If f:R \to R is a function defined by f( x ) = [ x ]\cos ( 2x - 1 \over 2 )\pi , where [x] denotes the greatest integer function, then f is

discontinuous only at non-zero integral values of x

Limits Continuity and Differentiability PYQ — JEE Main 2012 · Shift 0 · Q25 with Solution

View written solutionFree

Correct answer: A

Interpret the function carefully

Given

f(x)=[x]\cos\left(\frac{2x-1}{2}\pi\right),

where $[x]$ is the greatest integer function.

First simplify the cosine term:

\cos\left(\frac{2x-1}{2}\pi\right)=\cos\left(\pi x-\frac\pi2\right)=\sin(\pi x).

So the function becomes

f(x)=[x]\sin(\pi x).

We must check continuity of

f(x)=[x]\sin(\pi x)

at all real $x$ .

Where can discontinuity occur?

$[x]$ is discontinuous at every integer.
$\sin(\pi x)$ is continuous everywhere.

Hence, away from integers, both factors are continuous, so their product is continuous.

Thus we only need to check continuity at integers.

Check continuity at a non-integer point

If $a\notin \mathbb Z$ , then in some neighborhood of $a$ , $[x]$ is constant. Since $\sin(\pi x)$ is continuous, the product

f(x)=[x]\sin(\pi x)

is continuous at every non-integer $a$ .

Check continuity at an integer $n$

Let $n\in \mathbb Z$ .

At $x=n$ ,

f(n)=[n]\sin(n\pi)=n\cdot 0=0.

Now compute one-sided limits.

Left-hand limit

If $x\to n^-$ , then $[x]=n-1$ . Hence

f(x)=(n-1)\sin(\pi x).

As $x\to n$ ,

\sin(\pi x)\to \sin(n\pi)=0.

Therefore,

\lim_{x\to n^-} f(x)=(n-1)\cdot 0=0.

Right-hand limit

If $x\to n^+$ , then $[x]=n$ . Hence

f(x)=n\sin(\pi x).

As $x\to n$ ,

\sin(\pi x)\to 0,

\lim_{x\to n^+} f(x)=n\cdot 0=0.

Thus,

\lim_{x\to n^-} f(x)=\lim_{x\to n^+} f(x)=0=f(n).

So $f$ is continuous at every integer $n$ .

Conclusion

Continuous at every non-integer.
Continuous at every integer.

Hence $f$ is continuous for every real $x$ .

So the correct option is:

\boxed{\text{A}}

Comparison with stored correct answer

Stored correct answer: A

My derived answer: A

They agree.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: A

Interpret the function carefully

Given

f(x)=[x]\cos\left(\frac{2x-1}{2}\pi\right),

where $[x]$ is the greatest integer function.

First simplify the cosine term:

\cos\left(\frac{2x-1}{2}\pi\right)=\cos\left(\pi x-\frac\pi2\right)=\sin(\pi x).

So the function becomes

f(x)=[x]\sin(\pi x).

We must check continuity of

f(x)=[x]\sin(\pi x)

at all real $x$ .

Where can discontinuity occur?

$[x]$ is discontinuous at every integer.
$\sin(\pi x)$ is continuous everywhere.

Hence, away from integers, both factors are continuous, so their product is continuous.

Thus we only need to check continuity at integers.

Check continuity at a non-integer point

If $a\notin \mathbb Z$ , then in some neighborhood of $a$ , $[x]$ is constant. Since $\sin(\pi x)$ is continuous, the product

f(x)=[x]\sin(\pi x)

is continuous at every non-integer $a$ .

Check continuity at an integer $n$

Let $n\in \mathbb Z$ .

At $x=n$ ,

f(n)=[n]\sin(n\pi)=n\cdot 0=0.

Now compute one-sided limits.

Left-hand limit

If $x\to n^-$ , then $[x]=n-1$ . Hence

f(x)=(n-1)\sin(\pi x).

As $x\to n$ ,

\sin(\pi x)\to \sin(n\pi)=0.

Therefore,

\lim_{x\to n^-} f(x)=(n-1)\cdot 0=0.

Right-hand limit

If $x\to n^+$ , then $[x]=n$ . Hence

f(x)=n\sin(\pi x).

As $x\to n$ ,

\sin(\pi x)\to 0,

\lim_{x\to n^+} f(x)=n\cdot 0=0.

Thus,

\lim_{x\to n^-} f(x)=\lim_{x\to n^+} f(x)=0=f(n).

So $f$ is continuous at every integer $n$ .

Conclusion

Continuous at every non-integer.
Continuous at every integer.

Hence $f$ is continuous for every real $x$ .

So the correct option is:

\boxed{\text{A}}

Comparison with stored correct answer

Stored correct answer: A

My derived answer: A

They agree.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

Left-hand limit

Right-hand limit

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

Left-hand limit

Right-hand limit

More from Limits Continuity and Differentiability