Let $T = \left\{x \mid \dfrac{x+5}{x-7} - 5 = \dfrac{4x-40}{13-x}\right\}$. Is $T$ an empty set? Justify your answer.

> [!TLDR] > The problem asks if the set $T$ of solutions to the given rational equation is empty. By solving the equation and checking for domain restrictions, we find that $x=10$ is the only valid solution, meaning $T = \{10\}$, and thus **$T$ is not an empty set**. **Concept and Intuition:** The set $T$ is defined by an algebraic equation involving rational expressions. To determine if $T$ is an empty set, we need to find all values of $x$ that satisfy this equation. If no such values exist, or if all values we find are invalid due to domain restrictions (like making a denominator zero), then $T$ is empty. Otherwise, if at least one valid solution exists, $T$ is not empty. The process involves three main stages: 1. **Identify domain restrictions:** Before any algebraic manipulation, determine the values of $x$ for which any denominator in the original equation becomes zero. These values are excluded from the domain of $x$. 2. **Solve the equation:** Manipulate the equation algebraically to find potential solutions for $x$. 3. **Verify solutions:** Check if the potential solutions found in step 2 violate any of the domain restrictions identified in step 1. Any solution that violates a restriction is an extraneous solution and must be discarded. 4. **Form the set $T$:** The set $T$ consists only of the valid solutions. If this set is empty, then the answer is yes; otherwise, no. Let's proceed with these steps. 1. **Identify Domain Restrictions:** The given equation is $\dfrac{x+5}{x-7} - 5 = \dfrac{4x-40}{13-x}$. For the expressions to be defined, the denominators cannot be zero. * The first denominator is $x-7$. So, $x-7 \neq 0 \implies x \neq 7$. * The second denominator is $13-x$. So, $13-x \neq 0 \implies x \neq 13$. These are the critical restrictions. Any solution we find must not be $7$ or $13$. 2. **Simplify the Equation:** First, let's combine the terms on the left side of the equation by finding a common denominator:

Let T = \x mid (x+5)/(x-7) - 5 = (4x-40)/(13-x)\. – Class 11 Mathematics

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T = \left\{x \mid \dfrac{x+5}{x-7} - 5 = \dfrac{4x-40}{13-x}\right\}

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Concept understanding — Null Set Identification

Null Set Identification

1. The intuition: "Is there anything inside?"

Imagine a bag. You look inside.

If it has no objects at all, it's an empty bag.
In mathematics, a set with no elements is called a null set (or empty set), written $\emptyset$ or $\{\}$ .

Null set identification is simply the process of checking whether a given set has any elements at all.

Everyday examples

Situation	Set	Null?
Students in a class who are 200 years old	$\{x \mid x \text{ is a student aged } 200\}$	✓ Yes — no such student exists
Even numbers strictly between 2 and 4	$\{x \mid x \text{ is even}, 2 < x < 4\}$	✓ Yes — only 3 lies strictly between 2 and 4, and 3 is odd
Prime numbers less than 2	$\{x \mid x \text{ is prime}, x < 2\}$	✓ Yes — the smallest prime number is 2

2. The precise statement

Definition: A set $A$ is a null set (or empty set) if it contains no elements. We write $A = \emptyset$ or $A = \{\}$ .

To identify whether a given set $S$ is null:

List the elements, if you can — if the list has nothing in it, it's null.
Test the defining condition — if no object can ever satisfy it, the set is null.
Compare against a known-empty pattern — e.g. $\{x \mid x \neq x\}$ is always empty, for any universe.

Key properties to remember

Uniqueness: there is exactly one null set; $\emptyset$ and $\{\}$ name the same set.
Subset of everything: $\emptyset \subseteq A$ for every set $A$ .
Cardinality: $|\emptyset| = 0$ .
Not the same as $\{0\}$ : $\emptyset$ has zero elements; $\{0\}$ has one element, the number 0.

3. Worked examples

Example 1 — direct check. Is $A = \{\text{months with 32 days}\}$ null?

No month has 32 days, so $A = \emptyset$ . ✓ Null.

Example 2 — using a condition. Is $B = \{x \in \mathbb{N} \mid x^2 = -1\}$ null?

Squares of natural numbers are always positive, so no $x$ satisfies $x^2=-1$ . $B = \emptyset$ . ✓ Null.

Example 3 — the tricky case. Is $C = \{\emptyset\}$ null?

$C$ contains exactly one element — the empty set itself. Having an "empty thing" as your one element still makes the set non-empty. $|C| = 1$ , so $C$ is not null.

4. Why it matters for exams

Set operations: $A \cup \emptyset = A$ , $A \cap \emptyset = \emptyset$ , $A - \emptyset = A$ .
Proofs: showing two sets are disjoint is usually done by showing $A \cap B = \emptyset$ . …

Why this formula?

Null Set Identification: Understanding the "Why" Behind the Definition

What is a Null Set?

A null set (also called an empty set) is a set that contains no elements. It is denoted by the symbol $\emptyset$ or $\{ \}$ .

The key idea is not a "formula" in the traditional sense, but a definitional truth that flows from the very nature of what a set is.

The Core Principle: Why $\emptyset$ Exists

The Axiom of Empty Set (Zermelo-Fraenkel Set Theory)

In formal set theory, the existence of the null set is an axiom — a starting assumption we accept without proof. But we can understand why it must be true:

Intuition: If you can have a set of apples, you can also have a set of nothing. The concept of "collection" does not require that the collection actually contain anything.

The Defining Property

The null set is the unique set that satisfies:

\forall x \; (x \notin \emptyset)

This reads: "For every object $x$ , $x$ is not an element of the empty set."

Why this holds:

If there were any element in $\emptyset$ , it would violate the definition.
The statement is vacuously true — there is no counterexample because there is nothing to check.

