\(\frac { { x }^{ 2 } }{ 52 } +\frac { { y }^{ 2 } }{ 13 } =1\)

0.97

The equation of given ellipse is
16x² + y² =16
i.e. \(\frac { { 16x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 16 } \)=1 ⇒ \(\frac { { x }^{ 2 } }{ 1 } +\frac { { y }^{ 2 } }{ 16 } \)=1
Now 16> 1 ⇒ a² = 16 and b² = 1
So the equation of ellipse in standard form is
\(\frac { { y }^{ 2 } }{ { a }^{ 2 } } +\frac { { x }^{ 2 } }{ { b }^{ 2 } } \)=1
∴ a² = 16 ⇒ a = 4 and b² = 1⇒ b = 1
We know that c=\(\sqrt { { a }^{ 2 }-b^{ 2 } } \)
∴ c=\(\sqrt { 16-1 } =\sqrt { 15 } \)
∴ Coordinates of foci are (0,± c) i.e. (0, ±\(\sqrt { 15 } \))
Coordinates of vertices are (0, ±a) i.e. (0, ±4)
Length of major axis = 2a = 2 x 4 = 8
Length of minor axis = 2b = 2 x 1 = 2
Eccentricity (e) =\(\frac { c }{ a } =\frac { \sqrt { 15 } }{ 4 } \)
Length of latus rectum =\(\frac { 2{ b }^{ 2 } }{ a } =\frac { 2\times 1 }{ 4 } =\frac { 1 }{ 2 } \) .

Here, diameter AB=40cm
Radius OA=20cm
Chord AC = 20cm
\(\therefore\) \(\triangle\) AOC is an equilateral triangle

So \(\angle\)AOC = 60° =\(({60\times \pi \over 180})^C=({\pi\over 3})^C\)
We know that \(\theta ^c={l\over r}\)
\(\frac{\pi}{3}=\frac{\widehat{\mathrm{AB}}}{20} \Rightarrow \widehat{\mathrm{AB}}=\frac{20 \pi}{3} \mathrm{~cm}\)
\(\therefore {\pi\over 3}={l\over 20}\Rightarrow l={20\pi\over 3}cm\)

(1-i)⁴ =[(1-i)²]² = (1+i2-2i)²
= (1-1-2i)² = (-2i)²
= 4i²=4\(\times\)-1 =-4

- 47° 30' = -\((47{30\over 60})^c=-({95\over 2})^o\)
\(=-({95\over 2}\times{\pi\over 180})^c=-({19\pi\over 72})^c\)

Here \(\lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\)\([\frac{0}{0} form]\)
= \(\overset{lim}{z\rightarrow 1} \frac{(z^{1/6})^{2}-(1)^{2}}{z^{1/6}-1}\)
=\(\overset{lim}{z\rightarrow 1}=\frac{(z^{1/6}+1)(z^{1/6}-1)}{(z^{1/6}-1)}\)
=\(\overset{lim}{z\rightarrow 1} z^{1/6}+1=(1)^{1/6}+1\)
=1+1=2.

Here 5x - 3 < 7
5x < 7 + 3  \(\Rightarrow\) 5x < 10
Dividing both sides by 5, we have x < 2
(i) When x is an integer then values of x that make the statement true are ..., -3, -2, -1, 0, 1
The solution set of inequality is {..., -3, -2, -1,0, I}.
(ii) When x is a real number. The solution set of inequality is x \(\epsilon \) (-\(\infty\),2)

Let x and x + 2 be two consecutive even positive integers.
Then x > 10 and x + x + 2 < 28 \(\Rightarrow\) x > 10 and 2x + 2 < 28  \(\Rightarrow\) x > 10 and 2x < 28 -2
x > 10 and x < 13  \(\Rightarrow\) 10 < x < 13 \(\Rightarrow\) x = 12
Thus required pair of even integers is (12, 14).

Here, Total frequnecy=Total number of times a die is thrown
=500
and a number of observed frequency, A
= Number of time the outcome 3 is thrown =100
Probability of getting outcome 3,
\(P(A) = \frac { Number\quad of\quad observed\quad frequency }{ Total\quad frequency } \)
\(=\frac { 100 }{ 500 } \)
\(=\frac { 1 }{ 5 } \)

E₁={(1,3),(2,2), (3,1)}
E₂={(1,4),(2,3),(3,2),(4,1)}
E₃={(1,5),(2,4),(3,3),(5,1),(4,2)}
No event

Let us make the following table from the given data.

