\(\frac { 5 }{ 18 } \)

B = {3,4, 5, 6, 7, 8}

\(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } =1\)

19 = a + ( 6 - 1 ) d and 41 = a + ( 17 - 1 ) d = 87

\(\frac { 1 }{ \sqrt [ 3 ]{ \left( 27-6x \right) ^{ 2 } } } =\left( 27-6x \right) ^{ -\frac { 2 }{ 3 } }=\left( 27 \right) ^{ -\frac { 2 }{ 3 } }(1-\frac { 2 }{ 9 } x)^{ -\frac { 2 }{ 3 } }\)
\(=\frac { 1 }{ 9 } (1+\frac { 4 }{ 27 } x+\frac { 20 }{ 729 } { x }^{ 2 }+\frac { 320 }{ 59049 } { x }^{ 3 }+...),\quad when\quad \left| x \right| <\frac { 9 }{ 2 } \)

0 cannot be placed at thousand's place. So, thousand's place can be filled in 9 ways. Since repitition of digits is allowed, therefore each of the remaining 3 places can be filled in 10 ways.
Ans. 9000

Let x be the marks obtained by student in the annual examination. Then
\(\frac{62+48+x}{3} \geq 60\)
or 110 + x  \(\geq\) 180
or x \(\geq\)70
Thus, the student must obtain a minimum of 70 marks to get an average of at least 60 marks.

Consider P(k):2+4+6+8+.....+2k = k(k+1)
Now, P(k+1):2+4+6+.....2(k)+2(k+1)
=k(k+1)+2(k+1)=k² + 3k+2
=(k+1)(k+1)

All rectangles, Rhombus and square are parallelograms because its opposite sides are equal and parallel.
Therefore \({ F }_{ 2 }\subset { F }_{ 1 },{ F }_{ 3 }\subset { F }_{ 1 }\) and \({ F }_{ 4 }\subset { F }_{ 1 }\)
\({ F }_{ 1 }={ F }_{ 2 }\cup { F }_{ 3 }\subset { F }_{ 4 }\)

Given, A = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Since, -4 is not an element of A, therefore -4 \(\notin \) A

The given equations are x - y - 7 = 0 and 2x + y - 2 = 0
Equation of any line which passes through the intersection of the given lines is
(x - y - 7) + k (2x + y - 2) = 0 ... (i)
If equation (i) also passes through the origin
i.e., (0, 0), we get
-7 -2k= 0
\(∴\ k={7\over 2}\)
Putting the value of k is equation (i), we get
(x - y - 7 ) - \(7\over2\) (2x + y. - 2) = 0
2x - 2y - 14 - 14x - 7y + 14 = 0
⇒ -12x- 9y = 0
⇒ 4x + 3y = 0

Favourable outcomes are 1,3,5,7,9,11,13,15,17 and 19
\(\frac { 1 }{ 2 } \)

Possible outcomes are (1,2),(2,1),(1,5),(2,4),(3,3),(4,2),(5,1),(3,6),(4,5),(5,4),(6,6)
\(\frac { 1 }{ 3 } \)

Let \(\bar { x } \) be the mean of given data.
Then, \(\bar { x } \) =\(\frac { 6+5+5.25+5.5+4.75+4.5+6.25+7.75+9 }{ 9 } =\frac { 54 }{ 9 } =6\)
We make the table from the given data.

\(\therefore \) Mean deviation from mean,
\(MD(\bar { x } )=\frac { \sum { |x_{ i }-\bar { x } | } }{ n } =\frac { 10 }{ 9 } =1.1\)
Hence, mean deviation form mean is 1.1.

The vertices of the triangle are A(1, 2, 3), B(3, 4, 5) and C(-1, 6, 7). Also, centroid of the triangles is G(1, 4, 1/3).

Here, foci are at \((\pm 5,0)\)
\(\therefore \quad (\pm c,0)=(\pm 5,0)\Rightarrow c=5\)

\(Also,\quad we\quad know\quad that,\quad { c }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\)
\(\Rightarrow 25=16+{ b }^{ 2 }\)
\( \Rightarrow { b }^{ 2 }=9\)
\(Since,\quad the\quad foci\quad lies\quad on\quad x-axis.\)
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(On\quad putting\quad the\quad values\quad of\quad { a }^{ 2 }\quad and\quad { b }^{ 2 },\quad we\quad get\)
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 9 } =1\)
\(Which\quad is\quad the\quad required\quad equation\quad of\quad hyperbola.\) 

Here, focus is (4,0) and directrix is x + 4 = 0
Let p(x,y) be any moving point, then draw 
 from p to the directrix.

