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The function $f(x)=[x]\cos\big[\large\frac{2x-1}{2}\big]$$\pi$ where [.] denotes the greatest integer function, is discontinuous at

$\begin{array}{1 1}(a)\;\text{All x}&(b)\;\text{All integer points}\\(c)\;\text{No x}&(d)\;\text{x which is not an integer}\end{array}$

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When $x$ is not an integer,both the functions [x] and $\cos\big[\large\frac{2x-1}{2}\big]$$\pi$ are continuous.
$\therefore f(x)$ is continuous on all non integral points.
For $x=n\in I$
$\lim\limits_{x\to n^-}f(x)=\lim\limits_{x\to n^-}[x]\cos\big[\large\frac{2x-1}{2}\big]$$\pi$
$\Rightarrow (n-1)\cos\big[\large\frac{2n-1}{2}\big]$$\pi=0$
$\lim\limits_{x\to n^+}f(x)=\lim\limits_{x\to n^+}[x]\cos\big(\large\frac{2x-1}{2}\big)$$\pi$
$\Rightarrow n\cos\big(\large\frac{2n-1}{2}\big)$$\pi=0$
$f(n)=n\cos\big(\large\frac{2n-1}{2}\big)$$\pi=0$
$\therefore f$ is continuous at all integral points as well.
Thus f is continuous everywhere.
Hence (c) is the correct answer.
answered Jan 3, 2014 by sreemathi.v
 

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