JEE MAIN PART TEST 1

Correct Answer: (B)

Detailed Verification:

Solving for Set A:
|x² – 17| ≤ 8
⇒ -8 ≤ x² – 17 ≤ 8
⇒ 9 ≤ x² ≤ 25

This implies two conditions:
1) x² ≥ 9 ⇒ x ∈ (-∞, -3] ∪ [3, ∞)
2) x² ≤ 25 ⇒ x ∈ [-5, 5]

Taking the intersection of these two conditions:
A = [-5, -3] ∪ [3, 5]

Solving for Set B:
|x – 2| > 2
⇒ x – 2 > 2 or x – 2 < -2
⇒ x > 4 or x < 0
So, B = (-∞, 0) ∪ (4, ∞)

Evaluating the Options:

Option (A) check (A ∪ B):
A ∪ B = ([-5, -3] ∪ [3, 5]) ∪ (-∞, 0) ∪ (4, ∞)
Since [-5, -3] is completely contained within (-∞, 0), we get (-∞, 0).
Since (4, ∞) overlaps with [3, 5], we get [3, ∞).
Therefore, A ∪ B = (-∞, 0) ∪ [3, ∞). Option (A) is incorrect.

Option (B) check (B – A):
B – A consists of elements in B that are not in A.
B = (-∞, 0) ∪ (4, ∞)
Removing the intervals of A ([-5, -3] and [3, 5]):
From (-∞, 0), we subtract [-5, -3], leaving (-∞, -5) ∪ (-3, 0).
From (4, ∞), we subtract [3, 5], leaving (5, ∞).
Thus, B – A = (-∞, -5) ∪ (-3, 0) ∪ (5, ∞). Option (B) is correct.

Option (C) check (A – B):
A – B consists of elements in A that are not in B.
This is equivalent to A ∩ B^c.
B^c = [0, 4].
A ∩ B^c = ([-5, -3] ∪ [3, 5]) ∩ [0, 4] = [3, 4].
Option (C) says [3, 4), missing the element 4. It is incorrect.

Option (D) check (A ∩ B):
A ∩ B = ([-5, -3] ∪ [3, 5]) ∩ ((-∞, 0) ∪ (4, ∞))
[-5, -3] ∩ (-∞, 0) = [-5, -3]
[3, 5] ∩ (4, ∞) = (4, 5]
So, A ∩ B = [-5, -3] ∪ (4, 5]. Option (D) gives [4, 5] instead of (4, 5]. It is incorrect.

Correct Answer: (B)

Detailed Verification:

Solving for Set A:
The set A represents a rectangular region. Let’s find its boundaries:
|α – 3| ≤ 5
⇒ -5 ≤ α – 3 ≤ 5
⇒ -2 ≤ α ≤ 8

|β – 4| ≤ 7
⇒ -7 ≤ β – 4 ≤ 7
⇒ -3 ≤ β ≤ 11

Thus, Set A defines the rectangular region: α ∈ [-2, 8] and β ∈ [-3, 11].

Solving for Set B:
The set B is given by the inequality:
25(α – 4)² + 16(β – 5)² ≤ 400

Dividing the entire inequality by 400, we get the standard form of an ellipse:
(α – 4)² / 16 + (β – 5)² / 25 ≤ 1

This represents a solid elliptical region with its center at (4, 5).
Let’s find the bounding box of this ellipse to see its maximum extents:

• For the maximum horizontal extent (along the α-axis), we set (β – 5)² = 0:
  (α – 4)² / 16 ≤ 1 ⇒ -4 ≤ α – 4 ≤ 4 ⇒ 0 ≤ α ≤ 8.
• For the maximum vertical extent (along the β-axis), we set (α – 4)² = 0:
  (β – 5)² / 25 ≤ 1 ⇒ -5 ≤ β – 5 ≤ 5 ⇒ 0 ≤ β ≤ 10.

So, every point (α, β) in the ellipse B is bounded within: α ∈ [0, 8] and β ∈ [0, 10].

Evaluating the Options:

By comparing the boundaries:
The α-bounds for B [0, 8] are completely inside the α-bounds for A [-2, 8].
The β-bounds for B [0, 10] are completely inside the β-bounds for A [-3, 11].

Since the bounding box of ellipse B is entirely contained within rectangle A, it mathematically guarantees that every single point in B is also in A.
Therefore, B ⊂ A.

Option (D) check:
Since B ⊂ A, their union is simply A (i.e., A ∪ B = A).
The true bounds for A are α ∈ [-2, 8] and β ∈ [-3, 11].
Option (D) incorrectly lists the α-bounds as [-3, 7]. Therefore, (D) is incorrect.

