Set Theory Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

Given that system of equations has no solution

⇒ |A| = 0

|A| = – 30 + 6 k

But |A| = 0

⇒ 6k – 30 = 0

⇒ k = 5

∴ Option (b) is correct.

Answer: (c)

$x^2 – y^2$ = 0

(Equation of y = |x|)

$(x – a)^2 + y^2$ = 1

[Equation of a circle with centre (a, 0) and radius 1]

These have single positive solution

⇒ y = |x| to be tangent to the circle

AB = radius of circle

∠AOB = 45°

∠OAB = 90°

[tangent to the circle]

∠ABO = 45°

Length of OB = $√{AB^2 + OA^2} = √{1^2 + 1^2}$

OB = $√2$

Answer: (c)

65x – 33y = 97 .... (i)

(given)

33x – 65y = 1 ....(ii)

(given)

From eq (i) + eq (ii)

98x – 98y = 98

98(x – y) = 98 ∴ x – y = 1 ....(iii)

Now

From eq (i) × 1 – eq (iii) × 33

65x - 33y = 97

- 33x + 33y = - 33

32x = 64

∴ $x = 64/32$ = 2

From eq (iii)

x – y = 1

2 – y = 1

∴ y = 2 – 1 = 1

Hence, xy = 2 × 1 = 2

Answer: (a)

Here, P = {..., –6, –3, 0, 3, 6, ...}

Q = {..., –8, – 4, 0, 4, 8, ...}

and R = {..., –36, –6, 0, 6, 36, ...}

I. P ∪ Q = {..., –8, –6, – 4, –3, 0 3, 4, 6, 8, ...} ≠ R

II. Here, P ⊄ R

III. Here, R ⊂ (P ∪ Q) is true.

Answer: (d)

Number of students registered for

Physics n(P) = 50

Mathematics n(M) = 75

Number of students registered for both subjects n(P ∩ M) = 35.

Number of students, registered for either physics or mathematics

n(P ∪ M) = n(P) + n(M) – n(P ∩ M).

= 50 + 75 – 35 = 90

∴ Number of students registered neither for physics nor for mathematics

n($\ov{P ∪ M}$) = 250 – 90 = 160.

Answer: (b)

not possible

(2, 1) → $2^2$ – 4 × 1 ≥ 0 possible

(3, 1) → possible

(3, 2), (4, 1), (4, 2) (4, 3) (4, 4) → possible

So → possible values can be possible