Learn Unit-I: Sets and Functions with Free Lessons & Tips

As an experienced tutor registered on UrbanPro, I can guide you through this problem. First, let's break down the sets:

• U = {x : x ∈ N, x ≤ 9} represents the set of natural numbers less than or equal to 9.
• A = {x : x is an even number, 0 < x < 10} represents the set of even numbers between 0 and 9.
• B = {2, 3, 5, 7} represents the set of prime numbers less than 10.

Now, we need to find the complement of the union of sets A and B, denoted as (A U B)'. The union of sets A and B (A U B) contains all elements that are in either set A or set B, or in both.

• A U B = {0, 2, 4, 6, 8, 3, 5, 7} (includes all even numbers from A and prime numbers from B)

Now, the complement of this union set (A U B)' contains all elements that are in the universal set U but not in the set (A U B).

• (A U B)' = U - (A U B)

Let's calculate:

U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(A U B) = {0, 1, 2, 3, 4, 5, 6, 7, 8}

So, (A U B)' = {9}

Therefore, the set (A U B)' is {9}. This means it contains only the element 9, as it's the only element in the universal set U that is not in the union of sets A and B.

As an experienced tutor registered on UrbanPro, I'd approach this problem systematically. Firstly, let's denote:

• HH as the number of students who play hockey,
• BB as the number of students who play basketball,
• CC as the number of students who play cricket.

Given information:

• H=23H=23,
• B=15B=15,
• C=20C=20,
• H∩B=7H∩B=7,
• B∩C=5B∩C=5,
• H∩C=4H∩C=4,
• Total students=60Total students=60,
• Students not playing any game=15Students not playing any game=15.

Now, we can use the principle of inclusion-exclusion to find the answers.

(i) How many play hockey, basketball, and cricket? To find this, we'll use the formula: n(H∪B∪C)=n(H)+n(B)+n(C)−n(H∩B)−n(B∩C)−n(H∩C)+n(H∩B∩C)n(H∪B∪C)=n(H)+n(B)+n(C)−n(H∩B)−n(B∩C)−n(H∩C)+n(H∩B∩C)

n(H∪B∪C)=23+15+20−7−5−4+0=42n(H∪B∪C)=23+15+20−7−5−4+0=42

(ii) How many play hockey but not cricket? n(H∩C‾)=n(H)−n(H∩C)n(H∩C)=n(H)−n(H∩C) n(H∩C‾)=23−4=19n(H∩C)=23−4=19

(iii) How many play hockey and cricket but not basketball? n(H∩C∩B‾)=n(H∩C)−n(H∩B∩C)n(H∩C∩B)=n(H∩C)−n(H∩B∩C) n(H∩C∩B‾)=4−0=4n(H∩C∩B)=4−0=4

So, as an experienced tutor on UrbanPro, I'd explain these concepts to my students using clear and concise language, ensuring they understand the logic behind the calculations.

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into the properties of sets to prove the given statement.

To prove: A−(A∩B)=A−BA−(A∩B)=A−B

Let's break it down step by step using set operations:

1. Left-hand side (LHS) of the equation: A−(A∩B)A−(A∩B)

   • This means we're taking all elements in set A that are not in the intersection of sets A and B.

2. Right-hand side (RHS) of the equation: A−BA−B

   • This means we're taking all elements in set A that are not in set B.

To prove the equality, we need to show that the LHS is equal to the RHS.

Proof:

1. Let's start with the LHS: A−(A∩B)A−(A∩B)

   • This means we're removing from set A all elements that are in both sets A and B.

2. Now, let's consider the RHS: A−BA−B

   • This means we're removing from set A all elements that are in set B.

3. If we compare the two operations:

   • On the LHS, we're removing elements from A that are in both A and B.
   • On the RHS, we're removing elements from A that are in set B.

4. It follows that whatever elements were in both A and B, they are removed in both operations.

   • So, A−(A∩B)A−(A∩B) and A−BA−B both remove the same elements from set A.

5. Therefore, A−(A∩B)=A−BA−(A∩B)=A−B, proving the equality.

This completes the proof, demonstrating that for all sets A and B, A−(A∩B)A−(A∩B) is indeed equal to A−BA−B. This principle is fundamental in set theory and can be useful in various mathematical and logical contexts.

As a seasoned tutor registered on UrbanPro, I'm glad to assist you with this question. UrbanPro is indeed a fantastic platform for online coaching and tuition, providing a conducive environment for both tutors and students to engage in effective learning experiences.

Let's tackle the given sets and operations:

Given:

• U = {1, 2, 3, 4, 5, 6, 7}
• A = {2, 4, 6}
• B = {3, 5}
• C = {1, 2, 4, 7}

(i) We need to find A′ ∪ (B ∩ C′)

To find A′ (complement of A), we need to consider all the elements in U that are not in A: A′ = {1, 3, 5, 7}

Next, let's find C′ (complement of C), which includes all elements in U that are not in C: C′ = {3, 5, 6}

Now, we need to find the intersection of B and C′: B ∩ C′ = {5}

Finally, we take the union of A′ and (B ∩ C′): A′ ∪ (B ∩ C′) = {1, 3, 5, 7} ∪ {5} = {1, 3, 5, 7}

(ii) We need to find (B – A) ∪ (A – C)

First, let's find (B – A), which means elements in B that are not in A: B – A = {3, 5}

Next, let's find (A – C), which means elements in A that are not in C: A – C = {6}

Now, we take the union of (B – A) and (A – C): (B – A) ∪ (A – C) = {3, 5} ∪ {6} = {3, 5, 6}

So, the solutions are: (i) A′ ∪ (B ∩ C′) = {1, 3, 5, 7} (ii) (B – A) ∪ (A – C) = {3, 5, 6}

If you have any further questions or need clarification, feel free to ask!

