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Animated Solution for Mathematics - Functions: If $S = {a \in R : ∣2 a - 1∣ = 3 [a] + 2 {a}}$ , where $[t]$ denotes the greatest integer less than or equal to $t$ and ${t}$ represents the fractional part of $t$ , then $72 \sum_{a \in S} a$ is equal to ______

Q: Wait, why do we split the modulus at $1/2$?

That is a fantastic question! The modulus function $|x|$ is defined as $x$ if $x \ge 0$ and $-x$ if $x < 0$. In our equation, the expression inside the modulus is $2a - 1$. This expression is zero when $2a = 1$, or $a = 1/2$. For any value of $a$ greater than or equal to $1/2$, $2a - 1$ is non-negative, so the modulus is just $2a - 1$. For $a < 1/2$, $2a - 1$ is negative, so the modulus is $-(2a - 1) = 1 - 2a$. Splitting the problem at this critical point is the only way to remove the modulus sign and proceed with standard algebra.

Q: Why did we substitute $a = I + f$ instead of just solving for $a$ directly?

Solving for $a$ directly is difficult because $[a]$ is not a continuous function; it jumps at every integer. By substituting $a = I + f$, where $I = [a]$ is an integer and $0 \le f < 1$, we transform the equation into one involving an integer $I$ and a bounded variable $f$. This allows us to use the property $0 \le f < 1$ to create inequalities that restrict $I$ to specific integer values. It is a powerful technique for any problem involving the greatest integer function!

Q: Wait, why did we reject the case $a \ge 1/2$?

In the case $a \ge 1/2$, our equation simplified to $[a] = -1$. However, if $a \ge 1/2$, then $a$ is at least $0.5$. The greatest integer of any number greater than or equal to $0.5$ must be at least $0$. Since $-1$ is not greater than or equal to $0$, this is a logical contradiction. Therefore, no value of $a$ in the region $a \ge 1/2$ can satisfy the equation. It is not that we 'rejected' it arbitrarily; the math itself told us that no solutions exist there!

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. The equation

∣2 a - 1∣ = 3 [a] + 2 {a}

looks intimidating because of the mixed functions. The key is to unify the variables using the fundamental identity

a = [a] + {a}

. This allows you to express the fractional part

{a}

in terms of

a

and

[a]

, effectively reducing the number of unknowns.

Hint 2

Hint 2: The Arsenal. Once you substitute

{a} = a - [a]

, the equation becomes

∣2 a - 1∣ = [a] + 2 a

. The modulus

∣2 a - 1∣

is the real hurdle. Remember that the behavior of the modulus changes at the critical point

a = 1/2

. You must split the problem into two distinct cases:

a \geq 1/2

and

a < 1/2

Hint 3

Hint 3: The Execution. In the case

a < 1/2

, you will arrive at an equation involving

[a]

and

a

. Express

a

I + f

(where

I = [a]

and

0 \leq f < 1

). Substitute this into your equation and use the bounds

0 \leq f < 1

to create a compound inequality for

I

. This will constrain

I

to a single integer value, leading you directly to the solution.

Visualized Solution

Analyze the Equation

Given: $∣2 a - 1∣ = 3 [a] + 2 {a}$
Recall: $a = [a] + {a}$
Therefore: ${a} = a - [a]$

Simplify the RHS

Substitute ${a}$ in the RHS:
$RHS = 3 [a] + 2 (a - [a])$
$RHS = 3 [a] + 2 a - 2 [a] = [a] + 2 a$

Simplified Equation

The equation becomes: $∣2 a - 1∣ = [a] + 2 a$
We need to solve this by removing the modulus.
The critical point for the modulus is $a = \frac{1}{2}$ .

Case 1: $a \geq \frac{1}{2}$

Case 1: Assume $a \geq \frac{1}{2}$
The modulus opens positively: $∣2 a - 1∣ = 2 a - 1$
Equation becomes: $2 a - 1 = [a] + 2 a$

Solving Case 1

Cancel $2 a$ from both sides:
$- 1 = [a]$
But if $a \geq \frac{1}{2}$ , then $[a] \geq 0$ .
This is a contradiction! No solution in this case.

Case 2: $a < \frac{1}{2}$

Case 2: Assume $a < \frac{1}{2}$
The modulus opens negatively: $∣2 a - 1∣ = - (2 a - 1) = 1 - 2 a$
Equation becomes: $1 - 2 a = [a] + 2 a$

Isolating $[a]$

Rearrange the equation: $1 - 2 a = [a] + 2 a$
$[a] = 1 - 4 a$
Let $a = I + f$ , where $I = [a]$ and $0 \leq f < 1$ .

Substituting $a = I + f$

Substitute $a = I + f$ and $[a] = I$ :
$I = 1 - 4 (I + f)$
$I = 1 - 4 I - 4 f$
$5 I = 1 - 4 f ⟹ 4 f = 1 - 5 I$

Bounding the Fractional Part

We know the property of fractional part: $0 \leq f < 1$
Multiply by 4: $0 \leq 4 f < 4$
Substitute $4 f = 1 - 5 I$ :
$0 \leq 1 - 5 I < 4$

Solving for Integer $I$

Solve the inequality $0 \leq 1 - 5 I < 4$ :
Subtract 1: $- 1 \leq - 5 I < 3$
Divide by -5 (flip signs): $\frac{1}{5} \geq I > - \frac{3}{5}$
$- 0.6 < I \leq 0.2$
Since $I \in Z$ , the only possible value is $I = 0$ .

