Let $[x]$ denote the greatest integer $\le x$, where $x \in \mathbf{R}$. If the domain of the real valued function $f(x) = \sqrt{\frac{|[x]|-2}{|[x]|-3}}$ is $(-\infty, a) \cup [b, c) \cup [4, \infty)$, $a < b < c$, then the value of $a + b + c$ is: Options: (A) 8 (B) 1 (C) -2 (D) -3

### Analyzing the Setup Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a problem that involves finding the domain of the function: $$f(x) = \sqrt{\frac{|[x]|-2}{|[x]|-3}}$$ The presence of the greatest integer function $[x]$ adds a layer of discrete complexity to our continuous world. ### The Square Root Constraint The fundamental rule for this function is that the expression inside the square root must be non-negative. Therefore, we require: $$\frac{|[x]|-2}{|[x]|-3} \ge 0$$ Additionally, we must ensure the denominator is not zero. This implies $|[x]| - 3 \neq 0$, or simply $|[x]| \neq 3$. ### The Substitution Strategy To simplify the inequality, let us introduce the substitution $y = |[x]|$. Our inequality transforms into: $$\frac{y-2}{y-3} \ge 0$$ The critical points for this rational inequality are $y=2$ and $y=3$. Applying the wavy curve method, we observe that the expression is positive for $y \le 2$ and $y > 3$. ### The Floor Function Trap Now, we translate these conditions back to $x$. **Case 1:** $|[x]| \le 2$. This implies $-2 \le [x] \le 2$. Since $[x]$ must be an integer, the possible values are $\{-2, -1, 0, 1, 2\}$. The union of the corresponding intervals $[k, k+1)$ for these integers is: $$[-2, -1) \cup [-1, 0) \cup [0, 1) \cup [1, 2) \cup [2, 3) = [-2, 3)$$ **Case 2:** $|[x]| > 3$. This splits into two sub-cases: $[x] > 3$ and $[x] 3$, then $[x] \ge 4$, which implies $x \ge 4$. If $[x] < -3$, then $[x] \le -4$, which implies $x < -3$. ### Synthesis and Conclusion Combining all valid regions, the domain of the function is: $$(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$$ Comparing this with the form $(-\infty, a) \cup [b, c) \cup

JEE Main 2021 (20 July Shift 1)

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Animated Solution for Mathematics - Functions: Let $[x]$ denote the greatest integer $\leq x$ , where $x \in R$ . If the domain of the real valued function $f (x) = \frac{∣ [ x ] ∣ - 2}{∣ [ x ] ∣ - 3}$ is $(- \infty, a) \cup [b, c) \cup [4, \infty)$ , $a < b < c$ , then the value of $a + b + c$ is:

Q: Wait, why is the interval for $[x] \in \{-2, -1, 0, 1, 2\}$ equal to $[-2, 3)$ and not $[-2, 2]$?

That is a fantastic question! Remember the definition of the greatest integer function $[x]$. If $[x] = 2$, then $x$ can be any value from $2$ up to, but not including, $3$. If $x$ were $3$, then $[x]$ would be $3$. Therefore, the interval for $[x]=2$ is $[2, 3)$. When you combine all the intervals for $[x] = -2, -1, 0, 1, 2$, you get the union $[-2, -1) \cup [-1, 0) \cup [0, 1) \cup [1, 2) \cup [2, 3)$, which simplifies beautifully to $[-2, 3)$.

Q: Why do we reject $y=3$ in the inequality?

In the expression $\frac{|[x]|-2}{|[x]|-3}$, if $|[x]| = 3$, the denominator becomes zero. Division by zero is undefined in the real number system. Therefore, any value of $x$ that makes $|[x]| = 3$ must be excluded from the domain. This is why we use a strict inequality $y > 3$ instead of $y \ge 3$.

Q: How do we handle the modulus in $|[x]| > 3$?

The inequality $|[x]| > 3$ means the distance of $[x]$ from zero is greater than $3$. This splits into two distinct possibilities: either $[x] > 3$ or $[x] 3$ means $[x] \ge 4$, and $[x] < -3$ means $[x] \le -4$. Solving these gives you the intervals $[4, \infty)$ and $(-\infty, -3)$ respectively.

