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If $f:R\to R$ is a function defined by $f(x)=[x]\cos\big(\large\frac{2x-1}{2}\big)$$\pi$ where [x] denotes the greatest integer function then f is

$\begin{array}{1 1}(a)\;\text{continuous for every real x}\\(b)\;\text{discontinuous only at x=0}\\(c)\;\text{discontinuous only at non-zero integral values of x}\\(d)\;\text{Continuous only at x=0}\end{array}$

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$f(x)=[x]\cos\big(\large\frac{2x-1}{2}\big)$$\pi$
Doubtful points are $x=0,n\in I$
LHL=$\lim\limits_{x\to n^-}[x]\cos\big(\large\frac{2x-1}{2}\big)$$\pi$
$\Rightarrow (n-1)\cos\big(\large\frac{2n-1}{2}\big)$$\pi$
[x] is the greatest integer function.
RHL=$\lim\limits_{x\to n^+}[x]\cos\big(\large\frac{2x-1}{2}\big)$$\pi=n\cos\big(\large\frac{2n-1}{2}\big)$$\pi$
Now values of the function at x=0 is $f(n)=0$
Since LHL =RHL =f(n)
$f(x)=[x]\cos\big(\large\frac{2x-1}{2}\big)$ is continuous for every real x
Hence (a) is the correct answer.
answered Jan 3, 2014 by sreemathi.v
 

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