Finding $a$

Substitute $I = 0$ back into $4 f = 1 - 5 I$ :
$4 f = 1 - 5 (0) = 1 ⟹ f = \frac{1}{4}$
Calculate $a$ : $a = I + f = 0 + \frac{1}{4} = \frac{1}{4}$
Check condition: $a = \frac{1}{4} < \frac{1}{2}$ (Valid!)

Final Calculation

The set of solutions is $S = {\frac{1}{4}}$ .
We need to find $72 \sum_{a \in S} a$ .
$72 \times \frac{1}{4} = 18$ .
The final answer is 18.

00:00 / 00:00

The Sigma Insight: Classification of Functions

Solution Diagram

Analyzing the Setup

The given equation is

∣2 a - 1∣ = 3 [a] + 2 {a}

. To simplify this, we utilize the fundamental identity for any real number

a

a = [a] + {a} ⟹ {a} = a - [a]

Substituting this into the right-hand side of the equation, we obtain:

3 [a] + 2 (a - [a]) = 3 [a] + 2 a - 2 [a] = [a] + 2 a

The equation now simplifies to the more manageable form:

∣2 a - 1∣ = [a] + 2 a

The Modulus Barrier

The expression

∣2 a - 1∣

changes behavior at the critical point

a = 1/2

. We must analyze the equation by splitting it into two distinct cases based on this value.

Path 1

The Contradiction

Consider the region where

a \geq 1/2

. In this case, the modulus opens positively:

2 a - 1 = [a] + 2 a

Subtracting

2 a

from both sides yields:

[a] = - 1

However, if

a \geq 1/2

, then the greatest integer

[a]

must be at least

0

. Since

[a] = - 1

contradicts the condition

[a] \geq 0

, there are no solutions in this region.

Path 2

The Integer Constraint

Now, consider the region where

a < 1/2

. Here, the modulus opens negatively:

1 - 2 a = [a] + 2 a ⟹ [a] = 1 - 4 a

Let

I = [a]

and

f = {a}

, where

0 \leq f < 1

. Substituting

a = I + f

into the equation gives:

I = 1 - 4 (I + f)

I = 1 - 4 I - 4 f ⟹ 5 I = 1 - 4 f

Since

0 \leq f < 1

, it follows that

0 \leq 4 f < 4

. Substituting

4 f = 1 - 5 I

into this inequality:

0 \leq 1 - 5 I < 4

Solving for

I

- 1 \leq - 5 I < 3 ⟹ \frac{1}{5} \geq I > - \frac{3}{5}

Since

I

must be an integer, the only possible value is $I = 0$ .

Final Calculation

Using

I = 0

in the equation

4 f = 1 - 5 I

, we find:

4 f = 1 ⟹ f = \frac{1}{4}

Thus, the solution is

a = I + f = 0 + 1/4 = 1/4

. Since

1/4 < 1/2

, this is a valid solution.

The set of solutions

S

{1/4}

. The final required value is:

72 a \in S \sum a = 72 \times \frac{1}{4} = 18

The final answer is 18.

Common Traps & FAQs

Wait, why do we split the modulus at

1/2

That is a fantastic question! The modulus function

∣ x ∣

is defined as

x

x \geq 0

and

- x

x < 0

. In our equation, the expression inside the modulus is

2 a - 1

. This expression is zero when

2 a = 1

, or

a = 1/2

. For any value of

a

greater than or equal to

1/2

2 a - 1

is non-negative, so the modulus is just

2 a - 1

. For

a < 1/2

2 a - 1

is negative, so the modulus is

- (2 a - 1) = 1 - 2 a

. Splitting the problem at this critical point is the only way to remove the modulus sign and proceed with standard algebra.

Why did we substitute

a = I + f

instead of just solving for

a

directly?

Solving for

a

directly is difficult because

[a]

is not a continuous function; it jumps at every integer. By substituting

a = I + f

, where

I = [a]

is an integer and

0 \leq f < 1

, we transform the equation into one involving an integer

I

and a bounded variable

f

. This allows us to use the property

0 \leq f < 1

to create inequalities that restrict

I

to specific integer values. It is a powerful technique for any problem involving the greatest integer function!

Wait, why did we reject the case

a \geq 1/2

In the case

a \geq 1/2

, our equation simplified to

[a] = - 1

. However, if

a \geq 1/2

, then

a

is at least

0.5

. The greatest integer of any number greater than or equal to

0.5

must be at least

0

. Since

- 1

is not greater than or equal to

0

, this is a logical contradiction. Therefore, no value of

a

in the region

a \geq 1/2

can satisfy the equation. It is not that we 'rejected' it arbitrarily; the math itself told us that no solutions exist there!

