What does [x] mean in greatest integer function?

$[x]$ (floor) is the greatest integer less than or equal to $x$. It is the unique integer $n$ such that $n\le x < n+1$. This converts GIF questions into interval-based integer problems.

Is [x+3] always equal to [x]+3?

Yes. For any integer $k$, $[x+k]=[x]+k$ for all real $x$. Hence $[x+3]=[x]+3$ everywhere, which simplifies the function drastically.

Why does f(x) become constant on each interval [n,n+1)?

If $n=[x]$, then $[x]$ is fixed for all $x\in[n,n+1)$. Also $[x+3]=n+3$ stays fixed there. Therefore $f(x)$ depends only on $n$ and is constant on each such unit interval.

How to rewrite f(x) using n = [x]?

Put $n=[x]$. Then $[x+3]=n+3$, so $f(x)=n^2-(n+3)-3=n^2-n-6=(n-3)(n+2)$. This is an integer quadratic.

For which x is f(x) negative?

$f(x)<0$ means $(n-3)(n+2)<0$, which gives $-2<n<3$. With $n\in\mathbb{Z}$, $n\in\{-1,0,1,2\}$. Hence $x\in[-1,3)$.

Does f(x)=0 happen for finitely many values of x?

No. If $f(x)=0$ for some integer $n=[x]$, then it is zero on the whole interval $[n,n+1)$, which contains infinitely many real numbers. Here zeros occur on $[-2,-1)$ and $[3,4)$.

How to compute integrals of GIF functions like ∫ f(x) dx?

Split the integral into unit intervals where $[x]$ is constant. On each interval, $f(x)$ is constant, so the integral equals (constant value) × (interval length). Then add the pieces.

What is the value of ∫0^2 f(x) dx in this problem?

On $[0,1)$ and $[1,2)$, the value of $f(x)$ is $-6$. Therefore $\int_0^2 f(x)\,dx = -6-6=-12$, so the statement ‘= -6’ is false.

Is f(x)>0 only for x in [4,∞)?

No. Since $f(x)=(n-3)(n+2)$ with $n=[x]$, positivity occurs for $n\le-3$ or $n\ge4$, i.e. $x\in(-\infty,-2)\cup[4,\infty)$. So it’s not only $[4,\infty)$.

Let f(x) = [x]^2 − [x + 3] − 3, x ∈ R, where [.] is the greatest integer function. Then

Q: Is [x+3] always equal to [x]+3?

Yes. For any integer $k$, $[x+k]=[x]+k$ for all real $x$. Hence $[x+3]=[x]+3$ everywhere, which simplifies the function drastically.

Q: Why does f(x) become constant on each interval [n,n+1)?

If $n=[x]$, then $[x]$ is fixed for all $x\in[n,n+1)$. Also $[x+3]=n+3$ stays fixed there. Therefore $f(x)$ depends only on $n$ and is constant on each such unit interval.

Q: How to rewrite f(x) using n = [x]?

Put $n=[x]$. Then $[x+3]=n+3$, so $f(x)=n^2-(n+3)-3=n^2-n-6=(n-3)(n+2)$. This is an integer quadratic.

Q: For which x is f(x) negative?

$f(x)<0$ means $(n-3)(n+2)<0$, which gives $-2<n<3$. With $n\in\mathbb{Z}$, $n\in\{-1,0,1,2\}$. Hence $x\in[-1,3)$.

Q: Does f(x)=0 happen for finitely many values of x?

No. If $f(x)=0$ for some integer $n=[x]$, then it is zero on the whole interval $[n,n+1)$, which contains infinitely many real numbers. Here zeros occur on $[-2,-1)$ and $[3,4)$.

Q: How to compute integrals of GIF functions like ∫ f(x) dx?

Split the integral into unit intervals where $[x]$ is constant. On each interval, $f(x)$ is constant, so the integral equals (constant value) × (interval length). Then add the pieces.

Q: What is the value of ∫0^2 f(x) dx in this problem?