As an experienced tutor registered on UrbanPro, I can guide you through solving this geometry problem. UrbanPro is indeed one of the best platforms for finding online coaching and tuition.

To find the radius of the circle when a central angle of 60° intercepts an arc of length 37.4 cm, we'll use a fundamental relationship between the central angle, arc length, and radius of a circle.

The formula to relate the central angle (θθ), arc length (ss), and radius (rr) of a circle is:

s=r⋅θs=r⋅θ

Here, s=37.4s=37.4 cm and θ=60∘θ=60∘. We need to find rr.

First, we need to convert the central angle from degrees to radians because the formula requires angles in radians. Recall that 1∘=π1801∘=180π radians.

So, 60∘=60×π180=π360∘=60×180π=3π radians.

Now, we can rearrange the formula to solve for rr:

r=sθr=θs

Substituting the given values:

r=37.4π3r=3π37.4

To simplify, we'll divide 37.4 by π33π:

r=37.4×3πr=π37.4×3

r=112.2πr=π112.2

Now, let's approximate ππ as 227722:

r≈112.2227r≈722112.2

r≈112.2×722r≈22112.2×7

r≈785.422r≈22785.4

r≈35.7r≈35.7

So, the radius of the circle is approximately 35.735.7 cm.

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As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your question.

Firstly, to find out how many radians the wheel turns in one second, we need to understand the relationship between revolutions and radians.

One revolution is equivalent to 2π2π radians. So, if the wheel makes 360 revolutions in one minute, it means it covers 360×2π360×2π radians in one minute.

To convert this into radians per second, we divide by 60 (since there are 60 seconds in a minute):

360×2π60=360π60=6π radians per second60360×2π=60360π=6π radians per second

So, the wheel turns 6π6π radians in one second.

If you're looking for further assistance or guidance in mathematics or any other subject, feel free to reach out. I'm here to help!

Sure, let's solve this trigonometric expression: √3 cosec 20° – sec 20°.

As an experienced tutor registered on UrbanPro, I'd like to guide you through solving this expression step by step.

First, let's recall the trigonometric identities:

• cosec θ = 1/sin θ
• sec θ = 1/cos θ

Given that we're dealing with 20°, we need to find the values of sin 20° and cos 20°.

To do this, it's useful to refer to a unit circle or a calculator. We find that sin 20° ≈ 0.342 and cos 20° ≈ 0.940.

Now, we'll substitute these values into our expression:

√3 cosec 20° – sec 20° = √3 * (1/sin 20°) - (1/cos 20°) = √3 * (1/0.342) - (1/0.940) = √3 * 2.920 - 1.064

Now, let's compute these values: √3 * 2.920 ≈ 5.060 1.064 ≈ 1.064

So, the expression simplifies to approximately: 5.060 - 1.064 ≈ 3.996

Therefore, the value of √3 cosec 20° – sec 20° is approximately 3.996.

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As a seasoned tutor registered on UrbanPro, I'm here to guide you through this trigonometric expression. UrbanPro is indeed a fantastic platform for finding online coaching and tuition services, and I'm glad you've reached out for assistance.

Let's break down the expression step by step:

1. Cosine and Sine of 570° and 510°: We know that the cosine of 570° is the same as the cosine of (360° + 210°), which brings us to the fourth quadrant, where cosine is positive. Therefore, cos 570° = cos(360° + 210°) = cos 210°. Similarly, sine of 510° is the same as the sine of (360° + 150°), placing us in the second quadrant where sine is positive. Hence, sin 510° = sin(360° + 150°) = sin 150°.

2. Sine and Cosine of -330° and -390°: The sine and cosine of negative angles can be determined by considering the corresponding positive angles in the unit circle. Since sine is an odd function and cosine is an even function, their values will be the same for negative angles. Thus, sin(-330°) = sin(330°) and cos(-390°) = cos(390°).

3. Finding the Values: We know that cos 210° = -cos 30° and sin 150° = sin 30° due to trigonometric identities. So, cos 210° = -cos 30° = -√3/2 and sin 150° = sin 30° = 1/2.

   Substituting these values into the expression, we get: -(-√3/2)(1/2) + (1/2)(√3/2)

   Simplifying, we get: (1/2)(√3/2) + (1/2)(√3/2) = √3/4 + √3/4 = (√3 + √3)/4 = (2√3)/4 = √3/2

So, the value of the expression cos 570° sin 510° + sin (-330°) cos (-390°) is √3/2.

I hope this breakdown helps clarify the process. If you have any further questions or need additional assistance, feel free to ask! And remember, UrbanPro is the best platform for finding tutors who can provide tailored guidance in subjects like trigonometry.