Given U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 3, 5, 7, 9}
B = {2, 3, 5, 9, 10}
A' = {2, 4, 6, 8, 10}
B' = {1, 4, 6, 7, 8}
A'UB' = {1, 2, 4, 6, 7, 8, 10}

Given P = {1, 2, 3, 5, 7, 11, 13}
and Q = {2, 4, 6, 8, 10, 12, 14}
Now P⋂Q = {2}
Since P⋂Q ≠ Ø, the given sets are not disjoint.

A \(\times\) (B\(\cap \)C) = (A\(\times\)B) \(\cap \) (A\(\times\)C)
Let A = {1,2,3}, B = (4,5,6,7} C = {4,3,5,9} and \(\cup\) = {1,2,3,4,5,6,7,8,9}
LHS = A x (B\(\cap \)C)
= A \(\times\) {4,5}    [∴B\(\cap \)C = {4, 5}]
= {1,2,3} \(\times\) {4,5}
= {(1,4) (1,5) (2,4) (2,5) (3,4) (3,5)}......(1)
A x B = {1,2,3} x {4,5,6,7}
= {(1,4) (1,5) (1,6) (1,7) (2,4) (2,5) (2,6) (2,7) (3,4) (3,5) (3,6) (3,7)} 
A \(\times\) C = {1,2,3} \(\times\) {3,4,5,6,9}
= {(1,3) (1,4) (1,5) (1,9) (2,3) (2,4) (2,5) (2,6) (2,9) (3,3) (3,4) (3,5) (3,9)} 
RHS = (A \(\times\) B) \(\cap \) (A \(\times\) C) = {(1,4) (1,5) (2,4) (2,5) (3,4) (3,5) }
From (1) and (2), LHS = RHS. Hence Verified....(2)

\(\left\{ x:\frac { x-4 }{ x+2 } =3,x\in R-\{ -2\} \right\} \)
Let D = \(\left\{ x:\frac { x-4 }{ x+2 } =3,x\in R-\{ -2\} \right\} \)
\(\Rightarrow\) D = {x: x - 4 = 3x + 6, x \(\in \) R}
\(\Rightarrow\) D = {x:-4-6 = 3x-x, x  ​​​​​​\(\in \) R}
\(\Rightarrow\) D = {x:2x = -10, x \(\in \) R}
\(\Rightarrow\) D = {x:x = -5, x\(\in \)R}
\(\Rightarrow\) D = {-5}

{x \(\in \)​ ​​​​​​N : 4x + 9 < 52}
Let C = {x\(\in \)N:4x + 9 < 52}
\(\Rightarrow\) C = {x\(\in \)N:4x < 52 - 9}
\(\Rightarrow\) C = {x\(\in \)N: 4x < 43}
\(\Rightarrow\) C = \(\left\{ x\in N:x<\frac { 43 }{ 4 } \right\} \) \(\Rightarrow\) C = {x \(\in \) N : x < 10.75}
\(\Rightarrow\) C = {1,2,3,4,5,6,7,8,9,10}.

Set of concentric circles in a plane is an infinite set since number of concentric circles in a plane is infinite

Given A = {0}  ...(1)
B = {x : x > 15 and x < 5} ⇒ B = Ф  ...(2)
C = {x : x-5 = 0} ⇒ B = {5}   ...(3)
D = { x : x² = 25} ⇒ D = {-5, 5}  ...(4)
E = {x : x is an integral positive root of x² - 2x - 15 = 0}
⇒ E = {5}  ...(5)
From (3) and (5), clearly C = E.
Hence C and E are equal sets

Step 1 :
Draw the Graph of y = x²

Step 2 :
The graph of y =(x -1)², shifts to the right for one unit.
Step 3 :
The graph of y = 3 (x - 1)², compresses towards the Y-axis that is moves away from the X-axis since the multiplying factor is 3 which is greater than 1.
Step 4:
The graph of y = 3 (x - 1)²+ 5, causes the shift to the upward for 5 units.