Here,
Which item lies in the cumulative frequency 28. Therefore, 20-30 is the median class.
So, we have, l = 20, cf = 13, f = 15, h = 10 and N = 50
Now, median, M = \(l+\frac { \frac { N }{ 2 } -cf }{ f } \times h\)
= \(20+\frac { 25-13 }{ 15 } \times 10\)
= 20 + 8 = 28
Hence, the mean deviation about median is given by
MD(M) = \(\frac { 1 }{ N } \sum _{ i=1 }^{ 6 }{ { { f }_{ i } } } \left| { { x }_{ i } }{ -{ M } } \right| =\frac { 1 }{ 50 } \times 508=10.16\)
Hence, the mean deviation about median is 10.16.

The compound statement is '25 is a multiple of 5 and 8'.
Since, 25 is a multiple of 5 but it is not a multiple of 8.
Therefore, p is true but q is not true. Hence, the compound statement is not true. i.e. the statement 'p and q' is not a valid statement.

If PT Usha is not awarded, then she does not stand first in 200 m race.

\(\lim _{x \rightarrow 0} \frac{\log \left(x^{3}+1\right)}{x^{3}} \times \frac{x^{3}}{\sin ^{3} x} A n s .1\)

Let A(x, y, z) be any point which is equidistant from points A(1, 2, 3) and B (3, 2, -1).
Then AB = \(\sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z-3 \right) }^{ 2 } } \)
AC = \(\sqrt { { \left( x-3 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z+1 \right) }^{ 2 } } \)
It is given that AB = AC
\(\therefore\)  \(\sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z-3 \right) }^{ 2 } } \)= \(\sqrt { { \left( x-3 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z+1 \right) }^{ 2 } } \)
\(\Rightarrow { \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z-3 \right) }^{ 2 }\ =\ { \left( x-3 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( z+1 \right) }^{ 2 }\)
\(\Rightarrow\) x²+ 1 - 2x + z² + 9 - 6z = x² + 9 - 6x + z² + 1 + 2z
\(\Rightarrow\) - 2x - 6z + 10 = - 6x + 2z + 10
\(\Rightarrow\) -2x - 6z + 6x - 2z = 0
\(\Rightarrow\) 4x - 8z = 0
\(\Rightarrow\) x - 2z = 0.

Given points are (x,-8,4) and (3,-5,4) and distance between these points = 5
\(\therefore \sqrt { { (x-3) }^{ 2 }+{ { (-8+5) }^{ 2 } }+{ (4-4) }^{ 2 } } =5\)
\(\Rightarrow \sqrt { { (x-3) }^{ 2 }+{ { (-3) }^{ 2 } }+0 } =5\)
\(\Rightarrow\)(x - 3)² + 9 + 0 = 25 [ on squaring both sides]
\(\Rightarrow\)(x - 3)² = 16 \(\Rightarrow\) (x - 3)² = (4)²
\(\Rightarrow\) x - 3 = \(\pm \)4 [ taking square root on both sides]
\(\therefore \)  x = 7 or  -1

\(\text{Slope of line, }AB=\frac { 3-1 }{ 2+2 } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } ={ m }_{ 1 }(say)\)
\(\text{Slope of line, } BC=\frac { -4-3 }{ -2-2 } =\frac { 7 }{ 4 } ={ m }_{ 2 }\)
\(\therefore tanB=\left| \frac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| =\left| \frac { \frac { 7 }{ 4 } -\frac { 1 }{ 2 } }{ 1+\frac { 1 }{ 2 } .\frac { 7 }{ 4 } } \right| \)
\(Ans.\angle B={ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)

Given parallel lines are
lx + ly + h = 0 and lx + ly - r =0
On comparing with Ax + By + C₁ = 0
and  Ax + By + C₂ = 0 we get
A = l, B = l, C₁ = h, C₂ = -r
\(\therefore \)  Required distance \(=\frac { \left| { C }_{ 1 }-{ C }_{ 2 } \right| }{ \sqrt { { A }^{ 2 }+{ B }^{ 2 } } } =\left| \frac { h+r }{ \sqrt { l^{ 2 }+{ l }^{ 2 } } } \right| =\left| \frac { h\quad +\quad r }{ \sqrt { 21 } } \right| \)

We know that, AM \(\ge \) GM
\(\frac { x+y }{ 2 } >\sqrt { xy } ,\frac { y+z }{ 2 } >\sqrt { yz } and\quad \frac { z+x }{ 2 } >\sqrt { zx } \)
On multiplying the above inequalities, we get
\(\frac { x+y }{ 2 } .\frac { y+z }{ 2 } .\frac { y+z }{ 2 } >\sqrt { (xy)(yz)(zx) } \)
or (x + y)(y + z)(z + x) > 8xyz

