\(\frac { 3 }{ 8 } \)

Let us make the table from the given data.

\(\therefore Standard\ deviation,\sigma =\frac { 1 }{ N } \sqrt { N\sum { { f }_{ i }{ x }_{ i }^{ 2 } } -(\sum { { f }_{ i }{ x }_{ i }^{ 2 })^{ 2 } } } \)
Ans. 6.12

Let C (0, \(\frac{1}{3}\), 2) divides the joint of A (2, -3) and B(-1, 2,1) in the ratio k:1
Then coordinates of C are 
\(\left( \frac { -k+2 }{ k+1 } ,\frac { 2k-3 }{ k+1 } ,\frac { k+4 }{ k+1 } \right) \) [using internal ratio formula]
But coordinates of C are  (0, \(\frac{1}{3}\), 2)
On comparing Eqs.(i) and (ii) we get
\(\frac { -k+2 }{ k+1 } =0\Rightarrow -k+2=0\Rightarrow 2\)
\( \frac { 2k-3 }{ k+1 } =\frac { 1 }{ 3 } \Rightarrow 6k-9=k+1\Rightarrow 5k=10\Rightarrow k=2\)
\(and \quad \frac { k+4 }{ k+1 } =2\Rightarrow k+4=2k+2\Rightarrow k=2\)
From each of these equations, we get k=2 Since, from each equation, we get the same value of k. Therefore, the given points are collinear nad C divides AB internally in the ratio 2:1.

\({ P }_{ 1 }=\frac { \left| 0+0+3 \right| }{ \sqrt { { \left( -2 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 } } } =\frac { 3 }{ \sqrt { 5 } } \)
\({ P }_{ 2 }=\frac { \left| 0+0-7 \right| }{ \sqrt { { \left( 1 \right) }^{ 2 }+{ \left( -4 \right) }^{ 2 } } } =\frac { 7 }{ \sqrt { 17 } } \)
Again,   \({ P }_{ 1 }=\frac { 3 }{ \sqrt { 5 } } \times \frac { \sqrt { 17 } }{ \sqrt { 17 } } =\sqrt { \frac { 153 }{ 85 } } \)
and \({ P }_{ 2 }=\frac { 7 }{ \sqrt { 17 } } \times \frac { \sqrt { 5 } }{ \sqrt { 5 } } =\sqrt { \frac { 245 }{ 85 } } \)
\(\therefore \sqrt { \frac { 245 }{ 85 } } >\sqrt { \frac { 153 }{ 85 } } \Rightarrow { P }_{ 2 }>{ P }_{ 1 }\)
Ans. x - 4y - 7 = 0 is farther.

\(\left( 12 \right) ^{ 5 }+\left( 8 \right) ^{ 5 }=\left( 10+2 \right) ^{ 5 }+\left( 10-2 \right) ^{ 5 }\\ \left( 10+2 \right) ^{ 5 }=^{ 5 }{ C }_{ 0 }\times { 10 }^{ 5 }+^{ 5 }{ C }_{ 1 }\times { 10 }^{ 4 }\times 2+^{ 5 }{ C }_{ 2 }\times { 10 }^{ 3 }\times { 2 }^{ 2 }+.........(i)\\ \left( 10-2 \right) ^{ 5 }=^{ 10 }{ C }_{ 5 }\times 5-^{ 5 }{ C }_{ 1 }\times { 10 }^{ 4 }\times 2+^{ 5 }{ C }_{ 2 }\times { 10 }^{ 3 }\times { 2 }^{ 2 }+.........(ii)\\ Now\quad adding\quad equ(i)\quad and\quad (ii)\\ \left( 12 \right) ^{ 5 }+\left( 8 \right) ^{ 5 }=2\left[ ^{ 5 }{ C }_{ 0 }10^{ 5 }+^{ 5 }{ C }_{ 2 }\left( 10 \right) ^{ 3 }\left( 2 \right) ^{ 2 }+^{ 5 }{ C }_{ 4 }\left( 10 \right) ^{ 1 }\left( 2 \right) ^{ 4 } \right] \\ =281600\)

We have, \(\frac { 4 }{ x+1 } \le 3\le \frac { 6 }{ x+1 } \)
\(\Longrightarrow \)\(4\le 3\left( x+1 \right) \le 6\quad \left[ \because x+1\neq 0\Longrightarrow x\neq -1 \right] \)
\(\Longrightarrow \)\(\frac { 4 }{ 3 } \le x+1\le 2\)
\(\Longrightarrow \)\(\frac { 4 }{ 3 } -1\le x\le 2-1\Longrightarrow \frac { 1 }{ 3 } \le x\le 1\)
Ans. \(\left[ \frac { 1 }{ 3 } ,1 \right] \)

\(LHS={ m }^{ 2 }-{ n }^{ 2 }\)

