Functions 2 Question 13

14.

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: 14. (a,c)

Solution:

  1. 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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