We have, a_n = 2n²- n + 1
On putting n = 1, we get
a₁ = 2(1)²- 1 + 1 = 2 - 1 + 1 = 2
On putting n = 2, we get
a₂= 2(2)²- 2 + 1 = 8 - 2 + 1 = 7
On putting n = 3, we get
a₃ = 2(3)²- 3 + 1 = 18 - 3 + 1 = 16
On putting n = 4, we get
a₄ = 2(4)²- 4 + 1 = 32 - 4 + 1 = 29
On putting n = 5, we get
a₅ = 2(5)²- 5 + 1 = 50 - 5 + 1 = 46
Hence, the first five terms of the given sequence are 2, 7, 16, 29 and 46.

\((1+x+{ x }^{ 2 })^{ 3 }=[(1+x)+{ x }^{ 2 }]^{ 3 }\)
\(=^{ 3 }{ C }_{ o }(1+x)^{ 3 }+^{ 3 }{ C }_{ 1 }(1+x)^{ 2 }({ x }^{ 2 })^{ 1 }+^{ 3 }{ C }_{ 2 }(1+x)({ x }^{ 2 })^{ 2 }+^{ 3 }{ C }_{ 3 }({ x }^{ 2 })^{ 3 }\)
\(={ x }^{ 6 }+3{ x }^{ 5 }+6{ x }^{ 4 }+7{ x }^{ 3 }+6{ x }^{ 2 }+3x+1\)

Given equation of hyperbola is \(\frac { { y }^{ 2 } }{ 9 } -\frac { { x }^{ 2 } }{ 27 } =1\)
On comparing with \(\frac { { y }^{ 2 } }{ { a }^{ 2 } } -\frac { { x }^{ 2 } }{ { b }^{ 2 } } =1\) we get
a²=9, b²=27 or \(a=3,b=3\sqrt { 3 } \)
(ii) Centre (0,0)
(iii) Transverse axis, \(2a=2\times 3=6\)
(iv) Conjugate axis, \(2b=2\times 3\sqrt { 3 } =6\sqrt { 3 } \)
(v) Value of \(c=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =\sqrt { 9+27 } =\sqrt { 36 } =6\)
(vi) Vertices = \((0,\pm a)=(0,\pm 3)\)
(vii) Directions, \(y=\pm \frac { { a }^{ 2 } }{ c } =\pm \frac { 9 }{ 6 } =\pm \frac { 3 }{ 2 } \)
(viii) Foci = \((0,\pm c)=(0,\pm 6)\)
(ix) Eccentricity, \(e=\frac { c }{ a } =\frac { 6 }{ 3 } =2\)
(x) Length of latusrectum \(=\frac { { 2b }^{ 2 } }{ a } =\frac { 2\times 27 }{ 3 } =18\)

Let us first seat the 5 girls. This can be done in 5! ways. For each such arrangement, the three boys can be seated only at the cross marked places.
× G × G × G × G × G ×.
There are 6 cross marked places and the three boys can be seated in ⁶P₃ ways.
Hence, by multiplication principle, the total number of ways 
\(=5 ! \times{ }^{6} \mathrm{P}_{3}=5 ! \times \frac{6 !}{3 !}\)
= 4 × 5 × 2 × 3 × 4 × 5 × 6 = 14400.

Clearly, from Excercise 7, one question can be selected in 11 ways.
From Exercise 8, one question can be selected in 15 ways.
and from Exercise 9, one question can be selected in 10 ways.
Since, each operation is performed after performing the previous operation, so we use fundamendal principle of multiplication. 
\(\therefore \) Number of ways of selecting three questions
= 11 \(\times\)15 \(\times\)10 = 1650

Consider :\(P(k) : \sum _{ t=1 }^{ k-1 }{ t(t+1) } =\frac { k(k-1)(k+1) }{ 3 } ​​​​k\ge 2\) 
P(k):1.2+2.3+3.4+....+(k-1)k \(=\frac { k(k-1)(k+1) }{ 3 } \)
Now P(k+1):1.2+2.3+3.4+...+(k1)k+k(k+1)
\(=\frac { k(k-1)(k+1) }{ 3 } +k(k+1)\)
\(=\frac { k(k+1)(k+2) }{ 3 } k\ge 2\)

Clearly,f(x)=\(\lambda f,\lambda ^{ 2 }f\) and \(\left( \frac { 1 }{ \lambda } \right) f\)are function from R to itself
=\(\left( \frac { 1 }{ \lambda } f \right) (x)=\frac { 1 }{ \lambda } .f(x)=\frac { 1 }{ \lambda } \times \frac { x }{ \lambda } =\frac { x }{ \lambda ^{ 2 } } ,\forall x\epsilon R\)

