WeCP Sets and Function Quiz 1

y=f(x) = sin⁴x + 2 + cos²x =2+ (1 - cos²x)sin²x + cos²x
=2+ sin²x + cos²x - sin²x cos²x = 3- sin²x cos²x = 3- 1/4sin²x
∴ 2¾ ≤ f(x) ≤ 3 (∵ 0 ≤ sin²x ≤ 1)
∴ f(x) ∈ [2¾, 3]

X = {1, 7, 5, 6},   Y = {3, 2, 9, 8},   Z = {5, 9, 7, 6}

YΔ Z = (Y - Z) ∪ (Z - Y)

Y - Z= {3, 2, 9, 8} - {5, 9, 7, 6} = {2, 3, 8}

Z - Y= {5, 9, 7, 6} - {3, 2, 9, 8} = {5, 6, 7}

∴YΔ Z = {2, 3, 5, 6, 7, 8}

A= {1, 3, 5, 15}

B= {2, 4, 6, 8, 10}

∴A∩B = {}

Elements of A= {A, E, L, S, T}

Elements of B= {A, E, L, S, T}, which are same

Hence, A and B are equal.

R={(5,3),(4,2),(6,4),(2,3),(3,5),(2,6)}

Since (2,3) ∈ R and (2,6) ∈ R, therefore R is not a function.

Since (5,3) ∈ R and (3,5) ∈ R but (5,5) ∉ R, therefore R is not transitive.

Since (2,3) ∈ R but (3,2) ∉ R, therefore R is not symmetric.

Since (2,2) ∉ R, therefore R is not reflexive.

Here to find the range of f(x), we will first check the domain of {cos}^{-1}\left( \frac{1+{{x}^{2}}}{2x} \right)

{{\cos }^{-1}}\left( \frac{1+{{x}^{2}}}{2x} \right) is\ define\ if \left| \frac{1+{{x}^{2}}}{2x} \right|\le 1\ and\ [x\ne 0\]

\Rightarrow \[1+{{x}^{2}}-2\left| x \right|\le 0

\Rightarrow {{(\left| x \right|-1)}^{2}}\le 0\ \ or\ \ x=1,-1

Thus, the domain of f(x) is {1,-1}.

Hence, the range of f(x) = \{1,1+\pi \}

\because \{x\}\in [0,\ 1) \Rightarrow sin\{x\}\in (0,\ sin1) as\ f(x)\ is\ defined\ if\ \sin \{x\}\ne 0

i.e., \frac{1}{\sin \{x\}}\in \left( \frac{1}{\sin 1},\infty  \right) \Rightarrow \frac{1}{\sin \{x\}}-1 \in \left( \frac{1}{\sin 1}-1,\infty  \right)

Note: 1<\frac{\pi }{3}\ \Rightarrow  \sin 1<\sin \frac{\pi }{3}=0.866

\Rightarrow \frac{1}{\sin 1}>1.155\  \Rightarrow \frac{1}{\sin 1}-1>0.155

\therefore \left[ \frac{1}{\sin \{x\}}-1 \right]\in \{0,1,2,3...\} , which is the set of whole numbers.

In both the sets APPLE and PALE, they have the same element so option 1 is equal sets.

In both the sets ALLOY and LOYAL, they have the same element so option 2 is also equal sets.

Hence, both are equal sets.

P-B=  {4, 7, 9, 11, 12, 16, 18}- {7, 11, 12, 16,19} = {4,9,18}

AU(P-B)= {4, 5, 7, 9, 12} ∪ {4, 9, 18} = {4, 5, 7, 9, 12, 18}

∴ n(AU(P-B)) = 6

We have-

3³= 27,  4³= 64,  5³= 125,  6³= 216,  7³= 343,  .....

Hence, A = {x : x is the square of n ∀ n∈[3, ∞) and n is natural number}