LHS=RHS
\(=(tan\theta +sin\theta )^{ 2 }-({ tan\theta }-sin\theta )^{ 2 }\)
\(=(tan\theta +sin\theta +tan\theta -sin\theta )(tan\theta +sin\theta -tan\theta +sin\theta )\quad [\because { a }^{ 2 }-{ b }^{ 2 }=\left( a+b \right) \left( a-b \right) ]\)
\(=\left( 2tan\theta \right) \left( 2sin\theta \right)\)
\(\Rightarrow { m }^{ 2 }-{ n }^{ 2 }=4tan\theta sin\theta\)
\(RHS=4\sqrt { mn } \)
\(=4\sqrt { \left( tan\theta +sin\theta \right) (tan\theta -sin\theta ) }\)
\(=4\sqrt { { tan }^{ 2 }\theta -{ sin }^{ 2 }\theta } \quad [\because (a+b)(a-b)=({ a }^{ 2 }-{ b }^{ 2 })]\)
\(=4\sqrt { { sin }^{ 2 }\theta \left( \frac { 1 }{ { cos }^{ 2 }\theta } -1 \right) } \)
\(=4\sqrt { { sin }^{ 2 }\theta ({ sec }^{ 2 }\theta -1) } =4\sqrt { { sin }^{ 2 }\theta { tan }^{ 2 }\theta } \)
\(\Rightarrow 4\sqrt { mn } =4sin\theta tan\theta\)
\(\\ From\quad Eqs.(i)\quad and\quad (ii)\)
LHS=RHS
Hence proved.

\(Y-X=\{ f,g\} \)

Here ⁿP_r  = ⁿP_r+1
\(\Rightarrow \frac { n! }{ (n-r)! } =\frac { n! }{ (n-r-1)! } \)
\(\Rightarrow \frac { 1 }{ (n-r)(n-r-1)! } =\frac { 1 }{ (n-r-1)! } \)
\(\Rightarrow \frac { 1 }{ n-r } =1 \Rightarrow n-r=1\) ....(i)
Also ⁿC_r = ⁿC_r-1
\(\Rightarrow \frac { n! }{ (n-r)!r! } =\frac { n! }{ (n-r+1)!(r-1)! } \)
\(\Rightarrow \frac { 1 }{ (n-r)!(r-1)! } =\frac { 1 }{ (n-1+1)(n-r)!(r-1)! } \)
\(\Rightarrow \frac { 1 }{ r } =\frac { 1 }{ n-r+1 } \)
\(\Rightarrow \)  n-r+1=r \(\Rightarrow \) n - 2r = -1  ....(ii)
From (i) and (ii), we have
n = 3 and r = 2.

\(\frac{-1\pm\sqrt {19i}}{2\sqrt 5}\)

The equation of given hyperbola is 3x² - 6y²=-18
i.e \(\frac { 3{ x }^{ 2 } }{ -18 } -\frac { 6{ y }^{ 2 } }{ -18 } =1\)
⇒ \(\frac { { y }^{ 2 } }{ 3 } -\frac { { x }^{ 2 } }{ 6 } =1\) which is of the form \(\frac { { y }^{ 2 } }{ { a }^{ 2 } } -\frac { { x }^{ 2 } }{ { b }^{ 2 } } \)=1
The foci and vertices of the hyperbola lie on y-axis
a² = 3 ⇒ a =\(\sqrt { 3 } \) and b² = 6 ⇒ b = \(\sqrt6\)
Now c² = a² + b² = 3 + 6 = 9  ⇒ c = 3
.. Coordinates of foci are (0, ± c) i.e. (0, ± 3)
Coordinates of vertices are (0, ± a) i.e. (0, ± \(\sqrt3\))
Eccentricity (e) = \(\frac { c }{ a } =\frac { 3 }{ \sqrt { 3 } } =\sqrt { 3 } \)
Length of latus rectum =\(\frac { 2{ b }^{ 2 } }{ a } =\frac { 2\times 6 }{ \sqrt { 3 } } =4\sqrt { 3 } \).

\(\frac{-1}{9}\)

The collection of all even integers is {..., -4, -2, 0, 2, 4, } which is well defined and hence it forms a set.