\((f+g)(x)=x^{2}+2 x+1,(f-g)(x)=x^{2}-2 x-1,\)
\((\text { fg })(x)=x^{2}(2 x+1)=2 x^{3}+x^{2},\left(\frac{f}{g}\right)(x)=\frac{x^{2}}{2 x+1}, x \neq-\frac{1}{2} \)

Here,  A = {1,4,5} \(\Rightarrow \) n(A) = 3
Number of proper subsets = 2ⁿ - 1 = 7

[-23, 5)= {x:x \(\in\) R, -23\(\le\) x <5}

Here f(x)=\(\frac { { (px }^{ 2 }+qx+r) }{ ax+b } \)
\(\therefore f(x)=\frac { d }{ dx } \left[ \frac { { px }^{ 2 }+qx+r }{ ax+b } \right] \)
\( =\frac { (ax+b)\frac { d }{ dx } ({ px }^{ 2 }+qx+r)-({ px }^{ 2 }+qx+r)\frac { d }{ dx } (ax+b) }{ { (ax+b) }^{ 2 } } \)
\( =\frac { (ax+b)(2px+q)-({ px }^{ 2 }+qx+r)(a) }{ { (ax+b) }^{ 2 } } \)
\(=\frac { { 2px }^{ 2 }+aqx+2bpx+bq-{ apx }^{ 2 }-apx-ar }{ { (ax+b) }^{ 2 } } \)
\(=\frac { { apx }^{ 2 }+2bpx+bq-ar }{ { (ax+b) }^{ 2 } } \)

Here ²²P_r+1: ²⁰P_r+2=11:52
\(\Rightarrow \frac { 22! }{ (21-r)! } \times \frac { (18-r)! }{ 20! } =\frac { 11 }{ 52 } \)
\(\Rightarrow \frac { 22\times 21\times 20! }{ (21-r)(20-r)(19-r)(18-r)! } \times \frac { (18-r)! }{ 20! } =\frac { 11 }{ 52 } \)
\(\Rightarrow \frac { 22\times 21 }{ (21-r)(20-r)(19-r) } =\frac { 11 }{ 52 } \)
\(\Rightarrow (21-r)(20-r)(19-r)=2\times 21\times 52\)
\(\Rightarrow (21-r)(20-r)(19-r)=14\times 13\times 12\)
\(\Rightarrow (21-r)(20-r)(19-r)\) 
= (21-7)(20-7)(19-7)
\(\Rightarrow\)  r = 7

We have
\(L.H.S =tan \ 4x={2tan\ 2x\over 1-tan^2 \ 2x }\)
\([\because tan 2A={2tan A\over 1-tan^2A} ]\)
\(={2.{2tan \ x\over 1-tan ^2x}\over 1-({2tan \ x\over 1-tan^2 \ x})^2}\)
\(={{4 tan x\over 1-tan^2 x}\over {(1-tan^2 \ x)^2-4tan^2 \ x \over (1-tan^2 \ x)^2}}\)
\(={4tan \ x \over 1-tan^2 x} \times {(1-tan^2 x)^2\over 1+tan^4x-2tan ^2-4tan^2 x}={4tan \ x(1-tan^2 \ x)\over 1-6tan^2 \ x+tan ^4 \ x}\)
=R.H.S

\(\frac { 1 }{ 2 } \)

Ans : 9,15.08

\({ f }^{ ' }(x)=\lim_ { h\rightarrow 0 }{ lim } \frac { sin(x+h+1)-sin(x+1) }{ h } \)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { 2cos\left( \frac { x+h+1+x+1 }{ 2 } \right) sin\left( \frac { x+h+1-x-1 }{ 2 } \right) }{ h } \)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { 2cos\frac { 2x+h+2 }{ 2 } sin\frac { h }{ 2 } }{ 2\times \frac { h }{ 2 } } \)
\(cos(x+1)\)

Let D be the mid-point of BC and E be the centroid of \(\Delta ABC\), which divides AD in the ratio of 2:1
\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 2 } \frac { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } }{ 2 } \right) \)

Here, wire are vertical.
Let equation of the parabola be in the form
\({ x }^{ 2 }=4ay\)    ...(i)

