\(\log _{a} b \log _{b} c \log _{c} a=\log _{a} c \log _{c} a=\log _{a} a=1\)

\(\text { The domain is }[0, \infty)\)
\(\text { The codomain is also }[0, \infty)\)

\(\text { If } 4 \in X \text { then }(4,4) \notin \mathrm{R}\)
R is not reflexive
Symmetric can be easily checked

Let X = (a,b,c)
Then R = Universal relation
= {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}.
It is transitive

\(\mathrm{T}: x-y \text { is an integer } \Rightarrow x \mathrm{R} y\)
\(\text { i) } x-x=0 \text { is an integer }\)
\(\therefore \text { T is reflexive. }\)
\(\text { ii) }(x-y) \text { is an integer } \Rightarrow y-x \text { is also an integer }\)
\(\therefore \mathrm{T} \text { is symmetry }\)
iii) If (x - y)is an integer and y- z is also an integer, by adding
x - z is also an integer.
\(\therefore \mathrm{T} \text { is transitive }\)
Thus, T is an equivalence relation.
\(\mathrm{S}: \mathrm{y}=x+1 \Rightarrow x \mathrm{~S} y\)
\(\text { i) } x=x+1 \Rightarrow x S x \text { is not true. }\)
\(\therefore S \text { is not reflexive. }\)
Hence T is an equivalence relation but S is not an equivalence relation

\(\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R} \text { is defined by }\)
\(\mathrm{f}(x)=1-|x|\)
\(\text { The range is }(-\infty, 1] \text { as } f(-\infty)=-\infty\)
\(f(0)=1\)
\(f(\infty) =-\infty\)

The given inequality 3x+4y ≤ 60.
Draw the graph of the line 3x+3y =60
Table of values satisfying the equation
3x +4y= 60

Putting (0, 0) in the given inequation, we have 3\(\times\)0+4\(\times\)0≤60⇒ 0≤60
∴ Half plane of 3x+4y≤60 is towards origin.
Also the given inequality x+3y≤30.
Draw the graph of the line x + 3y = 30.
Table of values satisfying the equation
x+3y=30

Putting (0, 0) in the given inequation, we have 0+3\(\times\)0≤30 ⇒0≤30 which is true.
∴ Half plane of x+3y≤30 is towards origin.

Given h = 61.4 + 2.3 F
Given h = 160 ⇒160 = 61.4 + 2.3 F
⇒ 2.3 F = 160-61.4 = 98.6
F =\(\frac { 98.6 }{ 2.3 } \) = 42.87
Given h = 170,
170 = 61.4 + 2.3 F
⇒ 170 - 61.4 = 2.3 F
2.3 F = 108.6
⇒ F =\(\frac { 108.6 }{ 2.3 } \)= 47.23
So the ranges of values are
42.8 < x < 47.23

Observe that a straight line can be drawn if we identify any two points on it. For example, (3, 0) and (0, 3) can be easily identified as two .points on the straight line x + y = 3.
Draw the three straight lines x + y = 3, 2x - y = 5 and -x + 2y = 3.
Now (0, 0) does not satisfy x +y ≥ 3.  Thus, the half plane bounded by x + y = 3, not containing the origin, is the solution set of x + y ≥ 3.
Similarly, the half-plane bounded by 2x - y ≤ 5 containing the origin represents the solution set of the 2x - y ≤ 5

The region represented by -x + 2y ≤ 3 is the half space bounded by the straight line the line -x + 2y = 3 that contains the origin.
The region common to the above three half planes represents the solution set of the given linear inequalities.

Let \(\frac{x+1}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}\)
Then, x + 1 = Ax(x - 1) + B(x - 1) + Cx²
When x = 0, we have B = -1 and when x = 1, we get C = 2
When x = -1 we have 2A - 2B + C = 0 which gives A = -2
Thus, \(\frac{x+1}{x^2(x-1)}=\frac{-2}{x}-\frac{1}{x^2}+\frac{2}{x-1}\)

Given f(x) = \(\frac { \sqrt { 4-{ x }^{ 2 } } }{ \sqrt { { x }^{ 2 }-9 } } \)
When x = 2, f(x) = 0
When x = -2, f(x) = 0
For all the other values, we get negative value in the square root which is not possible.
\(\therefore\) Domain = {2, -2}