\(\frac { 6! }{ { 6 }^{ 6 } } \)

Let \(y=\frac { secx+tanx }{ secx-tanx } =\frac { \frac { 1 }{ cosx } +\frac { sinx }{ cosx } }{ \frac { 1 }{ cosx } -\frac { sinx }{ cosx } } \Rightarrow y=\frac { 1+sinx }{ 1-sinx }\) 
On differentiating both sides w.r.t. x,we get 
\(\frac { dy }{ dx } =\frac { d }{ dx } \left( \frac { secx+tanx }{ secx-tanx } \right) =\frac { d }{ dx } \left( \frac { 1+sinx }{ 1-sinx } \right)\) 
\(​​​​​​​=\frac { \left( 1-sinx \right) \frac { d }{ dx } \left( 1+sinx \right) -\left( 1+sinx \right) \frac { d }{ dx } \left( 1-sinx \right) }{ ({ 1-sinx) }^{ 2 } } \)
\(=\frac { \left( 1-sinx \right) \left( 0+cosx \right) -\left( 1+sinx \right) \frac { d }{ dx } \left( 1-sinx \right) }{ (1-sinx)^{ 2 } } \)  [using quotient rule of derivative] 
\(=\frac { \left( 1-sinx \right) \left( 0+cosx \right) -\left( 1+sinx \right) \left( 0-cosx \right) }{ ({ 1-sinx })^{ 2 } } \)
\(=\frac { 2cosx }{ (1-sinx)^{ 2 } } \)

Here the vertex is at (0, 0) and focus is at (0, -4) which lies on Y-axis. So, Y-axis is the axis of the parabola.
\(\therefore\) Equation of parabola is of the form
x² = -4ay \(\Rightarrow\) x² = -4(4)y  [\(\because\) a = 4]
\(\therefore\)  x² = -16y

Here, the man gets a fixed increment of Rs.320 each month.
Therefore, this forms an AP whose first term = 5200 and common difference (d) = 320
Salary for tenth month i.e. for n=10,
\({ a }_{ 10 }=a+(n-1)d\)
\(\Rightarrow { a }_{ 10 }=5200+(10-1)\times 320\)
\(\Rightarrow { a }_{ 10 }=5200+9\times 320\)
\(\Rightarrow { a }_{ 10 }=5200+2880\)
\(\therefore { a }_{ 10 }=8080\)

We know that, degree measure \(=\frac { 180 }{ \pi } \times \) Radian measure
\(\therefore \)  Required degree measure \(=\left( \frac { 180 }{ \frac { 22 }{ 7 } } \times 6 \right) \quad \left[ \therefore \pi =\frac { 22 }{ 7 } \right] \)
\(=\left( \frac { 1080\times 7 }{ 22 } \right) ^{ o }=\left( 343\frac { 7 }{ 11 } \right) ^{ 0 }\)
\(={ 343 }^{ o }+\frac { 7 }{ 11 } \times 60min\)  \(\left[ \because { 1 }^{ o }=60' \right] \)
\(={ 343 }^{ o }+38\frac { 2 }{ 11 } min={ 343 }^{ o }+38'+\frac { 2 }{ 11 } min\)
\(={ 343 }^{ o }+38'+\frac { 2 }{ 11 } \times 60''\)
\(={ 343 }^{ o }+38'+10.9''={ 343 }^{ o }38'11''(approx.)\)
Hence, the degree measure of 6 rad is 
\({ 343 }^{ o }38'11''(approx.).\)

Let given statement be false.i.e \(\sqrt { 11 } \) is rational.
It means \(\sqrt { 11 } \)=\(\frac { p }{ q } \), where p and q are coprime and q \(\neq \) 0
\(\Rightarrow \)   \(11=\frac { { p }^{ 2 } }{ { q }^{ 2 } } \Rightarrow { p }^{ 2 }=11{ q }^{ 2 }\)
It means 11 divides p.
Thus, there exists an integer r such that
p = 11r \(\Rightarrow \)p² = 121r²
It means 11 divides q.
It means 11 is a common factor of p and q which contradicts our assumption that p and q have no common factor.

Consider  \(P(k) : { 4 }^{ k }=3\lambda +1\)
Now,
P(k+1) :\({ 4 }^{ k+1 }={ 4 }^{ k }.4=(3\lambda +1)4\)
\(=12\lambda +4=3(4\lambda +1)+1\)

\(\frac { 1\times { 2 }^{ 2 }+2\times { 3 }^{ 2 }+...+n\times (n+1)^{ 2 } }{ { 1 }^{ 2 }\times 2+{ 2 }^{ 2 }3+...+{ n }^{ 2 }\times (n+1) } \)
=\(\frac { \Sigma n(n+1)^{ 2 } }{ \Sigma n^{ 2 }(n+1) } =\frac { \Sigma n({ n }^{ 2 }+2n+1) }{ \Sigma ({ n }^{ 3 }+n^{ 2 }) } \)
=\(\frac { \Sigma ({ n }^{ 3 }+2{ n }^{ 2 }+n) }{ \Sigma ({ n }^{ 3 }+{ n }^{ 2 }) } =\frac { \Sigma { n }^{ 3 }+2\Sigma { n }^{ 2 }+\Sigma n }{ \Sigma { n }^{ 3 }+\Sigma { n }^{ 2 } } \)
=\(\frac { \frac { { n }^{ 2 }(n+1)^{ 2 } }{ 4 } +\frac { 2n(n+1)(2n+1) }{ 6 } +\frac { n(n+1) }{ 2 } }{ \frac { { n }^{ 2 }(n+1)^{ 2 } }{ 4 } +\frac { n(n+1)(2n+1) }{ 6 } } \)
=\(\frac { \frac { n(n+1) }{ 2 } \left[ \frac { n(n+1) }{ 2 } +\frac { 2(2n+1) }{ 3 } +1 \right] }{ \frac { n(n+1) }{ 2 } \left[ \frac { n(n+1) }{ 2 } +\frac { 2n+1 }{ 3 } \right] } \)
=\(\frac { 3n^{ 2 }+11n+10 }{ 3{ n }^{ 2 }+7n+2 } =\frac { 3{ n }^{ 2 }+6n+5n+10 }{ 3{ n }^{ 2 }+6n+n+2 } \)
=\(\frac { 3n(n+2)+5(n+2) }{ 3n(n+2)+1(n+2) } \)
=\(\frac { (n+2)(3n+5) }{ (n+2)(3n+1) } =\frac { 3n+5 }{ 3n+1 } \).