Focus is at the middle of the cable and shortest and longest vertical supports are 6 m and 30 m and roadway in 100 m long.
Clearly, the coordinates of Q(50, 24) will satisfy Eq.(i)
\(\therefore \quad (50)^{ 2 }=4a\times 24\Rightarrow 2500=96a\Rightarrow a=\frac { 2500 }{ 96 } \)
\(Hence,from\quad Eq.(i),\quad { x }^{ 2 }=4\times \frac { 2500 }{ 96 } y\Rightarrow { x }^{ 2 }=\frac { 2500 }{ 24 } y\)
 Let PR=km
Then, point\ P(18,k) will satisfy the equation of parabola.
\(\therefore \quad From\quad Eq.(i),\quad (18)^{ 2 }=\frac { 2500 }{ 24 } \times k\)
\(\Rightarrow 324=\frac { 2500 }{ 24 } k\Rightarrow k=\frac { 324\times 24 }{ 2500 } =\frac { 324\times 6 }{ 625 } =\frac { 1944 }{ 625 } \)
\( \Rightarrow k=3.11\)
Therefore  Required length= 6+k=6+3.11=9.11m(approx.)

\({ P }_{ 1 }=\frac { \left| 0+0+3 \right| }{ \sqrt { { \left( -2 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 } } } =\frac { 3 }{ \sqrt { 5 } } \)
\({ P }_{ 2 }=\frac { \left| 0+0-7 \right| }{ \sqrt { { \left( 1 \right) }^{ 2 }+{ \left( -4 \right) }^{ 2 } } } =\frac { 7 }{ \sqrt { 17 } } \)
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.

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]

Coefficient of 2nd, 3rd and 4th  terms are ⁿC₁ , ⁿC₂ and ⁿC₃ respectively. Given ⁿC₁ , ⁿC₂ and ⁿC₃ are in AP.
2 ⁿC₂=ⁿC₁+ⁿC₃
\(\therefore \)  Ans: n=7

\(\frac { 2x+1 }{ 7x-1 } -5>0,\frac { x+7 }{ x-8 } -2>0 \Rightarrow \frac { -33x+6 }{ 7x-1 } >0\frac { -x+23 }{ (x-8) } >0\)
\(\Rightarrow (-33x+6)(7x-1)>0,(-x+23)(x-8)>0\)
\(\\ \Rightarrow x>7,x<23\)

We have, z₁ = 3+2i and z₂ = 2-4i
Now, LHS= \({ \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }+ { \left| { z }_{ 1 }-{ z }_{ 2 } \right| }^{ 2 }\)
On substituting the value of z₁ and z_{2, we get
LHS =} \({ \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }+ { \left| { z }_{ 1 }-{ z }_{ 2 } \right| }^{ 2 }=2({ \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 })\)
\(={ \left| 3+2i+2-4i \right| }^{ 2 }+\left| 3+2i+2-4i \right| \)
\( ={ \left| 5-2i \right| }^{ 2 }+{ \left| 1+6i \right| }^{ 2 }={ (5) }^{ 2 }+{ (-2) }^{ 2 }+{ (1) }^{ 2 }+{ (6) }^{ 2 }\)
\( [if\quad z=a+ib,\quad then\quad { \left| z \right| }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }]\)
\(=25+4+1+36\)
\(\because \quad LHS=66\quad ......(i)\)
\(and\ RHS=2({ \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 } =2({ \left| 3+2i \right| }^{ 2 }+{ \left| 2-4i \right| }^{ 2 })\)
\(=2[{ (3) }^{ 2 }+{ (2) }^{ 2 }+{ (2) }^{ 2 }+{ (-4) }^{ 2 }]\)
\(=2(9+4+4+16) =2\times 33\)
\(\because \quad RHS=66\quad ....(ii)\)
From Eq (i) and (ii), we get LHS = RHS.
Hence proved.

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

\(LHS={ K }^{ 3 }si{ n }^{ 3 }\quad A\quad sin(B-C)+k^{ 3 }si{ n }^{ 3 }\quad Bsin(C-A)+{ k }^{ 3 }si{ n }^{ 3 }Csin(A-B)\)
\(={ K }^{ 3 }[si{ n }^{ 2 }\quad AsinAsin(B-C)+{ Sin }^{ 2 }Bsin\quad Bsin(C-A)+{ Sin }^{ 2 }CsinCsin(C-A)]\)
\(={ K }^{ 3 }si{ n }^{ 3 }\quad A\quad sin(B+C)sin(B-C)+[{ Sin }^{ 2 }Bsin(C+A)+{ sin }(C-A)+{ Sin }^{ 2 }Csin(A+B)sin(A-B)]\)
\(={ K }^{ 3 }si{ n }^{ 3 }\quad A\quad sin(B+C)sin(B-C)+[{ Sin }^{ 2 }Bsin(C+A)+{ sin }(C-A)+{ Sin }^{ 2 }Csin(A+B)sin(A-B)]\)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}={ K }^{ 3 }\times 0=RHS\)