\({{{(x-1)}^{2}}\over{x^3+x}}={{{(x-1)}^{2}}\over{x(x^2+1)}}={{A}\over{x}}+{{Bx+C}\over{x^2+1}}\)
\(\Rightarrow\)  (x-1)² = A(x²+ 1) + (Bx + C)(x)
Putting x = 0 in (1) we get,
\(\boxed{1=A}\)
Putting x = 1 in (1) we get. 
0 = A(2) + (B + C)(1)
\(\Rightarrow\)   0 = 2A + B + C
Equating the Coefficient of x² in (1) we get,
1 = A + B \(\Rightarrow\) 1 = 1  B \(\Rightarrow\) \(\boxed{B=0}\)
Substituting A = 1, B = 0 in (2) we get,
0 = 2 + 0 + C \(\Rightarrow\)  \(\boxed{C=-2}\)
\(\therefore\)  \({{{(x-1)}^{2}}\over{x^3+x}}={{1}\over{x}}+{{0x-2}\over{x^2+1}}\)
\(\Rightarrow\)  \({{{(x-1)}^{2}}\over{x^3+x}}={{1}\over{x}}-{{2}\over{x^3+1}}\)

The relation is defined by aRb if a + b \(\le\) 6 for all a, b  \(\in \)N.
a+b \(\le\)6 \(\Rightarrow\) a \(\le\) 6 - b

\(\therefore\) The list of ordered pairs are (5,1) (4, 2) (3, 3) (2, 4) and (1, 5).
(4, 2) \(\in \) R and (2, 4) \(\in \) R \(\Rightarrow\) (4, 4) \(\notin \) R
\(\therefore\) R is not transitive.

Given equations are x² - ax + b = 0 .....(1)
and x² - ex + f = 0 .....(2)
Let \(\alpha\) be the common root.
Let \(\alpha ,\beta \) be the roots of x² - ax + b = 0
Then \(\alpha +\beta =a\) and \(\alpha \beta =b\) ....(3)
Let \(\alpha\), \(\alpha\) be the roots of x² - ex +f = 0 [\(\because\) the roots are equal]
\(\therefore\) \(\alpha\)+\(\alpha\) = e,   \(\alpha \times \alpha =f\)
\(\Rightarrow\)  \(2\alpha =e\)     and   \({ \alpha }^{ 2 }=f\) .....(4)
Now LHS = \(ae=\left( \alpha +\beta \right) 2\alpha \)
LHS = \(ae=2{ \alpha }^{ 2 }+2\alpha \beta \)
= 2(f) + 2b    [From (4)]
= 2 (b + f) = RHS. Hence proved.

Let x = \(\sqrt[3]{(45.4)^2\over (3.2)^2\times(6.5)^3}\)
Taking logarithms of both sides, we get
log x =  \(log\left[ (45.4)^2\over (3.2)^2\times(6.5)^3 \right]^{1/3}\)
\(={1\over 3}log\left(45.4)^2\over (3.2)^2\times(6.5)^2\right)\)
\(={1\over 3}[2log 45.4 - 2 log3.2 - 3 log 6.5]\)
\(={1\over 3}[2(1.6571) - 2(0.5052)- 3 (0.8129)]\)
\(={1\over 3}(-0.1349)= -0.0450\)
= -1 + 1 - 0.0450 = -1.9550
log x  = -1.9550
⇒ x = anti log (-1.9550)
⇒ x = 0.9016

Given A = {a, b, c, d}, B = {a, c, e}, C = {a, e}
B∩C = {a, c}
A∩(B∩C) = {a}
A ∩ B = {a, e}
(A ∩ B) ∩ C = {a}
From (1) and (2), it is clear that A ∩ (B ∩ C) = (A ∩ B) ∩ C.