\(\text { Let } f(x)=x+a \text { . Accordingly, } f(x+h)=x+h+a\)
\(\text { By first principle, }\)
\(f^{\prime}(x) =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \)
\(=\lim _{h \rightarrow 0} \frac{x+h+a-x-a}{h} \)
\(=\lim _{h \rightarrow 0}\left(\frac{h}{h}\right) \)
\(=\lim _{h \rightarrow 0}(1) \)
\(=1\)

The equations of lines AB, BC and AC are 
x + y - 6 = 0   ...(i)
x - 3y - 2 = 0  ...(ii)
5x- 3y + 2 = 0 ...(iii)
On solving the equation (i) and (ii), we have coordinates of point B (5,1).
On solving the equation (ii) and (iii), we have coordinates of point C (-1, -1).
On solving the equation (iii) and (i), we have coordinates of point A (2, 4).
Let D, E, F are mid points of BC, AC and AB respectively.
Coordinates of D are \(\left({5-1\over 2},{1-1\over 2}\right)\ i.e.(2,0)\)
Coordinates of E are \(\left({2-1\over 2},{4-1\over 2}\right)\ i.e.\left({1\over 2},{3\over 2}\right)\)
Coordinates of D are  \(\left({2+5\over 2},{4+1\over 2}\right)\ i.e.\left({7\over 2},{5\over 2}\right)\)
Equation of median AD is
\(y-4={(0-4)\over (2-2)}(y-2)\)
⇒  x - 2 = 0
Equation of median BE is
\(y-1={\left({3\over2}-1\right)\over \left({1\over2}-5\right)}(x-5)\)
\(⇒\ y-1={{1\over 2}\over -{9\over 2}}(x-5)\)
⇒ 9y - 9 = -x + 5
⇒ x + 9y - 14 = 0
Equation of CF is
\(y+1={\left({5\over 2}+1\right)\over \left({7\over 2}+5\right)}(x+1)\)
\(⇒\ y+1={{7\over 2}\over {9\over 2}}(x+1)\)
\(⇒\ y+1={7\over 9}(x+1)\)
⇒ 9y + 9 = 7x + 7 ⇒ 7x - 9y - 2 = 0
Thus equations of medians are x - 2 = 0, x + 9y - 14 = 0 and 7x - 9y - 2 = O.

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.

14 degree

(i) Here A = {1, 2, 3, 4} and B = {1, 5, 9, 11,15,16}.
\(\therefore\) A x B = {(1, 1), (1, 5), (1, 9), (1, 11),
(1,15), (1,16), (2,1), (2, 5),
(2, 9), (2, 11) (2, 15), (2, 16),
(3,1),(3,5),(3,9),(3,11),
(3,15), (3,16), (4,1),(4,5),
(4,9), (4,11), (4,15),(4,16)}
f = {1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
Now (1, 5), (2, 9), (3, 1), (4, 5), (2, 11) \(\in\) A x B
\(\therefore\) f is a relation from A to B.
(ii)Here f(2) = 9 and f(2) = 11
\(\therefore\) f is not a function from A to B.

The given inequality is 2x - y > l.
Draw the graph of the line 2x - y = l.

Table of value satisfying the equation 2x - y = 1

Putting (0, 0) in the given inequation, we have
2 x 0 - 0 > 1  \(\Rightarrow\) 0 > 1 , which is false
\(\therefore\) Half plane of 2x - y > 1 is away from origin
Also the given inequality is x - 2y < -l.
Draw the graph of the line x - 2y =-l.
Table of values satisfying the equation
x - 2y = -1

Putting (0, 0) in the given inequation, we have
0-2 x 0 < -1 \(\Rightarrow\) 0 < -1, which is false
\(\therefore\) Half plane of x - 2y < -1 is away from origin