On $[0,1)$ and $[1,2)$, the value of $f(x)$ is $-6$. Therefore $\int_0^2 f(x)\,dx = -6-6=-12$, so the statement ‘= -6’ is false.

Q: Which option is correct?

Option (B) is correct: $f(x) < 0$ only for $x \in [-1,3)$.

Let f(x) = [x]^2 − [x + 3] − 3, x ∈ R, where [.] is the greatest integer function. Then | JEE Main Mathematics

NEETJEE

Math GIF

Q MCQ Functions

Let $f(x) = [x]^2 - [x + 3] - 3,\; x \in \mathbb{R}$, where $[.]$ is the greatest integer function. Then

(A) $f(x)=0$ for finitely many values of $x$
(B) $f(x) < 0$ only for $x \in [-1, 3)$
(C) $\displaystyle \int_{0}^{2} f(x)\,dx = -6$
(D) $f(x) > 0$ only for $x \in [4, \infty)$

✅ Correct Answer

Option B

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Solution Steps

Freeze the floor value on a unit interval

Let $n=[x]$. Then:

$$n \le x < n+1,\quad n\in\mathbb{Z}.$$

Hence, for every $x\in[n,n+1)$, the value of $[x]$ is constant and equals $n$.

Simplify $[x+3]$ using integer shift

From $n \le x < n+1$, add $3$:

$$n+3 \le x+3 < n+4.$$

Therefore:

$$[x+3]=n+3=[x]+3.$$

Convert $f(x)$ to an integer quadratic

Using $[x]=n$ and $[x+3]=n+3$:

$$f(x)=n^2-(n+3)-3=n^2-n-6.$$

Factorize:

$$n^2-n-6=(n-3)(n+2).$$

Check option (A): zeros are not finite

$$f(x)=0 \iff (n-3)(n+2)=0 \iff n=3 \text{ or } n=-2.$$

If $n=3$, then $x\in[3,4)$ and $f(x)=0$ for infinitely many real $x$.

If $n=-2$, then $x\in[-2,-1)$ and $f(x)=0$ for infinitely many real $x$.

So option (A) is false.

Find exactly where $f(x)<0$ (tests option B)

For $(n-3)(n+2)<0$, $n$ lies strictly between the roots:

$$-2<n<3.$$

Since $n\in\mathbb{Z}$:

$$n\in\{-1,0,1,2\}.$$

Therefore:

$$x\in[-1,0)\cup[0,1)\cup[1,2)\cup[2,3).$$

So the negative set is:

$$x\in[-1,3).$$

Hence option (B) is correct.

Check option (C): compute the integral

On $[0,1)$, we have $[x]=0$. Then:

$$f(x)=0^2-[3]-3=0-3-3=-6.$$

On $[1,2)$, we have $[x]=1$. Then:

$$f(x)=1^2-[4]-3=1-4-3=-6.$$

So:

$$\int_0^2 f(x)\,dx=\int_0^1(-6)\,dx+\int_1^2(-6)\,dx=-6-6=-12\ne -6.$$

Therefore option (C) is false.

Check option (D): positivity is not only for $[4,\infty)$

We have $f(x)>0 \iff (n-3)(n+2)>0$, which gives:

$$n\le -3 \quad \text{or}\quad n\ge 4.$$

Hence:

$$x\in(-\infty,-2)\cup[4,\infty).$$

So $f(x)>0$ also holds for $x<-2$, not only for $x\in[4,\infty)$.

Option (D) is false, and the correct answer is $\boxed{\text{Option (B)}}$.

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Key Insight

Put $n=[x]$. Then $f(x)=(n-3)(n+2)$. Negativity occurs only for $n=-1,0,1,2$ ⇒ $x\in[-1,3)$.

