\(1\over8\)

\([{100^o,{5\pi\over9}}]\)

We have, A and B are mutually exclusive events. Also, it is given that P(A)=0.35 and P(B)=0.45
P(A\(\cap \)B) = 0

We have, A and B are mutually exclusive events. Also, it is given that P(A)=0.35 and P(B)=0.45
P(A\(\cup \)B)=P(A)+P(B)=0.35+0.45=0.8

Here, y - 0 = m(x - 1) \(\Rightarrow \) y - mx + m = 0
Since, the distance from origin is \(\frac { \sqrt { 3 } }{ 2 } \).
Then, \(\frac { \sqrt { 3 } }{ 2 } =\frac { \left| 0-0+m \right| }{ \sqrt { 1+{ m }^{ 2 } } } \)
Ans. \(\sqrt { 3x } -y-\sqrt { 3 } =0\) and \(\sqrt { 3x } +y-\sqrt { 3 } =0\)

a₂ - a₁ = a₃ - a₂ =....= a_n - a_n-1 = d (say)
If a₂ - a₁ = d, then \(\left( \sqrt { { a }_{ 2 } } \right) ^{ 2 }\) - \(\left( \sqrt { { a }_{ 1 } } \right) ^{ 2 }\) = d
\(\Longrightarrow \)\(\left( \sqrt { { a }_{ 2 } } -\sqrt { { a }_{ 1 } } \right) \) \(\left( \sqrt { { a }_{ 2 } } +\sqrt { { a }_{ 1 } } \right) \) = d
\(\Longrightarrow \) \(\frac { 1 }{ \sqrt { { a }_{ 1 }+\sqrt { { a }_{ 2 } } } } \) = \(\frac { \sqrt { { a }_{ 2 }-{ a }_{ 1 } } }{ d } \) 
On adding, we get
\(\frac { 1 }{ \sqrt { { a }_{ 1 }+\sqrt { { a }_{ 2 } } } } \) + \(\frac { 1 }{ \sqrt { { a }_{ 2 } } +\sqrt { { a }_{ 3 } } } \) + .....+ \(\frac { 1 }{ \sqrt { { a }_{ n-1 } } +\sqrt { { a }_{ n } } } \)
=  \(\frac { 1 }{ d } \) [\(\sqrt { { a }_{ 2 } } \) - \(\sqrt { { a }_{ 1 } } \) + \(\sqrt { { a }_{ 3 } } \) -\(\sqrt { { a }_{ 2 } } \) +...+ \(\sqrt { { a }_{ n } } \) - \(\sqrt { { a }_{ n-1 } } \) ]
=  \(\frac { 1 }{ d } \) [\(\sqrt { { a }_{ n } } \) - \(\sqrt { { a }_{ 1 } } \) ].......(i)
Also,  a_n - a₁ = (n - 1) d
\(\Longrightarrow \) \(\left( \sqrt { { a }_{ n } } \right) ^{ 2 }\) - \(\left( \sqrt { { a }_{ 1 } } \right) ^{ 2 }\)= (n - 1) d
\(\Longrightarrow \)  \(\sqrt { { a }_{ n } } \) - \(\sqrt { { a }_{ 1 } } \) = \(\frac { (n-1)d }{ \sqrt { { a }_{ n }+\sqrt { { a }_{ 1 } } } } \)

General term in \(\left( \frac { 3 }{ 2 } x^{ 2 }-\frac { 1 }{ 3x } \right) ^{ 9 }\)
\(T_{ r+1 }={ ^{ 9 }C }_{ r }\left( \frac { 3 }{ 2 } \right) ^{ 9-r }\left( -\frac { 1 }{ 3 } \right) ^{ r }x^{ 18-3r }\)
So, general form in the expansion of
\((1+x+{ 2x }^{ 3 })\left( \frac { 3 }{ 2 } x^{ 2 }-\frac { 1 }{ 3x } \right) ^{ 9 }\)
\(={ ^{ 9 }C }_{ r }\left( \frac { 3 }{ 2 } \right) ^{ 9-r }\left( -\frac { 1 }{ 3 } \right) ^{ r }x^{ 18-3r }+{ ^{ 9 }C }_{ r }\left( \frac { 3 }{ 2 } \right) ^{ 9-r }\left( -\frac { 1 }{ 3 } \right) ^{ r }x^{ 19-3r }+2{ ^{ 9 }C }_{ r }\left( \frac { 3 }{ 2 } \right) ^{ 9-r }\left( -\frac { 1 }{ 3 } \right) ^{ r }x^{ 21-3r }\)
For the term independent of x
18-3r=0, 19-3r=0 and 21-3r=0
\(\Rightarrow \ r=6,\quad \frac { 19 }{ 3 } and\quad 7\)
Thus,  posible value of r are 6 and 7
Term independent of x
\(={ ^{ 9 }C }_{ r }\left( \frac { 3 }{ 2 } \right) ^{ 9-6 }\left( -\frac { 1 }{ 3 } \right) ^{ 6 }+2.^{ 9 }C_{ 7 }\left( \frac { 3 }{ 2 } \right) ^{ 9-7 }\left( -\frac { 1 }{ 3 } \right) ^{ 7 }\)
\(=\frac { 84 }{ 8 } .\frac { 1 }{ 3^{ 3 } } -\frac { 36 }{ 4 } .\frac { 2 }{ 3^{ 5 } } =\frac { 17 }{ 54 } \)