Given equation is

\(8+9\sqrt{(3x-1)(x-2)}=3x^2-7x\)
⇒ \(9\sqrt{(3x-1)(x-2)}=3x^2-7x-8\)
⇒ \(9\sqrt{3x^2-7x+2}=(3x^2-7x-2)-10\)
Let \(\sqrt{3x^2-7x+2}=y\)
⇒ 9y = y² - 10 ⇒ y² - 9y - 10 = 0
⇒ (y-10)(y + 1) = 0   [By factorisation]
⇒ \(y=0\ or \ -1⇒\sqrt{3x^2-7x+2}=10\ or\ -1\)
Case (i) :
When \(\sqrt{3x^2-7x+2}=10\)
3x² - 7x + 2 = 100
3x² - 7 x - 98 = 0
(x - 7) (3x + 14) = 0
x = 7 or x = \(-14\over 3\)

Case(ii) :
When \(\sqrt{3x^2-7x+2}=-1\)
This is impossible
The roots are 7, \({{-14}\over{3}}\)

y = sin |x|

We know \(|x| =\begin{cases}x\ if\ x\ge0 \\-x\ if\ x<0 \end{cases}\)
∴ sin |x| = sin x if x > 0 and sin |x| = sin (-x) = -sin x if x < 0.
The graph of y = sin (-x) = - sin x is the reflection of the graph of sin x about Y-axis.

\({2x-1\over x}>-1\)
⇒ \({2x-1\over x}+1>-1+1\)
(i.e) \({2x-1+x\over x}>0\)
\(\frac { (3x-1) }{ x } >0\)
⇒ \(\frac { (3x-1)(x) }{ x } >0\times x\)
(i.e) x(3x-1) > 0
x>0 and x\(<\frac { 1 }{ 3 } \)
So  \((-∞,0) \cup \left( \frac { 1 }{ 3 } ,\infty \right) \)

Given that : f(x) =\(\frac { 1 }{ \sqrt { x-5 } } \)
Here, it is clear that/ex) is real when x - 5 > 0 ⇒ x > 5
Hence, the domain = \((5,\infty)\)
Now to find the range put
\(f(x)=y=\frac { 1 }{ \sqrt { x-5 } } \)
\(⇒ \sqrt { x-5 } =\frac { 1 }{ y } \Rightarrow x-5=\frac { 1 }{ { y }^{ 2 } } \)
\(⇒ x=\frac { 1 }{ { y }^{ 2 } } +5\)
For x \(\in \) \((5,\infty),\) y \(\in \) R⁺.
Hence, the range of f = R⁺

To check whether compositions can be defined, let us find the domain and range of these functions.
Domain of f = {1, 2, 3}, Range of f = {2, 4}, Domain of g = {2, 3, 4} and Range of g = {1, 2}. Since the range of f is contained in the domain of g we can define g o f, so as to find the image of 1 under g o f, we first find the image of 1 under f and then its image under g. The image of 1 under f is 2 and its image under g is 1. So (g o f) (1) = g(f(1)) = g(2) = 1.
Similarly we find that (g o f) (2) = 1 and (g o f) (3) = 2. So g o f = {(1, 1), (2, 1), (3, 2)}.
Similarly f o g = {(2, 2), (3, 2), (4, 2)}.

Observe that \(\sqrt{x+20}\) is defined only if x + 20 ≥ 0
By definition, \(\sqrt{x+20}\ge0\). So, x is positive.
Now squaring we get x²= x + 20. x²- x - 20 = 0
(x - 5)(x + 4) = 0, which gives x = 5, x = -4
Since, x is positive, the required solution is x = 5

Let f(x) = ax² + bx + c be the polynomial satisfying the given conditions.
f(0) = a(0)² + b(0) + c = 1, implies that c = 1. Now the other two conditions f(-2) = 0; f(1) = 0 give 4a - 2b + c = 0 and a + b + c = 0.
Using c = 1, we get 4a - 2b = -1 and a + b = -1. Solving these two equations we get a = b = \(-\frac{1}{2}\) and thus, we have f(x) =\(-\frac{1}{2}x^2-\frac{1}{2}x+1\)

Note that 3x - 9 ≥ 0 implies 3x ≥ 9, by multiplying both sides by \(\frac{1}{3}\) we get x ≥ 3.
Similarly, 4x - 10 ≤ 6 implies 4x ≤ 16 and hence x ≤ 4
So the solution set of 3x - 9 ≥ 0, 4x -10 ≤ 6 is the intersection of [3, ∞) and (-∞, 4]. Clearly, the intersection of these intervals give [3, 4].