Domain = {7, 9, 13, 17} Range = {8, 27, 125, 343}

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

Let P(n) = :a+ar+ar²+.......+ar^n-1 = \(\frac { a({ r }^{ n }-1) }{ r-1 } \)
For n =1
P(1) = ar^n-1 = \(\frac { a({ r }^{ 1 }-1) }{ r-1 } \)
⇒ a =a
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = a+ar+ar²+.....+ar^k-1
= \(\frac { a({ r }^{ k }-1) }{ r-1 } \)...(i)
For n = k+1
R.H.S = \(\frac { a({ r }^{ k+1 }-1) }{ r-1 } \)
L.H.S =\(\frac { a({ r }^{ n }-1) }{ r-1 } \)+αr^k [Using (i)]
=  \(\frac { { ar }^{ k } }{ r-1 } -\frac { a }{ r-1 } +{ ar }^{ k }\)
=  \({ ar }^{ k }\left( \frac { 1 }{ r-1 } +1 \right) -\frac { a }{ r-1 } \)
= \({ ar }^{ k }\left( \frac { r }{ r-1 } \right) -\frac { a }{ r-1 } \)
= \(\frac { { ar }^{ k+1 } }{ r-1 } -\frac { a }{ r-1 } =\frac { { ar }^{ k+1 }-a }{ r-1 } \) 
= \(\frac { { a(r }^{ k+1 }-1) }{ r-1 } \)
∴ 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

The three examples of sentences, which are not statements, are as follows.
(i) He is a doctor.
It is not evident from the sentence as to whom ‘he’ is referred to. Therefore, it is not a statement.
(ii) Geometry is difficult.
This is not a statement because for some people, geometry can be easy and for some others, it can be difficult.
(iii) Where is she going?
This is a question, which also contains ‘she’, and it is not evident as to who ‘she’ is. Hence, it is not a statement.

Step I Let P(n) be the given statement.
i.e P(n) :\(1+2+3......n<\frac { 1 }{ 8 } { (2n+1) }^{ 2 }\)
Step II For n=1 , we have
\(n<\frac { 1 }{ 8 } {[ (2(1)+1)] }^{ 2 }\) \(\Rightarrow \) 1< \(\frac { 9 }{ 8 } \) which is true.
Thus, P(1) is true.
Step III Let us assume that P(k) is true.
i.e. P(k):\(1+2+3......k<\frac { 1 }{ 8 } { (2k+1) }^{ 2 }\)........(i)
Step IV   Now we shall prove the statement for n=k+1.For this, we have to show that
\(1+2+3......(k+1)<\frac { 1 }{ 8 } { [2(k+1)+1] }^{ 2 }\)
from Eq.(i) , We have  
\(1+2+3......k<\frac { 1 }{ 8 } { (2k+1) }^{ 2 }\)
\(1+2+3......k+(k+1)<\frac { 1 }{ 8 } { (2k+1) }^{ 2 }+(k+1)\)   [adding (k+1) on both sides]
\(=\frac { 1 }{ 8 } { [(2k+1) }^{ 2 }+8(k+1)]\)
\(\frac { 1 }{ 8 } { [4k }^{ 2 }+1+4k+8k+8]=\frac { 1 }{ 8 } { [4k }^{ 2 }+12k+9]\)
\(\frac { 1 }{ 8 } { (2k+3) }^{ 2 }=\frac { 1 }{ 8 } { [2(k+1)+1] }^{ 2 }\)
\(\Rightarrow \)  \(1+2+3......k+1<\frac { 1 }{ 8 } { [2(k+1)+1] }^{ 2 }\)
Thus P(K+1) is true whenever P(k) is true.
Hence, by principle of mathematical induction P(n) is true for all natural numbers, \(n\ge 3\)

p: All the angles of a triangle are equal.
q: the triangles is an obtuse angled triangled.
We have to show that if p then \(\sim \) q. Take each angle equal to \({ 60 }^{ \circ }\), it means they are acute angle.
Hence, we conclude that the given statement is false.

