R = {(x, y): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)}
It is seen that (1, 1) ∉ R.
∴ R is not reflexive.
(1, 6) ∈R
But,
(6, 1) ∉ R.
∴ R is not symmetric.
Now, since there is no pair in R such that (x, y) and (y, z) ∈R, then (x, z) cannot belong to R.
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

R is reflexive, since every triangle is congruent to itself. Further, (T₁, T₂) \(\in\) R \(\Rightarrow\) T₁ is congruent to T₂ \(\Rightarrow\) T₂ is congruent to T₁ Þ (T₂, T₁) \(\in\) R. Hence, R is symmetric. Moreover, (T₁, T₂), (T₂, T₃) \(\in\) R \(\Rightarrow\) T₁ is congruent to T₂ and T₂ is congruent to T₃ \(\Rightarrow\) T₁ is congruent to T₃ \(\Rightarrow\) (T₁, T₃) \(\in\) R. Therefore, R is an equivalence relation.

f is not onto, as for some y ∈ R from co-domain, there is no x ∈ R from domain such that y = f{x), e.g. for - 2 ∈ R (co-domain) there is no x ∈ R (domain) such that f{x) = -2, i.e. |x|= -2. Hence, not onto.

\(f \circ g(x)=f(g(x))=f\left(\frac{3+x}{2}\right)=2\left(\frac{3+x}{2}\right)-3=3+x-3=x\)
\(g \circ f(x)=g(f(x))=g(2 x-3)=\frac{3+2 x-3}{2}=\frac{2 x}{2}=x\)
As fog and gof are identify functions, so one is inverse of the other

fog(x) = f(g(x))
= f(x^1/3) = 8(x^1/3)³
= 8x

Let \(y=\frac{3 x-4}{5} \Rightarrow 5 y=3 x-4\)
\(\Rightarrow x=f^{-1}(y)=\frac{5 y+4}{3} \Rightarrow f^{-1}(x)=\frac{5 x+4}{3}\)

\( a * b=a+3 b^2 \\ 2 * 4=2+3(4)^2 \\ =2+3 \times 16=50 \)

Here  \(f(x)=y=\frac { 1 }{ x } \Rightarrow x=\frac { 1 }{ y } .\)
\(\therefore \)   \(f^{ -1 }=\{ (y,x)\} =\left\{ \left( y,\frac { 1 }{ y } \right) \right\} \)
\(\Rightarrow \)  \(f^{ -1 }(x)=\frac { 1 }{ x } .\)
\(\therefore \)  \(fof^{ -1 }(x)=f\left( \frac { 1 }{ x } \right) =\frac { 1 }{ 1/x } =x.\)
Also     \(f^{ -1 }\quad of\quad (x)=f^{ -1 }\left( \frac { 1 }{ x } \right) =\frac { 1 }{ 1/x } =x.\)
Thus  \(fof^{ -1 }(x)=f^{ -1 }\quad of\quad (x)\)
Hence, \(fof^{ -1 }=f^{ -1 }\quad of.\)

We have: \(f(x)=\frac { (4x+3) }{ 6x-4) } ,x\neq \frac { 2 }{ 3 } .\)
(a) \(fof(x)=f(f(x))=f\frac { (4x+3) }{ 6x-4) } \)
\(=\frac { 4\left( \frac { (4x+3) }{ 6x-4) } \right) +3 }{ 6\left( \frac { (4x+3) }{ 6x-4) } \right) -4 } \)
\(=\frac { 16x+12+x8x-12 }{ 24x+18-24x+16 } =\frac { 34x }{ 34 } =x.\)
(b) Let \(y=\frac { (4x+3) }{ 6x-4) } \)
\(\Rightarrow \) \(6xy-4y=4x+3\Rightarrow (6y-4)x=4y+3\)
\(\Rightarrow \)  \(x=\frac { (4y+3) }{ 6y-4) } \)
\(\Rightarrow \)  \(g(y)=f^{ -1 }(y)=\frac { (4y+3) }{ 6y-4) } \)
\(\therefore \)   \(f^{ -1 }(x)=\frac { (4x+3) }{ 6x-4) } =f(x)\)
Hence, \(f^{ -1 }=f\) .