Now, FP=PM [by definition of parabola]
\(\Rightarrow \) FP²=PM²
\(\Rightarrow \) (x - 4)² + (y - 0)² = \(\frac { (x+4)^{ 2 } }{ (\sqrt { 1 } )^{ 2 } } \)
\(\Rightarrow \)  x²- 8x + 16 + y² = x² + 8x + 16
\(\Rightarrow \) y²=16x
which is the required equation of parabola.
Length of latusrectum=16

b,c,d are in GP.
\(\Rightarrow \) C² = bd...(ii)
Similarly, \(\frac { 1 }{ c } ,\frac { 1 }{ d } ,\frac { 1 }{ e } \)are in AP.
\(\Rightarrow \)  \(\frac { 2 }{ d } =\frac { 1 }{ c } +\frac { 1 }{ e } \)
\(\Rightarrow \)  \(d=\frac { 2ce }{ c+e } \)
On putting the values of b and d from Eq.(i) and (iii), in Eq.(ii), we get
\({ c }^{ 2 }=\left( \frac { a+c }{ 2 } \right) \times \left( \frac { 2ce }{ c+e } \right) \Rightarrow { c }^{ 2 }=ae\)
Therefore, a,c,e are in GP.

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

We have,
\(|w|=1\ \Rightarrow \frac { |1-iz| }{ |z-i| } =1\)
\(\Rightarrow |1-iz|=|z-i|\)
\(\Rightarrow |1+y-ix|=|x+i(y-1)|\)

(i) The contrapositive of the given statement is: If a positive integer has divisors other than 1 and itself then it is not prime. The converse of the given statement is: If a positive integer has no divisors other than 1 and itself, then it is a prime
(ii) The contrapositive of the given statement is: If! do not go to a beach then it is not a sunny day. The converse of the given statement is: If I go to beach, then it is a sunny day.
(iii) The contrapositive of the given statement IS: If you do not feel thirsty, then it is not hot outside. The converse of the given statement is: If you feel thirsty, then it is hot outside.

Let P(n) =\(\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +...+\frac { 1 }{ (3n-1)(3n+2) } \)=  \(\frac { n }{ (6n+4) } \)
For n =1
P(1)= \(\frac { 1 }{ (3\times 1-1)(3\times 1+2) } \)
=\(\frac { 1 }{ (6\times 1+4) } \)
⇒ \(\frac { 1 }{ 2\times 5 } =\frac { 1 }{ 10 } \Rightarrow \frac { 1 }{ 10 } =\frac { 1 }{ 10 } \)
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k) =\(\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +......+\frac { 1 }{ (3k-1)(3k+2) }\) 
\(\frac { k }{ (6k+4) } \)...(i)
For n = k + 1
P(k+1)= \(\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +......+\frac { 1 }{ (3k-1)(3k+2) } +\frac { 1 }{ [3(k+1)-1][2(k+1)+2] } \)
=\(\frac { k+1 }{ 6k+10 } \) 
= \(\frac { k }{ 6k+4 } +\frac { 1 }{ (3k+2)(3k+5) } \)
= \(\frac { 1 }{ (3k+2) } \left[ \frac { k }{ 2 } +\frac { 1 }{ 3k+5 } \right] \)
= \(\frac { 1 }{ (3k+2) } \left[ \frac { { 3k }^{ 2 }+5k+2 }{ 2(3k+5) } \right] \)
=\(\frac { 1 }{ (3k+2) } \left[ \frac { { 3k }^{ 2 }+3k+2k+2 }{ 2(3k+5) } \right] \) 
= \(\frac { 1 }{ (3k+2) } \left[ \frac { { 3k }(k+1)+2(k+1) }{ 2(3k+5) } \right] \)
= \(\frac { 1 }{ (3k+2) } \left[ \frac { (k+1)+(3k+2) }{ 2(3k+5) } \right] \)
= \(\frac { k+1 }{ 6k+10 } \)
∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n ∊ N.

