Knowledge in statistics provides us with the necessary tools and conceptual foundations in quantitative reasoning to extract information intelligently from this sea of data.

{tex}\sin 2x + \cos x = 0{/tex}

=>     {tex}\sin 2x = - \cos x{/tex}

=>     {tex}2\sin x.\cos x = - \cos x{/tex}

=>     {tex}\sin x = - {1 \over 2}{/tex}

=>     {tex}\sin x = \sin \left( {\pi + {\pi \over 6}} \right){/tex}

=>     {tex}\sin x = \sin {{7\pi } \over 6}{/tex}

=>     {tex}x = {{7\pi } \over 6} = {210^ \circ }{/tex}

Ans. Given :  {tex}sin X = {1\over \sqrt 5}{/tex}, {tex}sin Y = {1\over \sqrt {10}}{/tex}

X = {tex}sin^{-1}({1\over \sqrt 5}){/tex}

Y = {tex}sin^{-1}({1\over \sqrt {10}}){/tex}

=> X+Y = {tex}sin^{-1}({1\over \sqrt 5}) + sin^{-1}({1\over \sqrt {10}}){/tex}

{tex}[using \ \ sin^{-1} a + sin^{-1} b = sin^{-1}(a\sqrt{1-b^2}+b\sqrt{1-a^2})]{/tex}

{tex}=> X+Y = sin^{-1}\left ( {1\over \sqrt 5} \sqrt {1- {1\over 10}} + {1\over \sqrt {10}} \sqrt {1- {1\over 5}}\right ){/tex} 

{tex}=> X+Y = sin^{-1}\left ( {3\over \sqrt {50}} + {2\over \sqrt {50}} \right ){/tex}

{tex}=> X+Y = sin^{-1}\left ( {5\over 5\sqrt 2} \right ){/tex}

{tex}=> X+Y = sin^{-1}\left ( {1\over \sqrt 2} \right ){/tex}

{tex}=> X+Y = sin^{-1}\left ( sin {\pi \over 4} \right ){/tex}

{tex}=> X+Y = {\pi \over 4} {/tex}

Hence Proved

{tex}\cos ec{\rm{A}} + \sec {\rm{A}} = \cos ec{\rm{B}} + \sec {\rm{B}}{/tex}

=>     {tex}\cos ec{\rm{A}} - \cos ec{\rm{B}} = \sec {\rm{A}} - \sec {\rm{B}}{/tex}

=>     {tex}{1 \over {\sin {\rm{A}}}} - {1 \over {\sin {\rm{B}}}} = {1 \over {\cos {\rm{A}}}} - {1 \over {\cos {\rm{B}}}}{/tex}

=>     {tex}{{\sin {\rm{B}} - \sin {\rm{A}}} \over {\sin {\rm{A}}{\rm{.}}\sin {\rm{B}}}} = {{\cos {\rm{B}} - \cos {\rm{A}}} \over {\cos {\rm{A}}{\rm{.cos B}}}}{/tex}

=>     {tex}{{2\sin {{{\rm{B}} - {\rm{A}}} \over 2}\cos {{{\rm{B}} + {\rm{A}}} \over 2}} \over {\sin {\rm{A}}{\rm{.}}\sin {\rm{B}}}} = {{2\sin {{{\rm{B}} - {\rm{A}}} \over 2}\sin {{{\rm{B}} + {\rm{A}}} \over 2}} \over {\cos {\rm{A}}{\rm{.cos B}}}}{/tex}

=>     {tex}{{\cos {{{\rm{B}} + {\rm{A}}} \over 2}} \over {\sin {{{\rm{B}} + {\rm{A}}} \over 2}}} = {{\sin {\rm{A}}{\rm{.}}\sin {\rm{B}}} \over {\cos {\rm{A}}{\rm{.cos B}}}}{/tex}

=>     {tex}\cot {{{\rm{B}} + {\rm{A}}} \over 2} = \tan {\rm{A}}{\rm{.}}\tan {\rm{B}}{/tex}

=>     {tex}\tan {\rm{A}}{\rm{.}}\tan {\rm{B = }}\cot {{{\rm{B}} + {\rm{A}}} \over 2}{/tex}

L.H.S.

