- Home
- Standard 12
- Mathematics
વિધેય $f(x) = e^{x -[x]+|cos\, \pi x|+|cos\, 2\pi x|+....+|cos\, n\pi x|}$ નુ આવર્તમાન મેળવો, ( જ્યા $[.]$ એ મહત્તમ પુર્ણાક વિધેય છે.)
$1$
$n \pi$
$n$
$\frac{\pi}{n}$
Solution
$f(x)=e^{\{x\}+|\cos \pi x|+|\cos 2 \pi x|+\cdots+|\cos n \pi x|}$
$f(x)=e^{\{x\}} \cdot e^{|\cos \pi x|} \cdot e^{|\cos 2 \pi x|} \cdot \cdot e^{|\cos n \pi x|}$
$f(x)=g(x), h(x)$
Period $f(x)=2(m)$
Reriod $f \quad|\cos x|=\pi$
$y=|\cos x|=|\cos (\pi+x)|=|-\cos x|=|\cos x|=y$
Period of $e^{|\log \pi x|}=\frac{\pi}{\pi}=1$
Period of $e^{\left|(-3)^{2 \pi}\right| x \mid}=\frac{\pi}{2 \pi}=\frac{1}{2}$
Pesiod $f e^{1 \operatorname{los} 3 \pi x}=\frac{\pi}{3 \pi}=\frac{1}{3}$
Period $f e^{|\cos n \pi x|}=\frac{\pi}{n \pi}=\frac{1}{n}$
Perdod of $f(x)=\operatorname{Lcm}\left\{1,1, \frac{1}{2}, \frac{1}{3}, . . \quad \frac{1}{n}\right\}$
$y=\frac{1}{1}=1$
