Given equation of line is 3x + 4y - 6 = 0.
Any line perpendicular to 3x + 4y - 6 = 0 will be of the form 4x - 3y + k = 0 ....(1)
Given perpendicular distance is 4 units from (2,1) to line (1)
\(\therefore\)4 = \(\pm \frac { (4(2)-3(1)+k) }{ \sqrt { { 4 }^{ 2 }+{ \left( -3 \right) }^{ 2 } } } \)
\(\Rightarrow\) 4 = \(\pm \left( \frac { 8-3+k }{ \sqrt { 16+9 } } \right) \)
\(\Rightarrow\) 4 = \(\pm \left( \frac { 5+k }{ 5 } \right) \)
\(\therefore\) 20 = +(5 + k) or 20 = -(5+ k)
\(\Rightarrow\) k = 20 - 5  or k = -(20 + 5)
\(\Rightarrow\) k = 15 or k = -25
\(\therefore\) Required equation of the lines are 4x - 3y + 15 = 0 and 4x - 3y - 25 = 0.

\({{1}\over{x^2-a^2}}={{1}\over{(x+a)(x-a)}}-={{A}\over{x+a}}+{{B}\over{x+a}}\)
\({{1}\over{x^2-a^2}}={{A(x-a)B(x+a)}\over{(x+a)(x-a)}}\)
\(\Rightarrow\) 1 = A(x - a) + B(x + a)
Putting x = a in (1) we get,
\(1=B(2a)\Rightarrow A=-{{1}\over{2a}}\)
Putting x = -a in (1) we get,
\(1=A(-2a)\Rightarrow A=-{{1}\over{2a}}\)
\(\therefore{{1}\over{x^2-a^2}}={{-{{1}\over{2}}a}\over{x+a}}+{{{{1}\over{2}}a}\over{x-a}}={{-1}\over{2a(x+a)}}+{{1}\over{2a(x-a)}}\)

\(LHS=\frac { sin8x\quad cosx-sin6x\quad cos3x }{ cos2x\quad cosx-sin3x\quad sin4x } \)
\(=\frac { \frac { 1 }{ 2 } \left[ sin9x+sin(7x) \right] -\frac { 1 }{ 2 } \left[ sin9x+sin3x \right] }{ \frac { 1 }{ 2 } \left[ cos\quad 3x+cos\quad x) \right] -\frac { 1 }{ 2 } \left[ cos\quad (x)-cos7x) \right] } \quad \left[ \therefore sinAcosB=\frac { 1 }{ 2 } (sin(A+B)+sin(A-B)) \right] \)
\(=\frac { \frac { 1 }{ 2 } \left[ sin9x+sin7x-sin9x-sin3x \right] }{ \frac { 1 }{ 2 } \left[ cos3x+cosx-cosx+cos7x \right] } =\frac { sin7x-sin3x }{ cos3x+cos7x } \)
\(=\frac { 2cos\left( \frac { 7x+3x }{ 2 } \right) sin\left( \frac { 7x-3x }{ 2 } \right) }{ 2cos\left( \frac { 7x+3x }{ 2 } \right) .cos\left( \frac { 7x-3x }{ 2 } \right) } =\frac { 2cos5x.sin2x }{ 2cos5x.cos2x } \)
\(\left[ \because sinC-sinD=2cos\left( \frac { C+D }{ 2 } \right) sins\left( \frac { C-D }{ 2 } \right) andcosC+cosD=2cos\left( \frac { C+D }{ 2 } \right) coss\left( \frac { C-D }{ 2 } \right) \right] \)
= tan 2x = RHS

The relation R defined on the set of all positive integers by "mRn" if m divides n".
Given relation is "mRn if m divides n".
Reflexivity :  mRm since m divides m for all positive integers m.
\(\therefore\)  R is reflexive.
Symmetricity: mRn \(\Rightarrow\)  nRm.
m divides n \(\Rightarrow\)  n divides m but 'n' does not divide 'm'
\(\therefore\) R is not symmetric
Transitive : mRn and nRp \(\Rightarrow\)  mRp.
m divides n and n divides p \(\Rightarrow\) m divides p.
\(\therefore\) R is transitive.
\(\therefore\) R is reflexive, and transitive.