Select Answer:

Progressive Hints

Hint 1

The Spark: For the function

f (x) = g (x)

to be defined, the expression

g (x)

must be non-negative. Visualize the graph of the function inside the square root; where does the numerator and denominator change sign? Also, remember that the denominator cannot be zero.

Hint 2

The Arsenal: Use the substitution

y = ∣ [x] ∣

to simplify the inequality

\frac{∣ [ x ] ∣ - 2}{∣ [ x ] ∣ - 3} \geq 0

. This transforms the problem into a standard rational inequality in terms of

y

. Once you solve for

y

, you must translate these intervals back into conditions for

[x]

and finally for

x

Hint 3

The Execution: After solving the inequality for

y

, you will get

y \leq 2

y > 3

. For the first case,

∣ [x] ∣ \leq 2

implies

- 2 \leq [x] \leq 2

. List the possible integer values for

[x]

and determine the corresponding intervals for

x

for each integer value.

Visualized Solution

Condition for Domain

For $f (x) = g (x)$ to be defined, $g (x) \geq 0$ .
Here, $g (x) = \frac{∣ [ x ] ∣ - 2}{∣ [ x ] ∣ - 3}$ .
So, $\frac{∣ [ x ] ∣ - 2}{∣ [ x ] ∣ - 3} \geq 0$ .
Also, denominator $∣ [x] ∣ - 3 \neq = 0$ .

Substitution and Wavy Curve

Let $y = ∣ [x] ∣$ .
The inequality becomes $\frac{y - 2}{y - 3} \geq 0$ .
Critical points for $y$ are $2$ and $3$ .

Solving for $y$

Using the wavy curve method on $\frac{y - 2}{y - 3} \geq 0$ .
Solution: $y \leq 2$ or $y > 3$ .
Note: $y \neq = 3$ because it's in the denominator.

Case 1: $∣ [x] ∣ \leq 2$

Substitute $y$ back: $∣ [x] ∣ \leq 2$ .
This means $- 2 \leq [x] \leq 2$ .
Since $[x]$ is an integer, $[x] \in {- 2, - 1, 0, 1, 2}$ .

Interval for Case 1

If $[x] \in {- 2, - 1, 0, 1, 2}$ , what is $x$ ?
The smallest value is when $[x] = - 2 ⟹ x \geq - 2$ .
The largest value is when $[x] = 2 ⟹ x < 3$ .
So, $x \in [- 2, 3)$ .

Case 2: $∣ [x] ∣ > 3$

Substitute $y$ back: $∣ [x] ∣ > 3$ .
This splits into two sub-cases:
Sub-case A: $[x] > 3$ .
Sub-case B: $[x] < - 3$ .

Solving Sub-case A: $[x] > 3$

Since $[x]$ is an integer, $[x] > 3 ⟹ [x] \geq 4$ .
The condition $[x] \geq 4$ means $x \geq 4$ .
Interval: $[4, \infty)$ .

Solving Sub-case B: $[x] < - 3$

Since $[x]$ is an integer, $[x] < - 3 ⟹ [x] \leq - 4$ .
The condition $[x] \leq - 4$ means $x < - 3$ .
Interval: $(- \infty, - 3)$ .

Combining the Intervals

The complete domain is the union of all valid intervals.
Domain: $(- \infty, - 3) \cup [- 2, 3) \cup [4, \infty)$ .
The problem states the domain is $(- \infty, a) \cup [b, c) \cup [4, \infty)$ .

Finding $a, b, c$ and Final Answer

Comparing the intervals: $a = - 3$ , $b = - 2$ , $c = 3$ .
Check condition $a < b < c$ : $- 3 < - 2 < 3$ (True).
We need to find $a + b + c$ .
$a + b + c = - 3 + (- 2) + 3 = - 2$ .

00:00 / 00:00

The Sigma Insight: Domain and Range of a Function

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a problem that involves finding the domain of the function:

f (x) = \frac{∣ [ x ] ∣ - 2}{∣ [ x ] ∣ - 3}

The presence of the greatest integer function

[x]

adds a layer of discrete complexity to our continuous world.

The Square Root Constraint

The fundamental rule for this function is that the expression inside the square root must be non-negative. Therefore, we require:

\frac{∣ [ x ] ∣ - 2}{∣ [ x ] ∣ - 3} \geq 0

Additionally, we must ensure the denominator is not zero. This implies $|[x]| - 3 eq 0$, or simply $|[x]| eq 3$.

