When a die is rolled , then we get 1 or 2 or 3 or 4 or 5 or 6.
S = {1,2,3,4,5,6}
It is given that 
P(each odd number = 2 x P (each even number)
P(1) = 2x, P(2) = x, P(3) = 2x ,P(4) = x, P(5) =2x and P(6) = x
We know that , P(5) = 1
P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1
\(\Rightarrow \) 2x + x + 2x + x + 2x + x = 1 \(\Rightarrow \) \(x=\frac { 1 }{ 9 }\)
Thus P(1) = P(3) = P(5) = \(\frac { 2 }{ 9 } \)
and P(2) = P(4) = P(6) = \(\frac { 1 }{ 9 } \)
Now, P(G) = P (getting 4 or 5 or 6)
= P(4) + P(5) +P(6) = \(\frac { 4 }{ 9 } \)

Total number of symbols  = 36
There are 36 x 36 x 36 x 36 = (36)⁴ = 1679616 PINs in all.
[by fundamental principle of counting]
Note that, when repetition of symbols is not allowed, then there are  36 x 35 x 34 x 33 = 1413720  different  PINs
Now, the number of PINs that contains at least one repeated symbol = 1679616 - 1413720 = 265896
= 0.1583

0.556

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

\(Given,\ y=\frac { sinx+cosx }{ sinx-cosx } \)
\(\therefore \frac { \left[ (sinx-cosx)(cosx-sinx)-(sin+cosx)(cosx+sinx) \right] }{ \left( sinx-cosx \right) ^{ 2 } } \)
\( =\frac { -(sinx-cosx)^{ 2 }-(sinx+cosx)^{ 2 } }{ (sinx-cosx)^{ 2 } } =-2\)

1:3 externally

Let Since, it is passes through (3,0) and  \((3\sqrt { 2 } ,2)\)
\(\frac { 9 }{ { a }^{ 2 } } -0=1\)  and  \(\frac { 18 }{ { a }^{ 2 } } -\frac { 4 }{ { b }^{ 2 } } =1\)
a² = 9 and b² = 4
b² = a² (e²-1)
\(e=\frac { \sqrt { 13 } }{ 2 } \)

\(\left| x \right| +\left| y \right| =1\),
\(\Rightarrow \pm x\pm y=1\)  which forms a square
The locus of the point is a square.

Since, x , 2y and 3z are in AP.
\(\therefore \)  4y = x + 3z
And x, y, z are in GP.
 \(\therefore \)   y = rx and z = xr²
On putting the value of y and z in Eq. (i), we get
4xr = x + 3xr²
\(\Rightarrow \)  3r² - 4r + 1 = 0
r = \(\frac { 1 }{ 3 } \)  [\(\because \) r = 1 is not possible]

Here, r>1, n>2
\(\therefore \)  T_3r = ²ⁿC_3r-1 x^3r-1; T_{r+2 =} ²ⁿC_r+1 x^r+1
Then, ²ⁿC_3r-1= ²ⁿC_r+1 
\(\Rightarrow \) 3r - 1 + r + 1 = 2n \(\Rightarrow \) n = 2r

Total nuber of 4 letters code = ¹⁰C₄

7200

Let us make the following cases and find the number of possible choices in each case
Case I Baoy borrows mathematics part II. In this case, boy borrows mathematics part I also. so number od possible chocies are \(^{6}{C}{_1}\)
Case II boy dose not borrow mathamatics part II In this case passible chioices are \(^{7}{C}{_3}\)
Ans. 41

We know that, Profit=Revenue- Cost
In order to realise some profit, revenue should be greater than the cost
Thus, we should have R(x)>C(x)
\(\Rightarrow 60x+2000>20x+4000\)
\( \Rightarrow 60x+2000-20x+4000-20x\quad [subtracting\quad 20x\quad from\quad both\quad sides]\)
\(\Rightarrow 40x+2000>4000\)
\(\Rightarrow 40x+2000-2000>4000-2000[subtracting\quad 2000\quad from\quad both\quad sides]\)
\(\Rightarrow 40x>2000\Rightarrow \frac { 40x }{ 40 } >\frac { 2000 }{ 40 } [dividing\quad both\quad sides\quad by\quad 40]\)
\( \Rightarrow x>50\)
Hence,the manufacturer must sell more than 50 times to realise some profit.
 