Let P(n) = \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ n(n+1) } =\frac { n }{ n+1 } \) 
Substituting the value of n = 1, in the statement we get 
P(1) = \(\frac { 1 }{ 1.2 } =\frac{1}{2}\)
Hence, P(1) is true. Let us assume that the statement is true for n = k. Then
P(k) = \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } =\frac { k }{ k+1 } \)
We need to show that P(k + 1) is true. Consider,
\(P(k+1) =\underbrace{\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } ...+\frac { 1 }{ k(k+1) }} +\frac { 1 }{ (k+1)(k+2) } \)

= \(P(k)+\frac{1}{(k+1)(k+2)}\)
= \(\frac{ k} {(k + 1)} + \frac{1}{ (k + 1)(k + 2)}\)
= \(\frac{1} {k + 1} (\frac{ k}{ 1} + \frac{1} {k + 2})\)
= \(\frac{1} {k + 1} (\frac{ k ^2 + 2k + 1}{ k + 2 })\)
= \(\frac{1}{(k+1) }\frac{(k+1)^2}{k+2}=\frac{k+1}{k+2}\)
This implies, P(k + 1) is true
The validity of P(k + 1) follows from that of P(k)
Therefore by the principle of mathematical induction, for any natural number n,
\(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ n(n+1) } =\frac { n }{ n+1 } \).

Shape of the hall in the shopping mall is cuboid. When you open out the cuboid, the not of the cuboid will be as shown in the following diagram.
 
The path of the escalator is from OA to AB to BC to CD
In OAE, OA² = AE² + OE²
\(\Rightarrow{ OA }^{ 2 }={ \left[ \frac { 1 }{ 4 } (720) \right] }^{ 2 }+{ OE }^{ 2 }\)
\(\Rightarrow\)   OA² = (180)² + (800)²
\(\Rightarrow\)  OA² = (20\(\times\)9)² + (20 \(\times\) 40)²
\(\Rightarrow\)   OA² = 20² + (9²+ 40²)
\(\Rightarrow\)  OA² = 20² \(\times\) 1681
\(\Rightarrow\)  OA² = 20² \(\times\) 41²
\(\Rightarrow\)  OA² = 20 \(\times\) 41 = 820
\(\therefore\)  Total length of the escalator = OA + A + BC + CD
= 4 \(\times\) OA    (since \(\Delta\)OAE \(\equiv \) \(\Delta\)ABB' \(\equiv \) \(\Delta\)BCC' \(\equiv \) \(\Delta\)CDD')
= 4 \(\times\) 820
The minimum length = 3280 units
(ii) The height at which the escalator changes its direction.
\(AE=\frac { 1 }{ 4 } (720)=180\quad units\)
\(BE=\frac { 1 }{ 2 } (720)=360\quad units\quad and\quad GE=\frac { 3 }{ 4 } (7200=540\quad units\)
(iii) Slope of the escalator at the turning points
Let \(\angle \)AOE = \(\theta\)
\(In\ \Delta OAE,\ tan\theta =\frac { opp }{ adj } =\frac { AE }{ DE } =\frac { 180 }{ 800 } =\frac { 9 }{ 40 } \)
\(\therefore \ Slope\ at\ the\ point\ A=\frac { 9 }{ 40 } \)
\(Since\ \Delta OAE\equiv \Delta ABB'\equiv \Delta BCC'\equiv \Delta CDD'\)
Slope at the points B, C will be \(\frac{9}{40}\)

Let = 6 + 66 + 666 + ... upto n terms
= 6 (I + 11 + 111+ ....) upto n terms
\(={6\over9}(9+99+999+ ...)\) upto n terms
\(={63\over 6}[(10 -1) + (10^2-1) + (10^3 -1) + ...]\) upto n terms
\(={6\over 9}[(10+ 10^2 + 10^3+ ...) - (1+ 1+1...)]\) upto n terms​​​​​​​
\(={6\over9}\left[{ 10(10^n-1)\over 10-1}n\right]\)[In a G.P with a = 10 r = 10, \(S_n={(r^n-1)\over r-1}\)]
\(={6\over9}\left[{ 10(10^n-1)\over 10-1}n\right]={6\over9}\left[ 10(10^n-1)-9n\over9\right]\)
\({ S }_{ n }=\frac { 6 }{ 81 } \left[ 10\left( { 10 }^{ n }-1 \right) -9n \right] \)

(i) Either starts with L or ends with S.

1	4	3	2	1
L

Since the words starts with L, the remaining 4 boxes can be filled in 4 x 3 x 2 x 1 ways by the remaining letters 0, T, U, S.
∴ Number of words starting with L
= 1 \(\times\) 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 24.

1	2	3	4	1
				S

Here also, the remaining 4 boxes can be filled in 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 ways = 24.......(1)
Number of words ending with S = 24 ....(2)
Number of words starting with L and end with S are 3 \(\times\) 2 \(\times\) 1 = 6...(3)
∴  By fundamental principle of addition, number of words either starts with L nor ends with S = 24 + 24 - 6 = 48 - 6 = 42

	3	2	1
F				S

(ii) Neither starts with L nor ends with S.
Total number of words formed by the letters of the word LOTUS is 5 \(\times\) 4 \(\times\) 3 \(\times\) 2\(\times\) 1 = 120.
Now, number of words neither starts with L nor end with S.
= (Total number of words) - (Number of words starts with either L nor ends with S)
= 120 - 42
= 78.