3|x - 2| + 7 = 19, So that we have, |x - 2| = \(\frac { 19-7 }{ 3 } =4\). Thus, we have either, x - 2 = 4 (or) x - 2 = -4
Therefore the solutions are x = -2 (or) x = 6

The venn diagram of A and B are disjoint sets but both are subsets of C is

The shaded region is (A ∩ B)\C.

The shaded region is (A ∩ B) U (A ∩ B)

Given a²+ b = c²
Taking power 3 both sides, we get
\(\Rightarrow\)  (a²+ b²) = (c²)³
\(\Rightarrow\)  (a²)³+ 3(a²)²(b²) + 3a²(b²)² + (b²)³ = c⁶  [ \(\because\) (a + b)³ = a³ + 3a²b  3ab² + b³ ]
\(\Rightarrow\)  a⁶+ 3a⁴b²+3a²b⁴+ b⁶= c⁶
\(\Rightarrow\) a⁶+ b⁶+ 3a²b²(a²+b²) = c⁶
\(\Rightarrow\)  a⁶+ b⁶+ 3a²b²c² = c⁶    \([\because a^2+b^2=c^2]\)
Hence proved.

LHS = \(log\frac { { a }^{ 2 } }{ bc } +log\frac { b^{ 2 } }{ ca } +log\frac { c^{ 2 } }{ ab } \)
= \(log\left( \frac { { a }^{ 2 } }{ bc } .\frac { b^{ 2 } }{ ca } .\frac { c^{ 2 } }{ ab } \right) \)
= \(log\left( \frac { { a }^{ 2 }b^{ 2 }c^{ 2 } }{ { a }^{ 2 }b^{ 2 }c^{ 2 } } \right) \)
= log 1 = 0 =RHS
Hence proved

\(\frac{1}{6(x-1)}-\frac{1}{2(x+1)}+\frac{1}{3(x+2)}\)

x>\(\frac{36}{7}\)

Given a³+b³ = ab (8-3a-3b)
⇒ a³+ b³ = 8ab - 3a²b - 3ab²
⇒  a³+b³+3a²b+3ab²=8ab
⇒ (a+b)³= 2³ ab
⇒ \(\left( \frac { a+b }{ 2 } \right) ^{ 3 }\)= ab
⇒  log\(\left( \frac { a+b }{ 2 } \right) \) = log ab    [Taking logarithm on both sides]
⇒ 3 log\(\left( \frac { a+b }{ 2 } \right) \)  log a + log b   [Using power rule and products rule]
⇒  log \(\left( \frac { a+b }{ 2 } \right) \) = \(\frac { 1 }{ 3 } \) [log a + log b]

f associates each students to his/her roll number.
Since, no two different students of the class can have the same roll number, f is one-one.
Here, co-domain = N
Range = {1,2,3,...50}
Since Range ≠ co-domain, f is not onto.

Since A = {1, 2, 3}
(1,1) (2, 2) (3, 3) ∈ R ⇒ is reflexive.
Also (1, 2) ∈ R but (2, 1) ∉ R ⇒ R is not symmetric.
And (1, 2) ∈ R, (2, 3) ∈ R but (1, 3) ∉ R ⇒ R is not transitive.
R is reflexive but neither symmetric nor transitive.

The relation R is defined by xRy if x + 2y = 1 for x, y \(\in \)  N.
Reflexivity : Let x, y \(\in \)  N
xRx \(\Rightarrow\)  x + 2x = 1 \(\Rightarrow\) 3x = 1 \(\Rightarrow\) x = \(\frac { 1 }{ 3 } \notin N\)
\(\therefore\)  R is reflexive.
Symmetricity: xRy \(\Rightarrow\) yRx for x, y \(\in \)  N
xRy \(\Rightarrow\)  x + 2y = 1 which is not possible for any values of x, Y \(\in \) N
\(\therefore\) R is not symmetric
Transitivity: xRy and yRz \(\Rightarrow\) xRz.
xRy and yRz are not possible for any values of x, y, z \(\in \) N
\(\therefore\)  R is not transitive.
\(\therefore\) R is neither reflexive, nor symmetric and not transitive.