\(8.2<\frac { 8.48+8.35+x }{ 3 } <8.5\)
Between 7.77 and 8.77

Let,
\(z=\frac { (1+i)(2+i) }{ (3+i) } =\frac { 2+i+2i+{ i }^{ 2 } }{ 3+i } =\frac { 2+3i-1 }{ 3+i } \)
\( \Rightarrow z=\frac { 1+3i }{ 3+i } \quad [\because { i }^{ 2 }=-1]\)
Now,
\(\left| z \right| =\left| \frac { 1+3i }{ 3+i } \right| =\frac { \left| 1+3i \right| }{ \left| 3+i \right| } \quad \left[ \left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| =\frac { \left| { z }_{ 1 } \right| }{ \left| { z }_{ 2 } \right| } \right] \)
\(=\frac { \sqrt { { 1 }^{ 2 }+{ 3 }^{ 2 } } }{ \sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 } } } =1\)
Hence, \(\left| (1+i)\frac { (2+i) }{ (3+i) } \right| \)

Let z = x + iy, then
\(\frac { z-1 }{ z+1 } =\frac { ({ x }^{ 2 }-1)+{ y }^{ 2 }+i[y(x+1)-y(x-1)] }{ ({ x }^{ 2 }+1)^{ 2 }+y^{ 2 } } \\ \)
\(\because \frac { z-1 }{ z+1 } \)  is purely imaginary.
\(\therefore Re(\frac { z-1 }{ z+1 } ) =0\ i.e.\ \frac { ({ x }^{ 2 }-1)+{ y }^{ 2 } }{ ({ x }^{ 2 }+1)^{ 2 }+y^{ 2 } } =0\)
\({ x }^{ 2 }-1-{ y }^{ 2 }=0\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1= |z|=1\)

\(\frac { 2-i }{ 2 } \)

3i³- 2ai²+(1-a)i + 5
= 3( - i) + 2a + (1 - a)i+5  [i³ = -i and i² = -1]
= (2a + 5) + i(1-a-3), which will be real,
if 1 - a - 3 =0,
i.e, a = - 2

Domain = \((-\infty ,-1)\cup (1,4)\)

Given S = Set of points inside the square
T = Set of points inside the triangle
and C = Set of points inside the circle.
According to the given condition, the Venn diagram is given below

It is clear from the Venn diagram that, \(S\cup T\cup C=S\)

Given E = set of students studying English
M =Set of students studying Mathematics
U= Set of all students
Since, all of the students who study Mathematics also study English, but some students who study English do not study Mathematics
\(\therefore \quad \quad M\subset E\subset U\)
Through Venn diagram, we represent it as

The quantifier is 'for all' and the negation is There exists an even integer x such that x² is not even

2k<(k+2)!\(\Rightarrow \)2k+2(k+2)!+2]
\(\Rightarrow \) (k+1)² < 2^k + 2^k    [(2k+1)<2^k for k\(\le \)3]
\(\Rightarrow \) (k+1)² < 2^k+1

Step I: Let P(n) be the given statement
i.e. P(n): \(1\times 1!+2\times 2!+3\times 3!+....+n\times n!=(n+1)!-1\)
Step II : For n=1, we have
LHS=\(1\times 1!\)=1
and RHS=(1+1)!-1=2!-1=2-1=LHS
\(\because \)  LHS=RHS
\(\therefore \)  P(1) is true
Step III  Let us assume that P(n) is true for n=k
Then, we have
P(k): \(1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)!-1\quad \quad ....(i)\)
Step IV  Now, we shall prove the statement for n=k+1. For this we have to show that
\(1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)\times (k+1)!=(k+1+1)!-1\)
Then, LHS \(=1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)\times (k+1)!\)
=(k+1)!-1+(k+1)!\(\times \)(k+1)      [from Eq.(1)]
=(k+1+1)(k+1)!-1=(k+2)(k+1)!-1
=(k+2)!-1   [\(\because \)n(n-1)!=n]
Thus, P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all natural numbers n.