The Substitution Strategy

To simplify the inequality, let us introduce the substitution

y = ∣ [x] ∣

. Our inequality transforms into:

\frac{y - 2}{y - 3} \geq 0

The critical points for this rational inequality are

y = 2

and

y = 3

. Applying the wavy curve method, we observe that the expression is positive for

y \leq 2

and

y > 3

The Floor Function Trap

Now, we translate these conditions back to

x

Case 1:

∣ [x] ∣ \leq 2

. This implies

- 2 \leq [x] \leq 2

Since

[x]

must be an integer, the possible values are

{- 2, - 1, 0, 1, 2}

. The union of the corresponding intervals

[k, k + 1)

for these integers is:

[- 2, - 1) \cup [- 1, 0) \cup [0, 1) \cup [1, 2) \cup [2, 3) = [- 2, 3)

Case 2:

∣ [x] ∣ > 3

. This splits into two sub-cases:

[x] > 3

and

[x] < - 3

[x] > 3

, then

[x] \geq 4

, which implies

x \geq 4

. If

[x] < - 3

, then

[x] \leq - 4

, which implies

x < - 3

Synthesis and Conclusion

Combining all valid regions, the domain of the function is:

(- \infty, - 3) \cup [- 2, 3) \cup [4, \infty)

Comparing this with the form $(-\infty, a) \cup [b, c) \cup

Common Traps & FAQs

Wait, why is the interval for

[x] \in {- 2, - 1, 0, 1, 2}

equal to

[- 2, 3)

and not

[- 2, 2]

That is a fantastic question! Remember the definition of the greatest integer function

[x]

. If

[x] = 2

, then

x

can be any value from

2

up to, but not including,

3

. If

x

were

3

, then

[x]

would be

3

. Therefore, the interval for

[x] = 2

[2, 3)

. When you combine all the intervals for

[x] = - 2, - 1, 0, 1, 2

, you get the union

[- 2, - 1) \cup [- 1, 0) \cup [0, 1) \cup [1, 2) \cup [2, 3)

, which simplifies beautifully to

[- 2, 3)

Why do we reject

y = 3

in the inequality?

In the expression

\frac{∣ [ x ] ∣ - 2}{∣ [ x ] ∣ - 3}

, if

∣ [x] ∣ = 3

, the denominator becomes zero. Division by zero is undefined in the real number system. Therefore, any value of

x

that makes

∣ [x] ∣ = 3

must be excluded from the domain. This is why we use a strict inequality

y > 3

instead of

y \geq 3

How do we handle the modulus in

∣ [x] ∣ > 3

The inequality

∣ [x] ∣ > 3

means the distance of

[x]

from zero is greater than

3

. This splits into two distinct possibilities: either

[x] > 3

[x] < - 3

. Since

[x]

must be an integer,

[x] > 3

means

[x] \geq 4

, and

[x] < - 3

means

[x] \leq - 4

. Solving these gives you the intervals

[4, \infty)

and

(- \infty, - 3)

respectively.

Select Answer:

Visualized Solution

Condition for Domain

Substitution and Wavy Curve

Solving for $y$

Case 1: $∣ [x] ∣ \leq 2$

Interval for Case 1

Case 2: $∣ [x] ∣ > 3$

Solving Sub-case A: $[x] > 3$

Solving Sub-case B: $[x] < - 3$

Combining the Intervals

Finding $a, b, c$ and Final Answer

The Sigma Insight: Domain and Range of a Function

Analyzing the Setup

The Square Root Constraint

The Substitution Strategy

The Floor Function Trap

Synthesis and Conclusion

Similar Questions

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Select Answer:

Visualized Solution

Condition for Domain

Substitution and Wavy Curve

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣[x]∣≤2

Interval for Case 1

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣[x]∣>3

Solving Sub-case A: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x]>3

Solving Sub-case B: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x]<−3

Combining the Intervals

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,c and Final Answer

The Sigma Insight: Domain and Range of a Function

Analyzing the Setup

The Square Root Constraint

The Substitution Strategy

The Floor Function Trap

Synthesis and Conclusion

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Solving for $y$

Case 1: $∣ [x] ∣ \leq 2$

Case 2: $∣ [x] ∣ > 3$

Solving Sub-case A: $[x] > 3$

Solving Sub-case B: $[x] < - 3$

Finding $a, b, c$ and Final Answer

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