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

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

Putting a²=x and \(\sqrt { { a }^{ 2 }-1 } \) =y, we have
\(\left( { a }^{ 2 }+\sqrt { { a }^{ 2 }-1 } \right) ^{ 4 }+({ a }^{ 2 }-\sqrt { { a }^{ 2 }-1)^{ 4 } } \)
=(x+y)⁴+(x-y)⁴
=[⁴C₀x⁴+⁴C₁x³y+⁴C₂x²y²+⁴C₃xy³+⁴C₄y⁴]+[⁴C₀x^4-4C₁x³y+⁴C₂x²y^2-4C₃xy^3-4C₄y⁴]
=2[⁴C₀x⁴+⁴C₂x²y²+⁴C₄y⁴]
=2[x⁴+6x²y²+y⁴]
=2[a²)⁴+6(a²)²\(\left( \sqrt { { a }^{ 2 }-1 } \right) ^{ 2 }\)+\(\left( \sqrt { { a }^{ 2 }-1 } \right) ^{ 4 }\)]
=2[a⁸+6a⁴(a²-1)+(a²-1)²]
=2[a⁸+6a⁶-6a⁴+a⁴-2a²+1]
=2[a⁸+6a⁶-5a⁴-2a²+1]

Here A(0,0, 6), B(0,4, 0) and C(6, 0, 0) are vertices of \(\triangle \)ABC

Now D is mid point of BC 
\(\therefore \) Coordinates of D is \(\left( \frac { 0+6 }{ 2 } ,\frac { 4+0 }{ 2 } ,\frac { 0+0 }{ 2 } \right) \)= (3,2,0)
\(\therefore \quad AD=\sqrt { (0-3)^{ 2 }+(0-2)^{ 2 }+(6-0)^{ 2 } } \)
=\(\sqrt { 9+4+36 } =\sqrt { 7 } \) units.
Also E is mid point of AC
\(\therefore \)  Coordinates of E is \(\left( \frac { 0+6 }{ 2 } ,\frac { 0+0 }{ 2 } ,\frac { 6+0 }{ 2 } \right) \)= (3,0,3)
\(\therefore \quad BE=\sqrt { (0-3)^{ 2 }+(4-0)^{ 2 }+(0-3)^{ 3 } } \)
\(=\sqrt { 9+16+9 } =\sqrt { 34 } \)units.
Also F is mid point of AB
\(\therefore \) Coordiates of F is \(\left( \frac { 0+0 }{ 2 } ,\frac { 0+4 }{ 2 } ,\frac { 6+0 }{ 2 } \right) \)=(0,2,3)
\(\therefore \) CF=\(\sqrt { (6-0)^{ 2 }+(0-2)^{ 2 }+(0-3)^{ 2 } } \)
\(=\sqrt { 36+4+9 } =7\)units.

Let AB be a rod of length 12 cm and P(x, y) be any point on the rod such that PA = 3 cm and PB = 9 cm.

Let AR = a and BQ = b
Then ΔARP ~ ΔPQB
∴ \(\frac { AR }{ PQ } =\frac { AP }{ PB } \)
∴ \(\frac { a }{ x } =\frac { 3 }{ 9 } \) ⇒ 9a=3a  ⇒ a=\(\frac { x }{ 3 } \)
and \(\frac { BQ }{ BP } =\frac { PR }{ PA } \)
∴  \(\frac { b }{ 9 } =\frac { y }{ 3 } \) ⇒ 3b=9y ⇒ b=3y
Now OR + AR = x + a = x+\(\frac { x }{ 3 } =\frac { 4x }{ 3 } \)
OB = OQ + BQ = y + b = y + 3y = 4y
In right angled ΔAOB
AB²= OA²+ OB²
∴ (12)² = \(\left( \frac { 4x }{ 3 } \right) ^{ 2 }\)+(4y)²
⇒ 144=\(\frac { 16{ x }^{ 2 } }{ 9 } \)+16y²
⇒ \(\frac { 16{ x }^{ 2 } }{ 9\times 144 } +\frac { 16{ y }^{ 2 } }{ 144 } \) =1 ⇒ \(\frac { { x }^{ 2 } }{ 81 } +\frac { { y }^{ 2 } }{ 9 } \)=1
which is required locus of point P and which represents an ellipse.

The given lines are x - 3y + 1 = 0 and 2x + 5y - 9 = 0
Equation ofany line that passes through the point ofintersection ofthe given lines is
(x - 3y + 1) + k (2x + 5y - 9) = 0 ...(i)
⇒ (1 + 2k) x + (5k - 3) y + 1 - 9k = 0
Now using the formula to find the perpendicular distance from (x₁, y₁) to ax + by + C = 0
\(d=\left| ax_1+by_1+c\over \sqrt{a^2+b^2}\right|\)
We get
\(\left| (1+2k)\times0+(5k-3)\times0+(1-k)\over \sqrt{(1+2k)^2+(5k-3)^21}\right|=\sqrt{5}\)
\(⇒{(1-9k)\over\sqrt{(1+2k)^2+(5k-3)^2}}=\sqrt{5}\)
squaring both sides, we get
(1-9k)² = 5[(1+2k)² + (5k-3)2]
1+81k² - 18k = 5[1+4k²+4k+25k² + 9 - 30k]
⇒ 1+81k² - 18k = 50 + 145k² - 130k
⇒ 64k² - 112k + 49 = 0
⇒ (8k - 7)² = 0
⇒ \(k={7\over8}\)
Now substituting the value of k in equation (i), we get
(x - 3y + 1) + \(7\over8\)(2x + 5y - 9) = 0
8x - 24y + 8 + 14x + 35y - 63 = 0
⇒ 22x + 11y - 55 = 0
⇒ 2x + y - 5 = 0

