec} 10\left(x+\frac{2 \pi}{10}\right)
Thus period of \operatorname{cosec} 10 \mathrm{x} is \frac{\pi}{5}

POPULAR LINKS

apkplot
apkplothttps://apkplot.com
Apkplot.com is an all-in-one educational site that has answers to math problems and free past papers and MCQs to help people prepare for the MDCAT and ECAT. The website is based on FSC and covers the subjects of math, physics, chemistry, and biology for Parts 01 and 02. Students can use the website's large database of answers to math problems and step-by-step explanations to help them. Students taking FSC Parts 01 and 02, as well as the MCAT and ECAT, can also find past papers on the website. These papers help students understand the format of the exam and prepare well for the real test. The MCQs are also a great way for students to test their knowledge and figure out what they need to learn more about.
RELATED ARTICLES

LEAVE A REPLY

Please enter your comment!
Please enter your name here

Most Popular

Recent Comments

We use cookies to ensure that we give you the best experience on our website.