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

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')=P((A\(\cup\)B)')=1-P(A\(\cup\)B)=1-0.8=0.2

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

We have, A and B are mutually exclusive events. Also, it is given that P(A)=0.35 and P(B)=0.45
P(B')=1-P(B)=1-0.45=0.55

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') = 1 - P(A) = 1 - 0.35 = 0.65

y+2=0 and \(\sqrt { 3x } \)-y-2-3\(\sqrt { 3x } \)=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
\(\Longrightarrow \)  \(\sqrt { { a }_{ n } } \) - \(\sqrt { { a }_{ 1 } } \) = \(\frac { (n-1)d }{ \sqrt { { a }_{ n }+\sqrt { { a }_{ 1 } } } } \)

We have, \((2+3x)^{ 9 }=2^{ 9 }\left( 1+\frac { 3x }{ 2 } \right) ^{ 9 }\)
Now,  \(\frac { T_{ r+1 } }{ T_{ r } } =\frac { { 2 }^{ 9 }\left[ { ^{ 9 }C }_{ r }\left( \frac { 3x }{ 2 } \right) ^{ r } \right] }{ { 2 }^{ 9 }\left[ { ^{ 9 }C }_{ r }\left( \frac { 3x }{ 2 } \right) ^{ r-1 } \right] } \)
[general term in the binomial expansion \((a+x)^{ n }\quad is\quad T_{ r+1 }={ ^{ n }C }_{ r }a^{ n-r }x^{ r }\)]
\(=\frac { { ^{ 9 }C }_{ r } }{ { ^{ 9 }C }_{ r-1 } } .\left( \frac { 3x }{ 2 } \right) =\frac { 9! }{ r!(9-r)! } .\frac { (r-1)!(10-r)! }{ 9! } .\left( \frac { 3x }{ 2 } \right) \)
\(=\frac { 10-r }{ r } .\left( \frac { 3x }{ 2 } \right) =\frac { 10-r }{ r } \left( \frac { 9 }{ 4 } \right) \quad \left[ \because \quad x=\frac { 3 }{ 2 } \right] \)
But \(\frac { { T }_{ r+1 } }{ { T }_{ r } } \ge \quad \Rightarrow \quad \frac { 90-9r }{ 4r } \ge 1\)
\(\Rightarrow 90-9r\ge 4r \Rightarrow r\le \frac { 90 }{ 13 } \Rightarrow r\le 6\frac { 12 }{ 13 } \)
Thus, the maximum value of r is 6. Therefore the greatest term is T_r+1=T₇
Hence, \(T_{ 7 }={ 2 }^{ 9 }\left[ { ^{ 9 }C }_{ r }\left( \frac { 3x }{ 2 } \right) ^{ 6 } \right] ={ 2 }^{ 9 }.{ ^{ 9 }C }_{ 6 }\left( \frac { 9 }{ 4 } \right) ^{ 6 }\)[putting \(x=\frac { 3 }{ 2 } \) ]
 \(T_{ 7 }={ 2 }^{ 9 }.\frac { 9\times 8\times 7 }{ 3\times 2\times 1 } .\left( \frac { { 3 }^{ 12 } }{ { 2 }^{ 12 } } \right) =\frac { 7\times 3^{ 13 } }{ 2 } \)

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.

(1.2)⁴⁰⁰⁰ = (1 +0.2)⁴⁰⁰⁰
= ⁴⁰⁰⁰C₀ +   ⁴⁰⁰⁰C₁(0.2) + sum of positive terms
= 1 + 4000(0.2) + a positive number
= 1 + 800 + positive number
= 800.

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 = \(\frac { 10 }{ 3!\times 2!\times 4! } \)

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

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]

\(z+\sqrt { 2 } \left| (z+1) \right| +i=0\)
Let \( z=x+iy\)
Then, \( (x+iy)+\sqrt { 2 } \left| (x+iy+1 \right| +i\ =0\)
\( \Rightarrow x+i(y+1)+\sqrt { 2 } \left| (x+1)+iy \right| =0\)
\(\Rightarrow x+i(y+1)+\sqrt { 2 } \sqrt { { (x+1) }^{ 2 }+{ y }^{ 2 } } =0\)
\([if\quad z=a+ib,\quad then\quad \left| z \right| =\sqrt { { a }^{ 2 }+{ b }^{ 2 } } ]\)
\(\Rightarrow x+\sqrt { 2 } \sqrt { { x }^{ 2 }+1+2x+{ y }^{ 2 } } +i(y+1)=0+0i\)
On equating real and imaginary part, we get
\( x+\sqrt { 2 } \sqrt { { x }^{ 2 }+1+2x+{ y }^{ 2 } } =0\quad .......(i)\)
\(and\quad y+1=0\quad ......(ii)\)
From Eq. (ii), we get 
y = -1
Now, on substituting y = - 1 in Eq. (i), we get
\( x+\sqrt { 2 } \sqrt { { x }^{ 2 }+1+2x+{ +1 }^{ } } =0\)
\(\Rightarrow x=-\sqrt { 2 } \sqrt { { x }^{ 2 }+2x+2 } \)
On squaring both sides, we get
\({ x }^{ 2 }=2({ x }^{ 2 }+2x+2)\)
\(\Rightarrow { x }^{ 2 }=2{ x }^{ 2 }+4x+4\Rightarrow { x }^{ 2 }+4x+4=0\)
\(\Rightarrow { (x+1) }^{ 2 }=0\Rightarrow x+2=0\Rightarrow x=-2\)
\(Hence,\ z=x+iy=-2-i\)

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)

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

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

\(P(A)\frac { 1 }{ 4 } \)

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

0.77