82. Let R be the set of all real numbers. Consider the sets

P = {x ∈ R : (x − 1)(x² + 1) = 0}

Q = {x ∈ R : x² − 9x + 2 = 0}

S = {x ∈ R : x = 5y for some y ∈ R}

Then the set

(P ∩ S) ∪ Q

contains

(A) exactly two elements

(B) exactly three elements

(C) exactly four elements

(D) infinitely many elements

Finding the Number of Elements in (P ∩ S) ∪ Q

The present question combines polynomial equations with set operations. Instead of directly evaluating unions and intersections, we must first determine the individual sets by solving the given equations. Once the elements of each set are identified, the required set operation becomes straightforward. This problem illustrates the importance of understanding both algebraic equations and the basic properties of sets.

Correct Answer

Option (B): Exactly three elements

Understanding the Given Sets

The problem involves three different sets, each defined in a different manner.

• P is obtained by solving a polynomial equation.
• Q is obtained by solving a quadratic equation.
• S is described using a property involving multiples of five.

Our objective is to determine each set separately and then perform the required intersection and union operations.

Step 1: Determine the Set P

The set is defined by

(x − 1)(x² + 1)=0.

A product is zero if at least one factor is zero.

The first factor gives

x−1=0

or

x=1.

The second factor gives

x²+1=0

or

x²=−1.

Since there is no real number whose square is negative, this equation has no real solution.

Therefore,

P={1}.

Step 2: Determine the Set Q

The set is defined by

x²−9x+2=0.

Using the quadratic formula,

x=(9±√(81−8))/2

=(9±√73)/2.

Thus,

Q={(9−√73)/2,(9+√73)/2}.

Hence, Q contains exactly two distinct real numbers.

Step 3: Determine the Set S

The set is defined by

x=5y

for some real number y.

Since y can be any real number, every real number x can be written as

y=x/5.

Because x/5 is always a real number whenever x is real, every real number belongs to S.

Therefore,

S=R.

This means S is simply the set of all real numbers.

Step 4: Find P ∩ S

Since

S=R,

every element of P automatically belongs to S.

Therefore,

P∩S=P={1}.

Step 5: Compute (P ∩ S) ∪ Q

We have

P∩S={1}

and

Q={(9−√73)/2,(9+√73)/2}.

The number 1 is different from both quadratic roots because

(9−√73)/2≈0.228

and

(9+√73)/2≈8.772.

Hence, the union is

{1,(9−√73)/2,(9+√73)/2}.

This set contains exactly

three distinct elements.

Mathematical Verification

P={1} ✔

Q has two distinct roots ✔

S=R ✔

P∩S={1} ✔

(P∩S)∪Q has three distinct elements ✔

The calculation is completely verified.

Explanation of Every Option

Option (A): Exactly two elements

This is incorrect because the union contains the element 1 in addition to the two distinct roots of the quadratic equation.

Option (B): Exactly three elements

This is correct because the resulting union contains one element from P and two distinct elements from Q.

Option (C): Exactly four elements

This is incorrect because the quadratic equation contributes only two real roots, not three.

Option (D): Infinitely many elements

This is incorrect because although S itself is infinite, the intersection P∩S equals the finite set P. The final union therefore remains finite.

Alternative Method

Observe immediately that

S={5y:y∈R}=R.

Hence,

P∩S=P.

The question then reduces simply to finding

P∪Q.

Since P contains one element and Q contains two different elements, the answer can be obtained almost instantly.

Related Practice Example

Let

A={x∈R:x²−4=0}

B={x∈R:x=2y,y∈R}

Find

A∩B.

Since

A={−2,2}

and

B=R,

we obtain

A∩B={−2,2}.

This follows exactly the same reasoning used in the present question.

Key Takeaways

Whenever sets are defined using equations, determine each set individually before performing any union or intersection. Also remember that every real number can be written as five times another real number, so the set {5y:y∈R} is simply the entire set of real numbers. Recognizing such properties greatly simplifies many set theory problems.

Final Answer

The sets are

P={1}

Q={(9−√73)/2,(9+√73)/2}

S=R.

Therefore,

(P∩S)∪Q={1,(9−√73)/2,(9+√73)/2},

which contains exactly three elements.

Correct Option: (B)