2⁴ⁿ⁺⁴ - 15n - 16 = (2)⁴ⁿ⁺¹ - 15(n+1) - 1
= 16ⁿ⁺¹ - 15(n+1) - 1
= (1+15)ⁿ⁺¹ - 15(n+1) - 1
={ⁿ⁺¹C₀ + ⁿ⁺¹C₁ (15) + ⁿ⁺¹C₂ (15)² + ⁿ⁺¹C₃ (15)³ +........+ ⁿ⁺¹C_n+1 (15)ⁿ⁺¹} - 15(n+1) - 1
= 1 + 15ⁿ⁺¹ + ⁿ⁺¹C₁ (15) + ⁿ⁺¹C₂ (15)² + ⁿ⁺¹C₃ (15)³ +........+ ⁿ⁺¹C_n+1 (15)ⁿ⁺¹} - 15(n+1) - 1
= 225 {ⁿ⁺¹C₂ + ⁿ⁺¹C₃ (15) +....+ ⁿ⁺¹C_n+1 (15)ⁿ⁺¹}
= 225 \(\times \) a natural number.

Number of four different digits formed = ⁷C₄
(i) Even number should be in unit's place and remaining any there digits, any different number can be placed. 
\(\therefore \)Total number of ways  = ⁶P₄ x ³C₁
(ii) Any four digit number is divisible by 25, if last two digits should be divisible by 25 i.e. 25,75.
(iii) Any number is divisible by 4,  if last two digits is divisible by 4 i.e. last digits can be 12, 16, 24, 32, 36, 52, 56, 54, 72, 76
\(\therefore \) Total number of ways = ⁵P₂ x ¹⁰C₁
\(\therefore \)Total number of ways  = ⁵P₂ x ²C₁
(iv) Total number of words = \(\frac { 10 }{ 3!\times 2!\times 4! } \)

The letters in the word INDEPENDENCE are (I,N,N,N,D,D,E,E,E,E,P,C).
\(\therefore \) Total number of arrangement = \(\frac { 12! }{ 3!\times 2!\times 4! } \)
(i) Total numbr of words start with P =\(\frac { 11! }{ 3!\times 2!\times 4! } \)
(ii) Total numbr of words , when all the vowels occur together =\(\frac { 8!\times 5! }{ 3!\times 2!\times 4! } \) [consider all vowels at one unit]
(iii)Total number of ways = Total arrangements-Number of arrangements when vowels occurs together =\(\frac { 8!\times 5! }{ 3!\times 2!\times 4! } \)-\(\frac { 8!\times 5! }{ 3!\times 2!\times 4! } \)

Number of ways = ⁶C₄ x ⁷C₆+ ⁶C₅x  ⁷C₅^{+ 6}C₆x ⁷C₄

In automobile licence plates, two different letters can be arranging im ²⁶P₂ways and three next different digits can be arranging in ¹⁰P₃ ways.
\(\therefore \) Total number of ways = ²⁶P₂  x ¹⁰P₃
= 26 x 25 x10 x 9 x 8 = 468000

Required number of ways= ⁿC₁ +ⁿC₂+....+ⁿC_n
=ⁿC₀ +ⁿC₁ +ⁿC₂+....+ⁿC_n-ⁿC₀=2ⁿ-1

(0, 1] \(\cup \) [3, 4)

Firstly, put \(x+1=0\) and  \(x=0\Longrightarrow x=-1\) and \(x=0\)
\(\therefore \)  \(x=-1\), 0 are critical points. So, we will consider three intervals (-\(\infty \), -1), [-1, 0), [0, \(\infty \))
Ans. (-\(\infty \), -2) \(​​\cup \) (1, \(\infty \) )

Write the given inequalities as \(|x-2|\ge 1\) and \(|x-2|\le 3\) Solve each inequality separately and then take the intersection of their solution sets
Ans. [-1, 1] \(\cup \) [3, 5]