The binary operation "*' on the set {0, 1, 2, 3, 4, 5} is defined as:
\(a*b=\left\{ a+b,\quad if\quad a+b<6\\ a+b-6,\quad if\quad a+b\ge 6. \right\} \)
Thus we have the operation table:
Operation Table

\(\left[ \because \quad 0*0=0,.......,5*5=5;0*1=1,...............,5*1=5+1-6=0;etc. \right] \)
Existence of Identity:
\(a*0=a=a+0\) for each \(a\in \{ 0,1,2,3,4,5\} \)
\(\left[ \because \quad 0+0=0,1+0=1,\quad ........,\quad 5+0=5 \right] \)
Hence, '0' is identity for the given operation.
Existence of Inverse:
If 'b' be the inverse of 'a', then:
a*b = 0 = b*a.
Now \(a*b=0\Rightarrow a+b-6=0\Rightarrow b=6-a.\)
\(\because \)  For each element 'a', '6-a' is the inverse of a.
When a = 1, then b = 6-1 = 5.
When a = 2, then b = 6-2 = 4.
When a = 3, then b = 6-3 = 3.
When a = 4, then b = 6-4 = 2.
When a = 5, then b = 6-5 = 1.
Hence, the inverse of 1, 2, 3, 4 and 5 is 5, 4, 3, 2 and 1 respectively

\(f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \) \(9x_{ 1 }^{ 2 }+6x_{ 1 }-5=9x_{ 2 }^{ 2 }+6x_{ 2 }-5\)
\(\Rightarrow \) \(9(x_{ 1 }^{ 2 }-x_{ 2 }^{ 2 })+6(x_{ 1 }-x_{ 2 })=0\)
\(\Rightarrow \) \((x_{ 1 }-x_{ 2 })\quad (9(x_{ 1 }+x_{ 2 })+6=0\)
\(\Rightarrow \) \(x_{ 1 }-x_{ 2 }=0\Rightarrow x_{ 1 }=x_{ 2 }\)
\(\Rightarrow \) \(f\) is one-one \(\Rightarrow \) \(f\) is invertible
Let \(y=f(x)=9x^{ 2 }+6x-5\)
\(\Rightarrow \) \(​​​​9x^{ 2 }+6x-(5+y)=0\)
\(\Rightarrow \) \(x=\frac { -6\pm \sqrt { 36+36(5+y) } }{ 18 } \)
\(=\frac { -6+6\sqrt { 6+y } }{ 18 } =\frac { -1\pm \sqrt { y+6 } }{ 3 } \)
Since \(x\ge 0,\)
\(\therefore \) \(x=\frac { -1-\sqrt { y+6 } }{ 3 } \) is not possible.
\(\therefore \) \(x=\frac { -1+\sqrt { y+6 } }{ 3 } \).
Hence \(f^{ -1 }(y)=\frac { \sqrt { y+6 } -1 }{ 3 } \)

Let A = {1, 2, 3}.
A relation R on A is defined as R = {(1, 2), (2, 1)}.
It is seen that (1, 1), (2, 2), (3, 3) ∉R.
∴ R is not reflexive.
Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.
Now, (1, 2) and (2, 1) ∈ R
However,
(1, 1) ∉ R
∴ R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.

We have gof (x) = g(f (x)) = g(cos x) = 3 (cos x)² = 3 cos² x. Similarly, fog(x) = f (g(x)) = f (3x²) = cos (3x²). Note that 3cos² x \(\ne\) cos 3x², for x = 0. Hence, gof \(\ne\) fog.

Suppose f is not one-one. Then there exists two elements, say 1 and 2 in the domain whose image in the co-domain is same. Also, the image of 3 under f can be only one element. Therefore, the range set can have at the most two elements of the co-domain {1, 2, 3}, showing that f is not onto, a contradiction. Hence, f must be one-one.

No two different students of the class can have same roll number. Therefore, f must be one-one. We can assume without any loss of generality that roll numbers of students are from 1 to 50. This implies that 51 in N is not roll number of any student of the class, so that 51 can not be image of any element of X under f. Hence, f is not onto.