Correct Answer: (B)

Detailed Verification:

Solving for Set A:
The equation is given as:
log_1/2 |sin x| + log_1/2 |cos x| = 2

Using the logarithmic property log(a) + log(b) = log(ab):
log_1/2 (|sin x| · |cos x|) = 2

Converting from logarithmic form to exponential form:
|sin x · cos x| = (1/2)² = 1/4

Multiplying both sides by 2 allows us to use the double-angle identity (2 sin x cos x = sin 2x):
(1/2) |2 sin x cos x| = 1/4
|sin 2x| / 2 = 1/4
|sin 2x| = 1/2

Given that x ∈ (0, π) and x ≠ π/2, the domain for 2x is (0, 2π) excluding π.
In the interval (0, 2π), the equation |sin 2x| = 1/2 yields the following solutions:
2x = π/6, 5π/6, 7π/6, 11π/6

Solving for x, we get:
x = π/12, 5π/12, 7π/12, 11π/12

All 4 solutions are valid and lie perfectly within the specified domain.
Thus, n(A) = 4.

Solving for Set B:
The equation is given as:
√x (√x – 5) – 3√x + 15 = 0

Expanding and grouping the terms algebraically:
x – 5√x – 3√x + 15 = 0
x – 8√x + 15 = 0

Treating this as a quadratic equation in terms of √x and factoring:
(√x – 3)(√x – 5) = 0

This gives two possible real values for √x:
√x = 3 ⇒ x = 9
√x = 5 ⇒ x = 25

Both solutions satisfy the condition x > 0.
Thus, n(B) = 2.

Evaluating n(A ∪ B):
The elements of set A are non-integer multiples of π (irrational numbers), while the elements of set B are the integers 9 and 25.
Because they share no common elements, the sets are disjoint (A ∩ B = ∅).

Therefore, the number of elements in their union is simply the sum of their individual elements:
n(A ∪ B) = n(A) + n(B) = 4 + 2 = 6.

This matches option (B).

Correct Answer: (C)

Detailed Verification:

Analyzing the conditions for Set C:
Set C contains points that belong to both A and B (intersection), with the added restriction that they must lie on the coordinate axes (x = 0 or y = 0).

Let us evaluate the conditions on each axis:

Case 1: Points on the y-axis (where x = 0)
For a point (0, y) to be in B, it must satisfy:
|0| + |y| ≤ 5 ⇒ |y| ≤ 5

For the same point (0, y) to be in A, it must satisfy:
|0 + y| ≥ 5 ⇒ |y| ≥ 5

The only way both conditions (|y| ≤ 5 and |y| ≥ 5) can be simultaneously true is if |y| = 5.
This gives two points: (0, 5) and (0, -5).

Case 2: Points on the x-axis (where y = 0)
For a point (x, 0) to be in B, it must satisfy:
|x| + |0| ≤ 5 ⇒ |x| ≤ 5

For the same point (x, 0) to be in A, it must satisfy:
|x + 0| ≥ 5 ⇒ |x| ≥ 5

Similarly, the only way both conditions (|x| ≤ 5 and |x| ≥ 5) can be true is if |x| = 5.
This gives two points: (5, 0) and (-5, 0).

Evaluating the Sum:
The complete Set C consists of exactly four points:
C = { (5, 0), (-5, 0), (0, 5), (0, -5) }

We are asked to find the sum of |x + y| for all points in C:
∑ |x + y| = |5 + 0| + |-5 + 0| + |0 + 5| + |0 + -5|
∑ |x + y| = 5 + |-5| + 5 + |-5|
∑ |x + y| = 5 + 5 + 5 + 5 = 20

This matches option (C).

Correct Answer: (C)

Detailed Verification:

Understanding Set B:
The set B consists of all proper fractions m/n. The condition gcd(m, n) = 1 ensures that every fraction generated is in its simplest (irreducible) form. Because they are in simplest form, no two pairs (m, n) will produce the same rational number.
Therefore, finding n(B) is equivalent to counting the number of pairs (m, n) such that 1 ≤ m < n ≤ 12 and m is coprime to n.

Calculating using Euler’s Totient Function:
For any given denominator n, the number of valid numerators m (where m < n and gcd(m, n) = 1) is given by Euler’s totient function, denoted as φ(n).
To find the total number of elements in B, we simply need to calculate the sum of φ(n) for all denominators n from 2 to 12.

