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

We have A = A ⋂ U
A = A ⋂ (B ⋂ B)'    [∵ B U B' =U]
A = (A ⋂ B) U (A ⋂ B')    [⋂ is distribute over union]
A = (A ⋂ B) U Φ    [given A ⋂ B' = Φ]
A = A ⋂ B
A ⊂ B. Hence proved

We have aN = {a x : x ∈ N}
3N = {3x : x ∈ N} = {3, 6, 9, 12, ...}
7N= {7x : x ∈ N} = {7, 14,21,28, ...}
Hence 3N ∩ 7N = {21, 42, ... } = {21x : x ∈ N}
⇒ 3N ∩ 7N = 21N.

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

Given f : R ⇾ R, defined by f(x) = cos x
We know f(0) = cos 0 = 1
and f(2π) = cos 2π = 1
∴ f(0) = f(2π) ⇒ 0 ≠ (2π)
∴ f is not one-one.
Since the values of cos x tie between -1 and 1, the range of f(x) is not equal to its co-domains.
∴ f is not onto.
Hence, f is neither one-one nor onto.

Let x₁, x₂ ∈ N such that f(x₁) = f(x₂). Then f(x₁) = f(x₂) ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂
∴ f is one-one.
Clearly, f takes even values.
∴ No odd number is N has its pre-image.
∴ f is not onto

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.

n(A\(\times\) A) = 16. \(\Rightarrow\) n(A) = 4
Given S = {a,b) \(\in \) A\(\times\)A : a
\(\therefore\) A = {0, 1, 2, -1}
(A\(\times\)A) = {(0, 0)(0, 1)(0, 2)(0, -1),(1, 1) (1, 2)(1, -1)(2, 0)(2, 1)(2, 2)(2, -1)(-1, 0)(-1, 1)(-1, 2)(-1, -1)}
Now, S = {(0, 1)(0, 2)(-1, 0)(-1, 1)(-1, 2)}
\(\therefore\) The remaining elements of S are (0, 2)(-1, 0)(-1, 1)(1, 2)(0, 1) (1, 2)(0, 2)

B-A = {4, 5, 6, 7}
LHS = C-(B-A) = {3, 9}...(1)
C\(\cap \)A = {3}
B' = {1,2,3,8,9}
C\(\cap \)B' = {3, 9}
RHS = (C\(\cap \)A)\(\cup \) (C\(\cap \)B') = {3, 9}...(2)
From (1) and (2), LHS = RHS

(A\(\times\) B) = {(1,4) (1,5) (1,6) (1,7) (2,4) (2,5) (2,6) (2,7) (3,4) (3,5) (3,6) (3,7)}
(B\(\times\)A) = {(4,1) (4,2) (4,3) (5,1) (5,2) (5,3) (6,1) (6,2) (6,3) (7,1) (7,2) (7,3)}
LHS = (A\(\times\)B)\(\cap \)(B\(\times\)A) = { }....(1)
(A\(\cap \)B) = { }, (B\(\cap \)A) = { }
\(\therefore\) RHS = (A\(\cap \)B) \(\times\) (B\(\cap \)A) = { }.....(2)
From (1) and (2), LHS = RHS