Question

If $f:R \to R$ is a function defined by $f\left( x \right) = \left[ x \right]\cos \left( {\dfrac{{2x - 1}}{2}\pi } \right)$, where [x] denotes the greatest integer function, then f is:
$\left( a \right)$ Continuous for every real x
$\left( b \right)$ Discontinuous only at x = 0
$\left( c \right)$ Discontinuous only at non-zero integral values of x
$\left( d \right)$ Continuous only at x = 0

Answer 1

In this particular question use the concept that if the left hand limit i.e. LHL is equal to the RHL at a particular point i.e. ${\left( {\mathop {\lim }\limits_{x \to {n^ - }} } \right)_{LHL}} = {\left( {\mathop {\lim }\limits_{x \to {n^ + }} } \right)_{RHL}}$ then the function is continuous at every value of x otherwise not, where n belongs to integer values so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given function
$f\left( x \right) = \left[ x \right]\cos \left( {\dfrac{{2x - 1}}{2}\pi } \right)$
Now the above function is also written as
$\Rightarrow f\left( x \right) = \left[ x \right]\cos \left( {x\pi - \dfrac{\pi }{2}} \right)$
Now as we know that cos (a – b) = cos (b – a), as cos (-x) = cos (x) so use this property in the above equation we have,
$\Rightarrow f\left( x \right) = \left[ x \right]\cos \left( {\dfrac{\pi }{2} - \pi x} \right)$
Now as we know that $\cos \left( {\dfrac{\pi }{2} - x} \right) = \sin x$ so use this property in the above equation we have,
$\Rightarrow f\left( x \right) = \left[ x \right]\cos \left( {\dfrac{\pi }{2} - \pi x} \right) = \left[ x \right]\sin \left( {\pi x} \right)$
Now consider the RHL we have,
$\Rightarrow RHL = \mathop {\lim }\limits_{x \to {n^ + }} f\left( x \right)$, where n belongs to real integer values
$\Rightarrow RHL = \mathop {\lim }\limits_{x \to {n^ + }} \left[ x \right]\sin \left( {\pi x} \right) = \left[ {{n^ + }} \right]\sin \left( {\pi {n^ + }} \right)$
Now as we know that $\sin \left( {{n^ + }\pi } \right) = 0$ for every integer real value of n so we have,
$\Rightarrow RHL = 0$
Now consider the LHL we have,
$\Rightarrow LHL = \mathop {\lim }\limits_{x \to {n^ - }} f\left( x \right)$, where n belongs to integer values
$\Rightarrow LHL = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\sin \left( {\pi x} \right) = \left[ {{n^ - }} \right]\sin \left( {\pi {n^ - }} \right)$
Now as we know that $\sin \left( {{n^ - }\pi } \right) = 0$ for every integer real value of n so we have,
$\Rightarrow LHL = 0$
So as we see that, LHL = RHL = 0, so the function f (x) continuous for every real value of x.
So this is the required answer.
Hence option (a) is the correct answer.

Note : Whenever we face such types of questions the key concept we have to remember is that always recall some of the trigonometric identities such as cos (a – b) = cos (b – a), as cos (-x) = cos (x), $\cos \left( {\dfrac{\pi }{2} - x} \right) = \sin x$, and the value of $\sin \left( {n\pi } \right)$ is always zero for every integer real value of n.