We have, z₁ =\(\sqrt { 3 } +i\sqrt { 3 } \)  and \({ z }_{ 2 }=\sqrt { 3 } +i\)
\(\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { \sqrt { 3 } (1+i) }{ \sqrt { 3 } +i } =\frac { \sqrt { 3 } (1+i) }{ (\sqrt { 3 } +i) } x \frac { \sqrt { 3 } -i }{ \sqrt { 3 } -i }\)
\( [by\quad rationalising\quad the\quad denominator]\)
\(=\frac { \sqrt { 3 } (1+i)(\sqrt { 3 } -i) }{ (\sqrt { 3 } { ) }^{ 2 }-(i)^{ 2 } } \quad[({ z }_{ 1 }+{ z }_{ 2 })({ z }_{ 1 }-{ z }_{ 2 })={ z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 }]\)
\(=\frac { \sqrt { 3 } (\sqrt { 3 } -i+i\sqrt { 3 } -{ i }^{ 2 } }{ 3-{ i }^{ 2\\ } } \)
\(=\frac { \sqrt { 3 } (\sqrt { 3 } +i(\sqrt { 3 } -1)+1) }{ 3+1 } \quad [\because { i }^{ 2 }=-1]\)
\(=\frac { \sqrt { 3 } }{ 4 } ((\sqrt { 3 } +1+i(\sqrt { 3 } -1))\)
\(=\frac { \sqrt { 3 } (\sqrt { 3 } +1) }{ 4 } +\frac { i\sqrt { 3 } (\sqrt { 3 } -1) }{ 4 } \)
which is represented by a point in first quadrant.

Consider, \(\frac { 1+i }{ 1-i } =\frac { 1+i }{ 1-i } \times \frac { 1+i }{ 1+i } \)
[by rationalising the denominator]
\(=\frac { { \left( 1+i \right) }^{ 2 } }{ 1-{ i }^{ 2 } } =\frac { 1+{ i }^{ 2 }+2i }{ 1+1 }\)
\(\Rightarrow \ \frac { 1+i }{ 1-i } =\frac { 1-1+2i }{ 2 } =i\quad \left[ \because \ { i }^{ 2 }=-1 \right] ...(i)\)
\( Now,\ \frac { 1-i }{ 1+i } =\frac { 1 }{ \left( \frac { 1+i }{ 1-i } \right) } =\frac { 1 }{ i } \)
\(=\frac { 1 }{ i } \times \frac { i }{ i } =\frac { i }{ { i }^{ 2 } } =\frac { i }{ (-1) } =-i\quad \left[ \because \ { i }^{ 2 }=-1 \right] ...(ii)\)
\(Hence,\ { \left( \frac { 1+i }{ 1-i } \right) }^{ 3 }-{ \left( \frac { 1-i }{ 1+i } \right) }^{ 3 }={ i }^{ 3 }-{ \left( -i \right) }^{ 3 }\)
\(={ i }^{ 3 }+{ i }^{ 3 }=2{ i }^{ 3 }=2(-i)=0-2i\ \left[ \because \ { i }^{ 3 }=-i \right] \)
\(\therefore \ x+iy=0-2i\)
On comparing real and imaginary parts on both sides, we get x=0 and y=-2
\(\therefore \)  (x,y) = (0,-2)

\(cos^{ 4 }\frac { \pi }{ 8 } +cos^{ 4 }\frac { 3\pi }{ 8 } +cos^{ 4 }\frac { 5\pi }{ 8 } +cos^{ 4 }\frac { 7\pi }{ 8 } \)
\(=cos^{ 4 }\frac { \pi }{ 8 } +cos^{ 4 }\frac { 3\pi }{ 8 } +cos^{ 4 }\left( \pi -\frac { 3\pi }{ 8 } \right) +cos^{ 4 }\left( \pi -\frac { \pi }{ 8 } \right) \)
\(= cos^{ 4 }\frac { \pi }{ 8 } +cos^{ 4 }\frac { 3\pi }{ 8 } +cos^{ 4 }\frac { \pi }{ 8 } \)
\(= 2\left[ cos^{ 4 }\frac { \pi }{ 8 } +cos^{ 4 }\frac { 3\pi }{ 8 } \right] =2\left[ cos^{ 4 }\frac { \pi }{ 8 } +cos^{ 4 }\left( \frac { \pi }{ 2 } -\frac { \pi }{ 8 } \right) \right] \)
\( =2-\left( sin\frac { 2\pi }{ 8 } \right) ^{ 2 }=\frac { 3 }{ 2 } \)

\(f(x)=\begin{cases} 2x\quad 2\le x\le 3 \\ 4,\quad -2\le x<2 \\ -2x,\quad -3\le x<-2 \end{cases}\)

P(John promoted)= \(\frac { 1 }{ 8 } \);P(Rita promoted)=\(\frac { 1 }{ 4 } \)
P(Aslam promoted)=\(\frac { 1 }{ 2 } \);P(Gurpreet promoted)=\(\frac { 1 }{ 8 } \)

0.57

0.51

0.77