It is given that T₄ = \(\frac{20}{27}\) 
Comparing\(\left( ax+\frac { 1 }{ x } \right) ^{ n }\) with (A+ B)ⁿ, we have
A=ax and B=\(\frac{1}{x}\)
∴ T₄ = ⁿC₃ A^n-3\((\frac{1}{x})^3\)
= ⁿC₃ (ax)^n-3\((\frac{1}{x})^3\)
∴ ⁿc₃ (ax)^n-3\((\frac{1}{x})^3=\frac{20}{27}\)
∴ = ⁿc₃ a^n-3 x^n-6=\(\frac{20}{27}...(i)\)
Now n - 6 = 0 ⇒ n = 6
Putting n = 6 in (i), we have
⁶C₃ a^{6- 3}x^{6 - 6}=​​​​​​​\(\frac{20}{27}\)
⇒ 20a³=\(\frac{20}{27}\)
⇒ a³=​​​​​​​\(\frac{1}{27}\)
⇒ a=\(\frac{1}{3}\)
Thus n = 6 and a =​​​​​​​\(\frac{1}{3}\) 

Let D(x, y, z) be the fourth vertex of parallelogram ABCD.
We know that diagonals of a parallelogram bisect each other. So the mid points ofAC and BD coincide.

\(\therefore \) Coordinates of mid point of AC \(\left( \frac { 3-1 }{ 2 } ,\frac { -1+1 }{ 2 } ,\frac { 2+2 }{ 2 } \right) \)=(1,0,2)
Also coordiantes of mid point of BD \(\left( \frac { x+1 }{ 2 } ,\frac { y+2 }{ 2 } ,\frac { z-4 }{ 2 } \right) \)
\(\therefore \quad \frac { x+1 }{ 2 } =1\Rightarrow x+1=2\Rightarrow x=1\)
\(\frac { y+2 }{ 2 } =0\Rightarrow y+2=0\Rightarrow y=-2\)
\(\frac { z-4 }{ 2 } =2\Rightarrow z-4=4\Rightarrow z=8\)
Thus the coordinates of point Dare (1, -2,8).

Let two remaining observations be x and y. Then
\(\frac { 6+7+10+12+12+13+x+y }{ 8 } =9\)
\(\therefore\) 60 + x + y = 72  \(\Rightarrow\) x + y = 12  ......(i)
Also \(\frac { 1 }{ 8 } ({ 6 }^{ 2 }+{ 7 }^{ 2 }+{ 10 }^{ 2 }+{ 12 }^{ 2 }+{ 12 }^{ 2 }+{ 13 }^{ 2 }+{ x }^{ 2 }+{ y }^{ 2 })-{ (9) }^{ 2 }=9.25\)
\(\Rightarrow \frac { 1 }{ 8 } (36+49+100+144+144+169+{ x }^{ 2 }+{ y }^{ 2 })-81=9.25\)
\(\Rightarrow\)  642 + x² + y² = 722
\(\Rightarrow\) x² + y² = 80 ......(ii)
Now (x+y)² + (x-y)² = 2(x²+y²)
\(\Rightarrow\) (12)² + (x - y)² = 2 x 80
\(\Rightarrow\) (x - y)² = 160 - 144
\(\Rightarrow\) (x - y)² = 16 \(\Rightarrow\) x - y = \(\pm \) 4
When x - y = 4
Solving x + y = 12 and x - y = 4 we get x = 8 and y = 4
When x - y = -4
Solving x + y = 12 and x - y = -4 we get x = 4 and y = 8.