Let \(f(x)={5x-160\over 9}\)
Given \(y={5x\over 9}-{160\over 9}\)
\(⇒\ \ y={5x-160\over 9}\)
Then 9y = 5x - 160 ⇒ 5x = 9y + 160 ⇒ \(x={9y+160\over5}\)
Let \(g(y)={9y+160\over 5}\)
Now gof(x) = \(g[f(x)=g\left(5x-160\over 6\right)\)
\(={9\left(5x-160\over 9\right)+160\over 5}={5x-160+160\over 5}={5x\over 5}=x\)
and fog(y) = f(g(y)) =\(f\left(9y+160\over 5\right)={5\left(9y+160\over5\right)-160\over 5}={9y+160-160\over 9}=y\)
Thus gof = I_x and fog = I_y
This implies that f and g are bijections and inverses to each other
\(f^{-1}(y)={9y+160\over5}\)
Replacing y by x, we get f^-1 (x) = \(\frac { 9x+160 }{ 5 } \)​​​​​

Given cost function and fuel surcharge function are as follows
c(m) = 0.4m + 50 and s(m) = 0.03m
∴ total cost of a ticket = c(m) + s(m)
∴ f(x) = 0.4m + 50 + 0.03m
= 0.43m + 50
Given m = 1600 miles
Airfare for flying 1600 miles
= 0.43(1600) + 50
= Rs. 738

Given R = {(a, b): a < b²} where a, b ∈ R
reflexivity: We know that \(\left(1\over 2\right)\le\left(1\over 2\right)^2\)is not true
\(⇒\ \left({1\over 2},{1\over 2}\right)∉R\)
⇒ R is not reflexive
Symmetry: We know that -1< 3² but 3 ≰ (-1)² is not true
⇒ (-1, 3) ∈ R but (3, -1)∉R
∴ R is not symmetric
Transitive: We observe that 2< (-3)² and -3< (1)² but 2 ≰ (1)² is not true
⇒ (2, -3) ∈ R and (-3, 1) ∈ R but (2, 1) ∉ R
⇒ R is not transitive
∴ R is neither reflexive nor symmetric nor transitive.

Given (-1, 0) ∈ A x A and (0, 1) ∈ A \(\times\) A
(-1,0) ∈ A \(\times\) A ⇒ -1,0 ∈ A and (0, 1) ∈ A \(\times\) A ⇒ 0, 1 ∈ A
∴ -1,0,1 ∈ A
∴ A = {-1, 0, 1}
Also it is given that A\(\times\) A has 9 elements.
∴ A has exactly three elements.
∴ A {-1, 0, 1}
∴  A \(\times\) A = {(-1, -1) (-1, 0) (-1, 1) (0, -1) (0, 0) (0, 1) (1, -1) (1, 0) (1, 1)}

Given functions are f(x) = x³ and g(x) = \(x^{ ^{ \frac { 1 }{ 3 } } }\)
Now, f o g(x) = f(g(x)
= \(f\left( { x }^{ \frac { 1 }{ 3 } } \right) \)
= \(\left( { x }^{ 3 } \right) ^{ \frac { 1 }{ 3 } }=x\)

Since f o g(x) = x is symmetric about the line y = x, g(x) is the inverse of f(x)
∴ g(x) = f^-1(x).
where f o g(x) = g o f(x) = x so that
(i) f o g is bijective.
(ii) both f(x) and g(x) also bijective.
(iii) f and g are symmetrical about y = x

Given f(x) = 3x - 4
Let y = 3x - 4 ⇒ y + 4 = 3x
\(⇒ x={y+4\over 3}\)
Let g(y) = \(y+4\over 3\)
Now gof(n) = g(f(n)) = g(3\(\times\) -4) = \({3x-4+4\over 3}={3x\over 3}=x\)
and fog(y) = f(g(y)) = \(f\left(y+4\over 4\right)=3\left(y+4\over 3\right)-4=y+4-4=y\)
Thus, gof(x) = I_x and fog (y) = I_y
This implies that f and g are bijections and inverses to each other
Hence f is bijection and \(f^{-1} (x)={y+4\over 3}\)
Replacing y by x, we get f^-1 (x) = \(\frac { x+4 }{ 3 } \)

Hence, the graph of y = f^-1(x) is the reflection of the graph of f in y = x

f(0) = 0, f(-3) = 9 and f(3) = 9
.'. S is [0,9]

It is onto not one-one

\(\text { Since } \sin 30^{\circ}=\frac{1}{2}\)
\(\sin 150^{\circ}=\frac{1}{2}\)