Volume of the cylindrical tank = Volume of the rectangle tank
πr²h = 28 x 16 x 11 m³

\(h=\frac { 16\times 11 }{ 88 } =2m\)

x	x2
30	900
80	6400
60	3600
70	4900
20	400
40	1600
50	2500
Σx = 350	Σx2 = 20300

σ =\(\sqrt { \frac { \Sigma x^{ 2 } }{ n } -\left( \frac { \Sigma x }{ n } \right) ^{ 2 } } \)
=\(\\ \sqrt { \frac { 20300 }{ 7 } -\left( \frac { 350 }{ 7 } \right) ^{ 2 } } \)
=\(\sqrt { 400 } \) = 20

It is not possible to add A and B because they have different orders.

f o f(k) = f(f(k))
= 2(2k - 1) -1 = 4k - 3
Thus, f o f(k) = 4k - 3
But, it is given that f o f(k) = 5
Therefore 4k - 3 = 5 ⇒ k = 2

In \(\triangle\)ABC, AD is the bisector of \(\angle\)A
Therefore by Angle Bisector of \(\angle\)A
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac{4}{3}=\frac{6}{A C}\) gives 4AC = 18. Hence, AC \(=\frac{9}{2}=4.5 \mathrm{~cm}\)

First term = a = 20; common difference = d = 8
Arithmetic Progression is a, a + d , a + 2d , a + 3d,....
In this case , we get 20, 20 + 8, 20 + 2 (8), 20 + 3(8),....
So, the required A.P is 20, 28, 36, 44,.... 

Pictorial representation of R . From the diagram, we see that for each x \(\in \) X, there exists only one y \(\in \) Y. Thus all elements in X have only one image in Y. Therefore R is a function Domain X = {1, 2, 3, 4}; Co-domain Y = {2, 3, 6, 8,10}; Range of f = {2, 4, 6, 8}.

(i) Set builder form of R = ((x, y) | y = x - 2, x \(\in \) P, y \(\in \) Q}
(ii) Roster form R = {(5 , 3),(6 , 4)(7 , 5)}
(iii) Domain of R = {5, 6, 7} and range of R = {3, 4, 5}

tan² \(\theta \) - sin ²\(\theta \) = tan² \(\theta \) -\(\frac { si{ n }^{ 2 }\theta }{ co{ s }^{ 2 }\theta } \),cos²\(\theta \) 
= tan² \(\theta \) (1-cos² \(\theta \) ) = tan² \(\theta \) sin^{2 \(\theta \)}

We observe that A is a 2 x 2 matrix and B is a 2 x 2 matrix, hence AB is defined and it will be of the order 2 x 2.
AB = \(\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 2 & 0 \\ 1 & 3 \end{matrix} \right] =\left[ \begin{matrix} 4+1 & 0+3 \\ 2+3 & 0+9 \end{matrix} \right] =\left[ \begin{matrix} 5 & 3 \\ 5 & 9 \end{matrix} \right] \)
BA = \(\left[ \begin{matrix} 2 & 0 \\ 1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix} \right] =\left[ \begin{matrix} 4+0 & 2+0 \\ 2+3 & 1+9 \end{matrix} \right] =\left[ \begin{matrix} 4 & 2 \\ 5 & 10 \end{matrix} \right] \)
Therefore, AB ≠ BA.

Given the 13^th term = 3 so, t₁₃ = a + 12d = 3..... (1)
Sum of first 13 terms = 234 gives \(\frac { 13 }{ 2 } \) [2a + 12d] = 234
2a + 12 = 36...(2)
Solving (1) and (2) we get , a = 33, d = \(\frac { -5 }{ 2 } \)
Therefore, common difference is \(\frac { -5 }{ 2 } \).
Sum of first 21 terms S₂₁ = \(\frac { 21 }{ 2 } \left[ 2\times 33+\left( 21-1 \right) \times \left( -\frac { 5 }{ 2 } \right) \right] =\frac { 21 }{ 2 }\)[66 - 50] = 168.

Let the three consecutive terms be a - d, a, a + d
Given their sum is 27
(a - d) + a + (a + d) = 27
a - d + a + a + d = 27
3a = 27
\(a=\frac{27}{3}=9\)
Product = 288
(a - d) a (a + d) = 288
a(a² - d²) = 288
9(9² - d²) = 288
\(81-d^{2}=\frac{288}{9}\)
81 - d² = 32
d² = 81 - 32 = 49
d x d = 7 x 7
d = \(\pm\)7
(i) a = 9, d = 7, The three terms are,
= 9 -7, 9, 9 + 7
= 2,9, 16
(ii) a = 9, d = -7, The three terms are
9-(-7), 9, 9-7 \(\Rightarrow\) 16,9,2
The required three consecutive terms of the A.P are 2,9,16.

f: A ⟶ B
f(x) =\(\frac { x }{ 2 } \)-1,
f(2) = \(\frac {2 }{ 2 } -1 = 0\)
f(2) = \(\frac {4 }{ 2 } -1 = 1\)
f(2) = \(\frac {6 }{ 2 } -1 = 2\)
f(2) = \(\frac {8 }{ 2 } -1 = 4\)
f(2) = \(\frac {12 }{ 2 } -1 = 5\)
A = {2,4,6,10,12} B = {0,1,2,4,5,9}
(i) Set of ordered pairs
= {(2,0),(4,1),(6,2),(10,4),(12,5)}
(ii) Table
x 2 4 6 10 12 f(x) 0 1 2 4 5 (iii) Arrow diagram

(iv) Graph
 

Let, x + 2y - z = 5... (1)  
x - y + z = -2....(2)  
-5x - 4y + z = -11... (3)


Here we arrive at an identity 0 = 0
Hence the system has an infinite number of solutions.