f(x) = 3 + x, g(x) = x - 4
fog(x) = f(g(x)) =f(x - 4) = 3 + x - 4  = x - 1
gof(x) = g(f(x)) = g(3 + x) = 3 + x - 4 = x - 1
f o g = g o f

x² - 27, (x - 3)², x² - 9
x² - 27 = (x - 3)(x² + 3x + 9); (x - 3)² = (x - 3)²; (x² - 9) = (x + 3)(x - 3)
Therefore, LCM [(x³ - 27), (x-3)², (x² - 9)] = (x - 3)²(x + 3)(x² + 3x + 9)

We observe that A is a 2 x 3 matrix and B is a 3 x 3 matrix, hence AB is defined and it will be of the order 2 × 3..
Given A = \({ \left[ \begin{matrix} 1 & 2 & 0 \\ 3 & 1 & 5 \end{matrix} \right] }_{ 2\times 3 }\), B = \({ \left[ \begin{matrix} 8 & 3 & 1 \\ 2 & 4 & 1 \\ 5 & 3 & 1 \end{matrix} \right] }_{ 2x3 }\)
AB = \({ \left[ \begin{matrix} 1 & 2 & 0 \\ 3 & 1 & 5 \end{matrix} \right] }\times \left[ \begin{matrix} 8 & 3 & 1 \\ 2 & 4 & 1 \\ 5 & 3 & 1 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 8+4+0 & 3+8+0 & 1+2+0 \\ 24+2+25 & 9+4+15 & 3+1+5 \end{matrix} \right] =\left[ \begin{matrix} 12 & 11 & 3 \\ 51 & 28 & 9 \end{matrix} \right] \)

2x² - x - 1 = 0
\({ x }^{ 2 }-\frac { x }{ 2 } -\frac { 1 }{ 2 } \) = 0   (÷2 make co-efficient of x² as 1)
\({ x }^{ 2 }-\frac { x }{ 2 } =\frac { 1 }{ 2 } \)
\({ x }^{ 2 }-\frac { x }{ 2 } +{ \left( \frac { 1 }{ 4 } \right) }^{ 2 }=\frac { 1 }{ 2 } +{ \left( \frac { 1 }{ 4 } \right) }^{ 2 }\)
\({ \left( x-\frac { 1 }{ 4 } \right) }^{ 2 }=\frac { 9 }{ 16 } ={ \left( \frac { 3 }{ 4 } \right) }^{ 2 }\)
\(x-\frac { 1 }{ 4 } =\pm \frac { 3 }{ 4 } \) ⇒ x = 1, \(\frac {-1}{2}\)

1² + 2² +...+ 19² = \(\frac { 19\times \left( 19+1 \right) \left( 2\times 19+1 \right) }{ 6 } =\frac { 19\times 20\times 39 }{ 6 } =2470\)

Slope of the straight line 2x + 3y - 8 = 0 is
m₁ = \(\frac { -coefficient\quad of\quad x }{ cofficient\quad of\quad y } \)
m₂ = \(\frac{-2}{3}\)
Slope of the straight line 4x + 6y + 18 = 0 is
m₂ = \(\frac { -4 }{ 6 } =\frac { -2 }{ 3 } \)
Here, m₁ = m₂
That is, slopes are equal. Hence, the two straight lines are parallel.

Let PQ the height of the tower.
Take PQ = h and QR is the distance between the tower and the point R.in right triangle PQR,\(\angle \)PRQ=30°
tan\(\theta =\frac { PQ }{ QR } \)
tan30° = \(\frac { h }{ 48 } \) gives,\(\frac { 1 }{ \sqrt { 3 } } =\frac { h }{ 48 } \) so, h =\(16\sqrt { 3 } \)
Therefore the height of the tower \(16\sqrt { 3 } \) m

Let r and h be the radius and height of the cylinder respectively.
Given that, height h = 2 m, base area = 250 m²
Now, volume of a cylinder = \(\pi\)r h ² cu. units
= base area x h
= 250 x 2 = 500 m³
Therefore, volume of the cylinder = 500 m³

Let r and h be the radius and height of the cone respectively.
Given that, radius r = 7 m and height h = 24 m
Hence, l = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { 49+576 } \)
\(l=\sqrt { 625 } =25m\)
C.S.A. of the conical tent = \(\pi\)rl sq. units
Area of the canvas \(=\frac { 22 }{ 7 } \times 7\times 25={ 550 }m^{ 2 }\)
Now, length of the canvas \(\frac{Area\ of\ the\ canvas}{width}=\frac{550}{4}=137.5m\)
Therefore, the length of the canvas is 137.5 m

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 \(=\frac{9}{2}=4.5 \mathrm{~cm}\)

A = { 1, 2, 3, 7}, B = { 3, 0, -1, 7}
A x B = {(1, 3), (1, 0), (1, - 1), (1, 7),(2,3), (2, 0), (2, -1), (2, 7), (3,3), (3, 0), (3, - 1), (3, 7), (7, 3)., (7, 0), (7, -1), (7 ,7)}
R₁ = {(2, 1), (7, 1)}
Since (2, 1) and (7,1) are not the elements of A x B, R₁ is not a relation from A to B. Moreover \(1 \notin B .\)

Since 6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
2 is a factor of 6ⁿ. So, 6 n is always even.
But any number whose last digit is 5 is always odd.
Hence, 6ⁿ cannot end with the digit 5 

A x B = {(3,1), (3,7), (3,10), (4,1), (4,7), (4,10), (7,1), (7,7), (7,10), (8,1), (8,7), (8,10)}
We note that, R₁ ⊆  A x B. Thus, R₁ is a relation from A to B.