Let’s evaluate φ(n) for each n ∈ {2, 3, …, 12}:

• n = 2: coprime is {1} ⇒ φ(2) = 1
• n = 3: coprimes are {1, 2} ⇒ φ(3) = 2
• n = 4: coprimes are {1, 3} ⇒ φ(4) = 2
• n = 5: coprimes are {1, 2, 3, 4} ⇒ φ(5) = 4
• n = 6: coprimes are {1, 5} ⇒ φ(6) = 2
• n = 7: coprimes are {1, 2, 3, 4, 5, 6} ⇒ φ(7) = 6
• n = 8: coprimes are {1, 3, 5, 7} ⇒ φ(8) = 4
• n = 9: coprimes are {1, 2, 4, 5, 7, 8} ⇒ φ(9) = 6
• n = 10: coprimes are {1, 3, 7, 9} ⇒ φ(10) = 4
• n = 11: coprimes are {1, 2, 3, …, 10} (prime) ⇒ φ(11) = 10
• n = 12: coprimes are {1, 5, 7, 11} ⇒ φ(12) = 4

Total Sum:
n(B) = 1 + 2 + 2 + 4 + 2 + 6 + 4 + 6 + 4 + 10 + 4 = 45

This exactly matches option (C).

Correct Answer: (B)

Detailed Verification:

Step 1: Find the total number of elements in the given range.
The set of all natural numbers in the interval [150, 850] is S = {150, 151, …, 850}.
Total number of elements in S, let’s call it N(S) = 850 – 150 + 1 = 701.

Step 2: Define sets for the multiples.
Let A₃ be the set of multiples of 3 in S.
Let A₄ be the set of multiples of 4 in S.
The intersection A₃ ∩ A₄ will be the set of multiples of lcm(3, 4) = 12 in S.

Step 3: Calculate the number of multiples of 3, 4, and 12 in the interval.
Using the formula for the number of multiples of k in an interval [a, b]: n = ⌊b/k⌋ – ⌊(a-1)/k⌋.

Number of elements in A₃:
n(A₃) = ⌊850/3⌋ – ⌊149/3⌋
n(A₃) = 283 – 49 = 234

Number of elements in A₄:
n(A₄) = ⌊850/4⌋ – ⌊149/4⌋
n(A₄) = 212 – 37 = 175

Number of elements in A₃ ∩ A₄ (multiples of 12):
n(A₃ ∩ A₄) = ⌊850/12⌋ – ⌊149/12⌋
n(A₃ ∩ A₄) = 70 – 12 = 58

Step 4: Use the Principle of Inclusion-Exclusion.
To find the total number of numbers that are multiples of 3 OR 4 (i.e., n(A₃ ∪ A₄)):
n(A₃ ∪ A₄) = n(A₃) + n(A₄) – n(A₃ ∩ A₄)
n(A₃ ∪ A₄) = 234 + 175 – 58
n(A₃ ∪ A₄) = 409 – 58 = 351

Step 5: Find the numbers that are NEITHER.
We are looking for the complement of the union set.
n(neither 3 nor 4) = Total elements – n(A₃ ∪ A₄)
n(neither 3 nor 4) = 701 – 351 = 350

This exactly matches option (B).

Correct Answer: (D)

Detailed Verification:

Step 1: Finding the values of m and n.
We know that the total number of subsets for a finite set with k elements is 2^k.
According to the problem, the number of subsets of A (which is 2^m) is 112 more than the number of subsets of B (which is 2ⁿ).

This gives the equation:
2^m – 2ⁿ = 112

Factor out 2ⁿ from the left side:
2ⁿ (2^m-n – 1) = 112

Now, find the prime factorization of 112 to identify an odd and an even factor:
112 = 16 × 7 = 2⁴ × (2³ – 1)

By comparing the components on both sides of the equation:
1) 2ⁿ = 2⁴   ⇒   n = 4
2) 2^m-n – 1 = 2³ – 1   ⇒   m – n = 3

Substituting n = 4 into the second equation:
m – 4 = 3   ⇒   m = 7

Step 2: Calculating the distance.
We now have the coordinates for point P(m, n) = (7, 4).
The given point Q is (-5, -1).

Using the distance formula d = √((x₂ – x₁)² + (y₂ – y₁)²):
d = √((7 – (-5))² + (4 – (-1))²)
d = √((12)² + (5)²)
d = √(144 + 25)
d = √169 = 13

This exactly matches option (D).

Correct Answer: (B)

Detailed Verification:

Step 1: Identify given information.
Let n(A), n(B), and n(C) be the number of medals awarded in events A, B, and C.
n(A) = 52
n(B) = 34
n(C) = 22

Total men who received medals: n(A ∪ B ∪ C) = 75
Men who received medals in all three events: n(A ∩ B ∩ C) = 4

Step 2: Use the Principle of Inclusion-Exclusion.
The formula for the union of three sets is:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – [n(A ∩ B) + n(B ∩ C) + n(C ∩ A)] + n(A ∩ B ∩ C)

Let S = n(A ∩ B) + n(B ∩ C) + n(C ∩ A). Substituting the known values:
75 = 52 + 34 + 22 – S + 4
75 = 108 – S + 4
75 = 112 – S
S = 112 – 75 = 37

Step 3: Relate ‘S’ to the exact number of medals.
The sum S = n(A ∩ B) + n(B ∩ C) + n(C ∩ A) counts the people who got exactly two medals once, but it counts the people who got all three medals exactly three times.

