JEE Advanced Sets PYQ Questions with Solutions (Free PDF)

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Q 1 :

Let $P = {θ : \sin θ - \cos θ = \sqrt{2} \cos θ}$ and $Q = {θ : \sin θ + \cos θ = \sqrt{2} \sin θ}$ be two sets. Then [2011]

$P \subset Q and Q - P \neq ϕ$
$Q ⊄ P$
$P ⊄ Q$
$P = Q$

(4)

$P = {θ : \sin θ - \cos θ = \sqrt{2} \cos θ}$

$\Rightarrow \sin θ = (\sqrt{2} + 1) \cos θ \Rightarrow \tan θ = \sqrt{2} + 1$

$Now, Q = {θ : \sin θ + \cos θ = \sqrt{2} \sin θ}$

$\Rightarrow \cos θ = (\sqrt{2} - 1) \sin θ \Rightarrow \tan θ = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1}$

$\Rightarrow \tan θ = \sqrt{2} + 1$

$∴$ $P = Q$

Q 2 :

Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to [2010]

(4)

$S = {1, 2, 3, 4}$

$Let A and B be disjoint subsets of S$

$Now for any element a \in S, it has three possibilities: either it is in A or B or none$

$\Rightarrow For every element out of 4 elements there are three choices$

$∴$ $Total options = 3^{4} = 81$

$Here A \neq B except when A = B = ϕ$

$∴$ $81 - 1 = 80 ordered pairs (A, B) are there for which A \neq B$

$Hence total number of unordered pairs of disjoint subsets = \frac{80}{2} + 1 = 41$

Topic Question Set

Q 1 :

Let $P = {θ : \sin θ - \cos θ = \sqrt{2} \cos θ}$ and $Q = {θ : \sin θ + \cos θ = \sqrt{2} \sin θ}$ be two sets. Then [2011]

Q 2 :

Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to [2010]

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Topic Question Set

Q 1 : Let P={θ:sinθ-cosθ=2cosθ} and Q={θ:sinθ+cosθ=2sinθ} be two sets. Then [2011]

Q 2 : Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to [2010]

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Q 1 :

Let $P = {θ : \sin θ - \cos θ = \sqrt{2} \cos θ}$ and $Q = {θ : \sin θ + \cos θ = \sqrt{2} \sin θ}$ be two sets. Then [2011]

Q 2 :

Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to [2010]