Let \(\frac { a }{ { r }^{ 3 } } ,\frac { a }{ { r } } \) ,ar,ar³ be the four numbers in G.P.
It is given that  \(\frac { a }{ { r }^{ 3 } } +\frac { a }{ { r } } \)+ar+ar³=85 \(\Rightarrow \) a\(\left( \frac { 1 }{ { r }^{ 3 } } ,\frac { 1 }{ { r } } +r+{ r }^{ 3 } \right) \)=85....(i)
and  \(\frac { a }{ { r }^{ 3 } } \times \frac { a }{ { r } } \)\(\times \)ar\(\times \)ar³=4096 \(\Rightarrow \)  a⁴4096   \(\Rightarrow \) a=8
Putting value of a in (i) 
\(\therefore \)  8\(\left( \frac { 1 }{ { r }^{ 3 } } +\frac { 1 }{ r } +r+{ r }^{ 3 } \right) \)  =85 \(\Rightarrow \)   8\(\left( { r }^{ 3 }+\frac { 1 }{ { r }^{ 3 } } \right) \)+8 \(\left( { r }+\frac { 1 }{ { r } } \right) \)=85
\(\Rightarrow \) 8 \(\left[ \left( r+\frac { 1 }{ r } \right) ^{ 3 }-3\left( r+\frac { 1 }{ r } \right) \right] \) \(\Rightarrow \)  8\(\left[ \left( r+\frac { 1 }{ r } \right) ^{ 3 }-16\left( r+\frac { 1 }{ r } \right) \right] \)-85=0
put r+\(\frac { 1 }{ r } \)=y then
8y³ - 16y - 85 = 0 \(\Rightarrow \)(2x-5)(4x²+10x + 17) = 0
\(\Rightarrow \) 2x-5=0 or (2x - 5)(4x2 +10x + 17) = 0
\(\Rightarrow \)  2x=5    
or  x=\(\frac { -10\pm \sqrt { \left( 10 \right) ^{ 2 }-4\times 4\times 17 } }{ 8 } =\frac { -10\pm \sqrt { 100-272 } }{ 8 } \)
\(\Rightarrow \) x=\(\frac { 5 }{ 2 } \)or \(\frac { -10\pm \sqrt { -172 } }{ 8 } \)(which is not possible)
\(\therefore \)  r+\(\frac { 1 }{ r } \)=\(\frac { 5 }{ 2 } \) \(\Rightarrow \) 2r²-5r+2=0 \(\Rightarrow \)(r-2)(2r-1)=0
\(\Rightarrow \)  r-2=0 or 2r-1=0
\(\Rightarrow \) r=2 or r= \(\frac { 1}{ 2 } \)     
Thus four numbers in G.P. are 1, 4,16,64 or 64,16,4, 1. 
 

Here \(\sum _{ i=1 }^{ 50 }{ { x }_{ i }=212,\sum _{ i=1 }^{ 50 }{ { x }_{ i }^{ 2 } } =902.8, } \) \(\sum _{ i=1 }^{ 50 }{ { y }_{ i } } =261,\sum _{ i=1 }^{ 50 }{ { y }_{ i }^{ 2 } } =1457.6\)
Now \(\bar { x } =\frac { 212 }{ 50 } =4.24\)
\({ \sigma }_{ x }^{ 2 }=\frac { 1 }{ 50 } \times \times 902.8-{ \left( \frac { 212 }{ 50 } \right) }^{ 2 }\)
= 18.056 - 17.978 = 0.078
\({ \sigma }_{ x }=\sqrt { 0.078 } =0.28\)
Also \(\bar { y } =\frac { 261 }{ 50 } =5.22\)
\({ \sigma }_{ y }^{ 2 }=\frac { 1 }{ 50 } \times 1457.6-{ \left( \frac { 261 }{ 50 } \right) }^{ 2 }\)
= 29.152-27.248 = 1.904
\({ \sigma }_{ y }=\sqrt { 1.904 } =1.38\)
C.V. of length=\(\frac { 0.28 }{ 4.24 } \times 100=6.6\)
C.V. of weight=\(\frac { 1.38 }{ 5.22 } \times 100=26.45\)
C.V. of weight > C.V. of length
Thus weight have more variability than length.

(i) Here X = {1, 3, 5}and Y = {1, 2, 3}
\(\therefore\) X \(\cap\)Y = {1, 3}
(ii) Here A = {a, e, i, o, u} and B = {a, b, c}
\(\therefore\) A \(\cap\) B= {a}
(iii) Here A = {x : x is a natural number and multiple of 3}
= {3, 6, 9, 12,...........} and
B = {x : x is a natural number less than 6}
= {1, 2, 3, 4, 5}
\(\therefore\) A \(\cap\) B = {3}
(iv) Here A = {x : x is a natural number and 1 < x \(\le\) 6}
= {2, 3, 4, 5, 6}
and B = {x : x is a natural number and 6 < x < 10}
= {7, 8, 9}
\(\therefore\) A \(\cap\) B=ф
(v) Here A = {1, 2, 3} and B =ф
\(\therefore\) A \(\cap\) B=ф