A x B = {(3,2), (3,4), (5,2), (5,4)}
We have A = {set of all first coordinates of elements of A x B}. Therefore, A = {3,5}
B = {set of all second coordinates of elements of A x B}. Therefore, B = {2,4}
Thus A = {3,5} and B = {2,4}.

xi	xi2
13	169
8	64
4	16
9	81
7	49
12	144
10	100
\({ \Sigma x }_{ i }\) = 63	\({ \Sigma x }_{ i }^{ 2 }\) = 623

Standard deviation
σ =\(\sqrt { \frac { \Sigma x_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma x }_{ i } }{ n } \right) ^{ 2 } } \)
=\(\sqrt { \frac { 623 }{ 7 } -\left( \frac { 63 }{ 7 } \right) ^{ 2 } } \)
=\(\\ \sqrt { 89-81 } =\sqrt { 8 } \)
Hence, σ ≃ 2.83
(ii) Mean method:
Another convenient way of finding standard deviation is to use the following formula.
Standard deviation (by mean method) σ = \(\sqrt { \frac { \Sigma ({ x }_{ i }-\bar { x } )^{ 2 } }{ n } } \)
If di = xi - \(\bar { x } \) are the deviations, then σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } } \).

If a and b are the intercepts then a + b = 7 or b = 7 − a
By intercept form \(\frac { x }{ a } +\frac { y }{ b } =1\)...(1)
We have \(\frac { x }{ a } +\frac { y }{ 7-a } =1\)
As this line pass through the point (-3,8), we have
\(\frac { -3 }{ a } +\frac { 8 }{ 7-a } =1\) gives –3(7–a) + 8a = a(7 – a)
−21 + 3a + 8a = 7a − a²
So, a² + 4a − 21 = 0
Solving this equation (a − 3) (a + 7) = 0
a = 3 or a = −7
Since a is positive, we have a = 3 and b = 7 – a = 7 – 3 = 4.
Hence \(\frac { x }{ 3 } +\frac { y }{ 4 } =1\)
Therefore, 4x + 3y − 12 = 0 is the required equation.

The natural numbers between 300 and 600 which are divisible by 7 are 301, 308, 315, …, 595.
The sum of all natural numbers between 300 and 600 is 301 + 308 + 315 +...+ 595
The terms of the above series are in A.P.
First term a = 301; common difference d = 7; Last term l = 595.
\(n=\left( \frac { l-a }{ d } \right) +1=\left( \frac { 595-301 }{ 7 } \right) +1=43\)
Since, \({ S }_{ n }=\frac { n }{ 2 } \left[ a+l \right] \), we have \({ s }_{43 }=\frac { 43 }{ 2 } \left[ 301+595 \right] \) = 19264 

Given cosec\(\theta \) + cot\(\theta \) = p   ...(1)
cosec²\(\theta \) - cot²\(\theta \) = 1 (identity)
\(\operatorname{cosec} \theta-\cot \theta=\frac{1}{\operatorname{cosec} \theta+\cot \theta}\)
cosec\(\theta \) - cot\(\theta \) =\(\frac { 1 }{ { p } } \)  .... (2)
Adding(1) and (2) we get, 2cosec\(\theta \) = \(p+\frac { 1 }{ p } \) 
2cosec\(\theta \)\(\frac { { p }^{ 2 }+1 }{ p } \) ....(3) 
Subtracting (2) from (1), we get, 2cot\(\theta \) = \(p-\frac { 1 }{ p } \) 
2cot\(\theta \) = \(\frac { { p }^{ 2 }-1 }{ p } \)   ...(4)
Dividing (4) by (3) we get,\(\frac { 2cot\theta }{ 2cosec\theta } =\frac { { p }^{ 2 }-1 }{ p } \times \frac { p }{ { p }^{ 2 }+1 } gives,cos\theta =\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)

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.

The points are P(-1, 5, 3), Q(6, -2) , R(-3, 4)
Area of Δ PQR = \(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) }
= \(\frac{1}{2}\) { (3 + 24 - 9) - (18 + 6 - 6) }
= \(\frac{1}{2}\) { 18 - 18 } = 0
Therefore, the given points are collinear.

To find HCF of three given numbers, first we have to find HCF of the first two numbers.
To find HCF of 396 and 504
Using Euclid’s division algorithm we get 504 = 396  x  1 + 108
The remainder is 108 \(\neq \) 0
Again applying Euclid’s division algorithm 396  = 108  x 3 + 72 
The remainder is 72 \(\neq \) 0
Again applying Euclid’s division algorithm 108  = 72 x 1 + 36 
The remainder is 36 \(\neq \) 0
Again applying Euclid division algorithm 72 = 36 x 2 + 0
Here the remainder is zero. Therefore HCF of 396 , 504 = 36, To find the HCF of 636 and 36
Using Euclid’s division algorithm we get 636 = 36 x 17 + 24
The remainder is 24 \(\neq \) 0
Again applying Euclid's division algorithm 36 = 24 x 1 + 12
The remainder is 12 \(\neq \) 0
Again applying Euclid's division algorithm 24 = 12 x 2 + 0
Here the remainder is zero. Therefore HCF of 636,36 = 12
Therefore Highest Common Factor of 396, 504 and 636 is 12. 

Given, radius = 4 cm
Construction
Step 1 : With O as the centre, draw a circle of radius 4 cm.
Step 2 : Take a point L on the circle. Through L draw any chord LM.
Step 3 : Take a point M distinct from L and N on the circle, so that L, M and N are in anti clockwise direction. Join LN and NM.
Step 4 : Through L draw a tangent TT' such that \(\angle\)TLM =\(\angle\)MNL
Step 5 : TT' is the required tangent.