Animated Solution for Mathematics - Functions: If $S = {a \in R : ∣2 a - 1∣ = 3 [a] + 2 {a}}$ , where $[t]$ denotes the greatest integer less than or equal to $t$ and ${t}$ represents the fractional part of $t$ , then $72 \sum_{a \in S} a$ is equal to ______

Enter Numerical Value:

Visualized Solution

Analyze the Equation

Simplify the RHS

Simplified Equation

Case 1: $a \geq \frac{1}{2}$

Solving Case 1

Case 2: $a < \frac{1}{2}$

Isolating $[a]$

Substituting $a = I + f$

Bounding the Fractional Part

Solving for Integer $I$

Finding $a$

Final Calculation

The Sigma Insight: Classification of Functions

Analyzing the Setup

The Modulus Barrier

Path 1

Path 2

Final Calculation

Similar Questions

The sum of the maximum and minimum values of the function $f (x) = ∣5 x - 7∣ + [x^{2} + 2 x]$ in the interval $[\frac{5}{4}, 2]$ , where $[t]$ is the greatest integer $\leq t$ is ________

For $x \in R$ , let $[x]$ denote the greatest integer $\leq x$ , then the sum of the series $[\frac{1}{3}] + [\frac{1}{3} + \frac{1}{100}] + [\frac{1}{3} + \frac{2}{100}] + \dots + [\frac{1}{3} + \frac{99}{100}]$ is

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Let $[t]$ denote the greatest integer $\leq t$ . Then the equation in $x$ , $[x]^{2} + 2 [x + 2] - 7 = 0$ has :

If $f (x) = \frac{2 ^{2 x}}{2 ^{2 x} + 2}, x \in R$ , then $f (\frac{1}{2023}) + f (\frac{2}{2023}) + \dots + f (\frac{2022}{2023})$ is equal to

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The absolute minimum value of the function $f (x) = ∣ x^{2} - x + 1∣ + [x^{2} - x + 1]$ , where $[t]$ denotes the greatest integer function, in the interval $[- 1, 2]$ , is :

If a function $f$ satisfies $f (m + n) = f (m) + f (n)$ for all $m, n \in N$ and $f (1) = 1$ , then the largest natural number $λ$ such that $\sum_{k = 1}^{2022} f (λ + k) \leq (2022)^{2}$ is equal to _________

Enter Numerical Value:

Visualized Solution

Analyze the Equation

Simplify the RHS

Simplified Equation

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a≥21​

Solving Case 1

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a<21​

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [a]

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=I+f

Bounding the Fractional Part

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

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

Final Calculation

The Sigma Insight: Classification of Functions

Analyzing the Setup

The Modulus Barrier

Path 1

Path 2

Final Calculation

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Case 1: $a \geq \frac{1}{2}$

Case 2: $a < \frac{1}{2}$

Isolating $[a]$

Substituting $a = I + f$

Solving for Integer $I$

Finding $a$

The sum of the maximum and minimum values of the function $f (x) = ∣5 x - 7∣ + [x^{2} + 2 x]$ in the interval $[\frac{5}{4}, 2]$ , where $[t]$ is the greatest integer $\leq t$ is ________

For $x \in R$ , let $[x]$ denote the greatest integer $\leq x$ , then the sum of the series $[\frac{1}{3}] + [\frac{1}{3} + \frac{1}{100}] + [\frac{1}{3} + \frac{2}{100}] + \dots + [\frac{1}{3} + \frac{99}{100}]$ is

If $[x]$ be the greatest integer less than or equal to $x$ , then $\sum_{n = 8}^{100} [\frac{( - 1 ) ^{n} n}{2}]$ is equal to:

Let $[α]$ denote the greatest integer $\leq α$ . Then $[1] + [2] + [3] + \dots + [120]$ is equal to

Let $f : R \to R$ be a function defined by $f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b, a \neq = 1$ If $f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y$ , then the value of $28 \sum_{i = 1}^{5} ∣ f (i) ∣$ is

Let $[t]$ denote the greatest integer $\leq t$ . Then the equation in $x$ , $[x]^{2} + 2 [x + 2] - 7 = 0$ has :

If $f (x) = \frac{2 ^{2 x}}{2 ^{2 x} + 2}, x \in R$ , then $f (\frac{1}{2023}) + f (\frac{2}{2023}) + \dots + f (\frac{2022}{2023})$ is equal to

If $f (x) = \frac{2 ^{x}}{2 ^{x} + 2}, x \in R,$ then $\sum_{k = 1}^{81} f (\frac{k}{82})$ is equal to

The absolute minimum value of the function $f (x) = ∣ x^{2} - x + 1∣ + [x^{2} - x + 1]$ , where $[t]$ denotes the greatest integer function, in the interval $[- 1, 2]$ , is :

If a function $f$ satisfies $f (m + n) = f (m) + f (n)$ for all $m, n \in N$ and $f (1) = 1$ , then the largest natural number $λ$ such that $\sum_{k = 1}^{2022} f (λ + k) \leq (2022)^{2}$ is equal to _________