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Greatest Integer Function Theory

1. Definition of the greatest integer function and unit intervals

The greatest integer function (GIF), written as $[x]$, maps a real number to the greatest integer less than or equal to it. The most useful working form is: $[x]=n$ if and only if $n\le x < n+1$, where $n$ is an integer. This statement partitions the real line into disjoint unit intervals $[n,n+1)$, and on each such interval the value of $[x]$ remains constant. That is why expressions involving $[x]$ behave like step functions: they do not change continuously, but jump at integers. In JEE, many “only for” statements depend on endpoints. Because the left endpoint is included and the right endpoint is excluded, values like $x=3$ lie in $[3,4)$ and not in $[2,3)$. A disciplined approach is to set $n=[x]$, solve conditions in integers, and then translate each valid $n$ to the corresponding real interval $[n,n+1)$ before merging intervals.

2. Integer shifts: why $[x+k]=[x]+k$ for integer $k$

A powerful identity for GIF is that for any integer $k$, we have $[x+k]=[x]+k$ for all real $x$. The logic is direct: if $[x]=n$, then $n\le x < n+1$. Adding $k$ gives $n+k\le x+k < n+k+1$, so the floor of $x+k$ must be $n+k$. This property removes most casework and is a frequent JEE trick. In expressions like $[x+3]$, since $3$ is an integer, it immediately becomes $[x]+3$ exactly (not approximately, and not conditionally). Many mistakes happen when students hesitate near integer boundaries, but integer shifts are always exact and safe. This identity often converts a complicated-looking function into a clean polynomial in $n=[x]$, after which factoring and sign analysis become straightforward.

3. Converting GIF expressions to integer algebra for sign questions

Once you substitute $n=[x]$, the expression becomes a function of an integer variable, so sign analysis reduces to checking an integer quadratic or product. For example, if $f(x)=n^2-n-6$, you factor it as $(n-3)(n+2)$. The sign is positive outside the roots and negative between the roots, but the important point is that $n$ is restricted to integers. So the inequality $-2

4. Integrals of step functions built from $[x]$

Integrating functions involving $[x]$ is easiest when you treat them as step functions. Since $[x]$ is constant on every interval $[n,n+1)$, any expression depending on $[x]$ is also constant there. Therefore, to compute $\int_a^b f(x)\,dx$, you split the integration range at all integers between $a$ and $b$, compute the constant value on each piece, and multiply by the piece length. This turns calculus into simple arithmetic. In this question, the interval $[0,2]$ splits into $[0,1)$ and $[1,2)$, and on each part the value is the same, so the integral is just the sum of two equal rectangles. This method is exact, fast, and avoids graphing. It also aligns perfectly with JEE expectations: correct partitioning, correct constants, and clean summation without missing jump points at integers.

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FAQs

What is the definition of $[x]$?

⌄

$[x]=n$ exactly when $n\le x < n+1$, with $n\in\mathbb{Z}$. It is the floor of $x$.

Why can we set $n=[x]$?

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Because $[x]$ always equals an integer. Writing $n=[x]$ converts the question into integer algebra plus interval mapping.

Is $[x+3]=[x]+3$ always valid?

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Yes. Since $3$ is an integer, integer shift property gives $[x+3]=[x]+3$ for every real $x$.

What is $f(x)$ in terms of $n=[x]$?

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$f(x)=n^2-(n+3)-3=n^2-n-6=(n-3)(n+2)$.

Where does $f(x)=0$ happen?

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When $n=[x]=3$ or $n=[x]=-2$, i.e., on $[3,4)$ and $[-2,-1)$. Hence zeros are infinite in count.

For which $x$ is $f(x)<0$?

⌄

$f(x)<0$ for $n=-1,0,1,2$, which corresponds to $x\in[-1,3)$.

How to compute $\int_0^2 f(x)\,dx$ quickly?

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Split into $[0,1)$ and $[1,2)$. On each, $f(x)=-6$. So the integral is $-6-6=-12$.

Why is option (C) false?

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Because the correct value is $\int_0^2 f(x)\,dx=-12$, not $-6$.

Is $f(x)>0$ only for $x\ge 4$?

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No. $f(x)>0$ also for $x<-2$, so positivity is on $(-\infty,-2)\cup[4,\infty)$.

Which option is correct finally?

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Option (B): $f(x) < 0$ only for $x \in [-1, 3)$.

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