# If [x] denote the greatest integer not exceeding the real number x, then the period of the function f(x)=sin [pi^2] x+sin [-pi^2]x, is

Solution:      $\pi^{2}=9.87\Rightarrow [\pi^{2}]=9.$   Similarly,

$-\pi^{2}=-9.87\Rightarrow [-\pi^{2}]=-10.$

$\therefore$        $f(x)=\sin 9\pi x-\sin 10 \pi x$      $.........(1)$

$pd(\sin 9\pi x)=\frac{2\pi}{9};$    and    $pd(\sin 10\pi x)=\frac{2\pi}{10}.$

$\therefore$        $pd(f)=\frac{LCM\{2\pi,2\pi\}}{HCM\{9,10\}}=2\pi$

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