Functions Ques 40
If $f(x)=\cos \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x$, where $[x]$ stands for the greatest integer function, then
(1991, 2M)
(a) $f(\pi / 2)=-1$
(b) $f(\pi)=1$
(c) $f(-\pi)=0$
(d) $f(\pi / 4)=1$
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Answer:
Correct Answer: 40.(a,c)
Solution:
- Since, $f(x)=\cos \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x$
$\Rightarrow \quad f(x)=\cos (9) x+\cos (-10) x$
$ \text { [using } \left.\left[\pi^{2}\right]=9 \text { and }\left[-\pi^{2}\right]=-10\right] $
$\therefore \quad f (\frac{\pi}{2})=\cos \frac{9 \pi}{2}+\cos 5 \pi=-1$
$f(\pi)=\cos 9 \pi+\cos 10 \pi=-1+1=0$
$f(-\pi)=\cos 9 \pi+\cos 10 \pi=-1+1=0$
$f (\frac{\pi}{4})=\cos \frac{9 \pi}{4}+\cos \frac{10 \pi}{4}=\frac{1}{\sqrt{2}}+0=\frac{1}{\sqrt{2}}$
Hence, (a) and (c) are correct options.
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