Let E₂ = Number of men receiving exactly two medals.
Let E₃ = Number of men receiving all three medals (which is 4).

Therefore, we can write the equation:
S = E₂ + 3(E₃)
37 = E₂ + 3(4)
37 = E₂ + 12
E₂ = 37 – 12 = 25

Thus, exactly 25 men received medals in exactly two of the three events. This matches option (B).

Correct Answer: (B)

Detailed Verification:

Step 1: Define the sets and their percentages.
Let the total number of patients be represented as a universal set U, where n(U) = 100%.
Let n(A) = 76% (patients with ailment A)
Let n(B) = 92% (patients with ailment B)
Let n(A ∩ B) = K% (patients with both ailments)

Step 2: Find the minimum possible value for K.
We know from set theory that the union of two sets cannot exceed the universal set:
n(A ∪ B) ≤ n(U)
n(A) + n(B) – n(A ∩ B) ≤ 100
76 + 92 – K ≤ 100
168 – K ≤ 100
K ≥ 68

Step 3: Find the maximum possible value for K.
The intersection of two sets cannot exceed the size of the smallest individual set:
n(A ∩ B) ≤ min(n(A), n(B))
K ≤ min(76, 92)
K ≤ 76

Step 4: Determine the valid range and check the options.
Combining the inequalities, the possible values for K lie in the closed interval [68, 76].
For K to “not belong to the set”, the given set must have no intersection with the interval [68, 76].

Let’s evaluate the options:
(A) {65, 70, 75, 80} — 70 and 75 are in [68, 76].
(B) {59, 61, 63, 65} — None of these values are in [68, 76].
(C) {66, 71, 76, 81} — 71 and 76 are in [68, 76].
(D) {68, 73, 78, 83} — 68 and 73 are in [68, 76].

Since none of the elements in option (B) can be a valid value for K, K cannot belong to this set.

Correct Answer: (A)

Detailed Verification:

Step 1: Set up the equation based on cases for x.
The given equation is (|x| – 5)|x + 2| = 12.
We analyze the critical points where expressions inside the absolute values change sign, namely at x = 0 and x = -2. This divides the real number line into three cases:

Case 1: x < -2
In this region, x is negative, so |x| = -x, and x + 2 is negative, so |x + 2| = -(x + 2) = -x – 2.
Substitute these into the equation:
(-x – 5)(-x – 2) = 12
(x + 5)(x + 2) = 12
x² + 7x + 10 = 12
x² + 7x – 2 = 0

Using the quadratic formula, the roots are:
x = (-7 ± √(49 – 4(1)(-2))) / 2 = (-7 ± √57) / 2

Let’s check if these roots satisfy the condition x < -2:
* Since √57 ≈ 7.55:
* x₁ = (-7 + 7.55) / 2 = 0.55 / 2 = 0.275 (Rejected, because 0.275 is not < -2)
* x₂ = (-7 – 7.55) / 2 = -14.55 / 2 = -7.275 (Valid, since -7.275 < -2)

Thus, we obtain 1 valid solution from Case 1.

Case 2: -2 ≤ x < 0
In this region, x is negative, so |x| = -x, but x + 2 is non-negative, so |x + 2| = x + 2.
Substitute these into the equation:
(-x – 5)(x + 2) = 12
-(x + 5)(x + 2) = 12
x² + 7x + 10 = -12
x² + 7x + 22 = 0

Checking the discriminant for this quadratic equation:
Discriminant D = 7² – 4(1)(22) = 49 – 88 = -39 < 0.
Since the discriminant is negative, this equation has no real roots.

Case 3: x ≥ 0
In this region, x is non-negative, so |x| = x, and x + 2 is positive, so |x + 2| = x + 2.
Substitute these into the equation:
(x – 5)(x + 2) = 12
x² – 3x – 10 = 12
x² – 3x – 22 = 0

Using the quadratic formula, the roots are:
x = (3 ± √(9 – 4(1)(-22))) / 2 = (3 ± √97) / 2

Let’s check if these roots satisfy the condition x ≥ 0:
* Since √97 ≈ 9.85:
* x₃ = (3 + 9.85) / 2 = 12.85 / 2 = 6.425 (Valid, since 6.425 ≥ 0)
* x₄ = (3 – 9.85) / 2 = -6.85 / 2 = -3.425 (Rejected, because -3.425 is not ≥ 0)

Thus, we obtain 1 valid solution from Case 3.

Conclusion:
The valid solutions for the equation are x = -7.275 and x = 6.425.
The total number of elements in the set is 2.

This matches option (A).