\(\therefore \frac { 5! }{ (5-r)! } =\frac { 6! }{ (7-r)! }\)
\( \Rightarrow \frac { 5! }{ (5-r)! } =\frac { 6\times 5! }{ (7-r)(6-r)(5-r)! }\)
\( \Rightarrow 1=\frac { 6 }{ (7-r)(6-r) } \)
\(\Rightarrow { r }^{ 2 }-13r+42=6\)
\(\Rightarrow { r }^{ 2 }-13r+36=0\)
\( \Rightarrow { r }^{ 2 }-{ 9r }-4r+36=0\)
\(\Rightarrow r(r-9)-4(r-9)=0\)
\(\Rightarrow (r-9)(r-4)=0\)
\( \Rightarrow r=9\quad or\quad r=4\)
\(Now\quad r=9\quad is\quad not\quad possible\quad because\quad r>n\)
\(Thus\quad r=4\)

Here \(\frac{1-ix}{1+ix}=a-ib\) By componendo and dividendo,
we have \(\frac{1-ix+1+ix}{1-ix-1-ix}=\frac{a-ib+1}{a-ib-1}\)
\(\Rightarrow \frac{2}{-2ix}=\frac{1+a-ib}{-(1-a+ib)}\)
\(\Rightarrow \frac{1}{ix}=\frac{1+a-ib}{1-a+ib}\)
\(\Rightarrow ix=\frac{1-a+ib}{1+a-ib}\times\frac{1+a+ib}{1+a+ib}\)
\(\Rightarrow ix=\frac{1-a^2-b^2+2ib}{(1+a)^2-i^2b^2}\)
\(\Rightarrow ix=\frac{1-a^2-b^2+2ib}{(1+a)^2+b^2}\)
\(=\frac{1-a^2-b^2}{(1+a)^2+b^2}+\frac{2b}{(1+a)^2+b^2}i\)
If a² + b² = 1 then
\(x=\frac{2b}{(1+a)^2+b^2}\) which is real.

\(\frac{N}{2}=\frac{100}{2}=50\)
∴ Median class is 35.5-40.5
∴ Median =\(35.5+\frac { 50-37 }{ 26 } \times 5=35.5+2.5=38\)
M.D. about median
\(=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-M \right| } =\frac { 1 }{ 100 } \times 735=7.35\)

Taking A, B, C events from question 6 above we have
\(i)\ A\cap B=\phi \)
Thus A and B are mutually exclusive and exhaustive events.
∴ True.
\( ii)\ A\cap B\ =\phi \ and\ A\cap B=S\)
Thus A and B are mutually exclusive and exhaustive events.
∴ True.
(iii) B′={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}=A.
∴ True.
(iv) A∩C = {(2,1)(2,2)(2,3)(4,1)}=ϕ
Thus A and C are not mutually exclusive events.
∴ False
(v) A∩B′ = A ≠ ϕ
Thus A and B are not mutually exclusive events.
(vi) Since A′ = B and B′ = A, A∩B = ϕ
B∩C = {(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)} ≠ ϕ
A∩C = {(2,1)(2,2)(2,3)(4,1)} = ϕ
Thus A′, B′ and C are not mutually exclusive.
∴ False.

The given inequality ix x + y \(\ge\) 4
Draw the gtaph of the line x + y = 4

Total of value satisfying the equation x +y = 4

Putting (0, 0) in the given inequation, we have 0 + 0 \(\ge\) 4 \(\Rightarrow\) 0 \(\ge\) 4 which is false
\(\therefore\) Half plane of x + y \(\ge\) 4 is away from origin
Also the given inequality is 2x - y > o.
Draw the graph of the line 2x - y = o.
Table of values satisfying the equation
2x -y =0

Putting (3, 0) in the given inequation, we have
2 x 3 - 0 > 0 => 6 > 0, which is true.
\(\therefore\) Half plane of 2x - y > 0 containing (3,0)

(-\(\infty\), -2) \(\cup\) (4,\(\infty\))