Key "Formulae" and Their Reasoning

1. Cardinality: $|\emptyset| = 0$

Why:

The cardinality of a set is the number of distinct elements it contains. Since $\emptyset$ has no elements, the count is zero.

|\emptyset| = 0

This is not a derived result — it is the definition of zero applied to set size.

2. Subset Relationship: $\emptyset \subseteq A$ for any set $A$

Why this is always true:

The definition of subset: $A \subseteq B$ means "every element of $A$ is also an element of $B$ ."

For $\emptyset \subseteq A$ :

We need to check: "Is every element of $\emptyset$ also in $A$ ?"
Since $\emptyset$ has no elements, there is nothing to check.
The statement is vacuously true — there is no counterexample.

Key insight: The empty set is a subset of every set, including itself.

3. Union with Null Set: $A \cup \emptyset = A$

Why:

The union of two sets contains all elements that belong to at least one of them.

A \cup \emptyset = \{ x \;|\; x \in A \text{ or } x \in \emptyset \}

Since $\emptyset$ contributes no new elements, the result is exactly $A$ .

Derivation:

If $x \in A$ , then $x \in A \cup \emptyset$ (by definition of union).
If $x \notin A$ , then $x \notin \emptyset$ (since nothing is in $\emptyset$ ), so $x \notin A \cup \emptyset$ .
Therefore, $A \cup \emptyset$ contains exactly the elements of $A$ .

4. Intersection with Null Set: $A \cap \emptyset = \emptyset$

Why:

The intersection of two sets contains elements that belong to both sets.

A \cap \emptyset = \{ x \;|\; x \in A \text{ and } x \in \emptyset \}

Since $\emptyset$ has no elements, the condition " $x \in \emptyset$ " is never true. Therefore, no element can satisfy both conditions.

Result: The intersection is empty — it is the null set itself.

--- …

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The problem asks if the set $T$ of solutions to the given rational equation is empty. By solving the equation and checking for domain restrictions, we find that $x=10$ is the only valid solution, meaning $T = \{10\}$ , and thus $T$ is not an empty set.

Concept and Intuition:

The set $T$ is defined by an algebraic equation involving rational expressions. To determine if $T$ is an empty set, we need to find all values of $x$ that satisfy this equation. If no such values exist, or if all values we find are invalid due to domain restrictions (like making a denominator zero), then $T$ is empty. Otherwise, if at least one valid solution exists, $T$ is not empty.

The process involves three main stages:

Identify domain restrictions: Before any algebraic manipulation, determine the values of $x$ for which any denominator in the original equation becomes zero. These values are excluded from the domain of $x$ .
Solve the equation: Manipulate the equation algebraically to find potential solutions for $x$ .
Verify solutions: Check if the potential solutions found in step 2 violate any of the domain restrictions identified in step 1. Any solution that violates a restriction is an extraneous solution and must be discarded.
Form the set $T$ : The set $T$ consists only of the valid solutions. If this set is empty, then the answer is yes; otherwise, no.

Let's proceed with these steps.

Identify Domain Restrictions:

The given equation is $\dfrac{x+5}{x-7} - 5 = \dfrac{4x-40}{13-x}$ .

For the expressions to be defined, the denominators cannot be zero.
- The first denominator is $x-7$ . So, $x-7 \neq 0 \implies x \neq 7$ .
- The second denominator is $13-x$ . So, $13-x \neq 0 \implies x \neq 13$ . These are the critical restrictions. Any solution we find must not be $7$ or $13$ .
Simplify the Equation:

First, let's combine the terms on the left side of the equation by finding a common denominator:

\dfrac{x+5}{x-7} - 5 = \dfrac{x+5 - 5(x-7)}{x-7}

= \dfrac{x+5 - 5x + 35}{x-7}

= \dfrac{-4x+40}{x-7}

Next, let's rewrite the right side to make the denominator $x-13$ instead of $13-x$. This often simplifies further steps by having consistent forms:

\dfrac{4x-40}{13-x} = \dfrac{4x-40}{-(x-13)} = -\dfrac{4x-40}{x-13}

Now, the original equation becomes:

\dfrac{-4x+40}{x-7} = -\dfrac{4x-40}{x-13}

Notice that $-4x+40 = -4(x-10)$ and $4x-40 = 4(x-10)$. Substituting these into the equation:

\dfrac{-4(x-10)}{x-7} = -\dfrac{4(x-10)}{x-13}

To solve, move all terms to one side to set the equation to zero:

\dfrac{-4(x-10)}{x-7} + \dfrac{4(x-10)}{x-13} = 0

Factor out the common term $4(x-10)$:

4(x-10) \left( \dfrac{-1}{x-7} + \dfrac{1}{x-13} \right) = 0

Solve the Simplified Equation:

For the product of two factors to be zero, at least one of the factors must be zero. This gives us two cases:

Case A: $4(x-10) = 0$

Dividing by $4$ (which is non-zero):

$x-10 = 0$

$x = 10$

Case B: $\dfrac{-1}{x-7} + \dfrac{1}{x-13} = 0$

Rearrange the terms:

$\dfrac{1}{x-13} = \dfrac{1}{x-7}$ …

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Q.Let T={x∣x+5x−7−5=4x−4013−x}T = \left\{x \mid \dfrac{x+5}{x-7} - 5 = \dfrac{4x-40}{13-x}\right\}T={x∣x−7x+5​−5=13−x4x−40​}. Is TTT an empty set? Justify your answer.

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Q.Let $T = \left\{x \mid \dfrac{x+5}{x-7} - 5 = \dfrac{4x-40}{13-x}\right\}$ . Is $T$ an empty set? Justify your answer.

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