{tex}\left( {1 + \cot {\rm{A}} + \tan {\rm{A}}} \right)\left( {\sin {\rm{A}} - \cos {\rm{A}}} \right){/tex}

= {tex}\left( {1 + {{\cos {\rm{A}}} \over {\sin {\rm{A}}}} + {{\sin {\rm{A}}} \over {\cos {\rm{A}}}}} \right)\left( {\sin {\rm{A}} - \cos {\rm{A}}} \right){/tex}

= {tex}{{\left( {\sin {\rm{A}} - \cos {\rm{A}}} \right)\left( {\sin {\rm{AcosA}} + {{\sin }^2}{\rm{A}} + {{\cos }^2}{\rm{A}}} \right)} \over {\sin {\rm{AcosA}}}}{/tex}

= {tex}{{{{\sin }^3}{\rm{A}} - {{\cos }^3}{\rm{A}}} \over {\sin {\rm{AcosA}}}}{/tex}

= {tex}{{{{\sin }^2}{\rm{A}}} \over {{\rm{cosA}}}} - {{{{\cos }^2}{\rm{A}}} \over {\sin {\rm{A}}}}{/tex}

= {tex}{1 \over {{\rm{cosA}}}}.{{{{\sin }^2}{\rm{A}}} \over 1} - {1 \over {{\rm{sinA}}}}.{{{{\cos }^2}{\rm{A}}} \over 1}{/tex}

= {tex}{{\sec {\rm{A}}} \over {\cos e{c^2}{\rm{A}}}} - {{\cos ec{\rm{A}}} \over {{{\sec }^2}{\rm{A}}}}{/tex}

= R.H.S.

Let the Sum of n^th term of first AP be {tex}{{\rm{S}}_n}{/tex} and that of second term be S'_n.

=>     {tex}{{{n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]} \over {{n \over 2}\left[ {2a' + \left( {n - 1} \right)d'} \right]}} = {{5n + 4} \over {9n + 6}}{/tex}

=>     {tex}{{2a + \left( {n - 1} \right)d} \over {2a' + \left( {n - 1} \right)d'}} = {{5n + 4} \over {9n + 6}}{/tex}

Putting n = 35, we get

=>     {tex}{{2a + \left( {35 - 1} \right)d} \over {2a' + \left( {35 - 1} \right)d'}} = {{5 \times 35 + 4} \over {9 \times 35 + 6}}{/tex}

=>     {tex}{{2a + 34d} \over {2a' + 34d'}} = {{175 + 4} \over {315 + 6}}{/tex}

=>     {tex}{{a + 17d} \over {a' + 17d'}} = {{179} \over {321}}{/tex}

=>     {tex}{{a + \left( {18 - 1} \right)d} \over {a' + \left( {18 - 1} \right)d'}} = {{179} \over {321}}{/tex}

=>     {tex}{{{a_{18}}} \over {a{'_{18}}}} = {{179} \over {321}}{/tex}

Therefore, the ratio of 18th term of two APs is 179 : 321.

Radian measure = {tex}{\pi\over 180}\times degree\ measure{/tex}

So, 

Degree measure = {tex}{18\over \pi}\times radian \ measure{/tex}

=> Degree measure = {tex}{180\over 22}\times 7\times 4{/tex}

= 229.09 

{tex}{\cos ^2}2{\rm{A}} + {\cos ^2}2\left( {{\rm{A}} + {{120}^ \circ }} \right) + {\cos ^2}2\left( {{\rm{A}} - {{120}^ \circ }} \right) = {3 \over 2}{/tex}

We know that

{tex}{\cos ^2}{\rm{A}} = {{1 + \cos 2{\rm{A}}} \over 2}{/tex}                                    ..............(i)

{tex}{\cos ^2}2\left( {{\rm{A}} + {{120}^ \circ }} \right) = {{1 + \cos \left( {{\rm{2A}} + {{240}^ \circ }} \right)} \over 2}{/tex}         ..............(ii)

And   {tex}{\cos ^2}2\left( {{\rm{A}} - {{120}^ \circ }} \right) = {{1 + \cos \left( {{\rm{2A}} - {{240}^ \circ }} \right)} \over 2}{/tex}..............(iii)

Adding all these three equations,

{tex}{\cos ^2}2{\rm{A}} + {\cos ^2}2\left( {{\rm{A}} + {{120}^ \circ }} \right) + {\cos ^2}2\left( {{\rm{A}} - {{120}^ \circ }} \right){/tex} = {tex}{3 \over 2} + {{1 + \cos 2{\rm{A}}} \over 2} + {{1 + \cos \left( {{\rm{2A + }}{{240}^ \circ }} \right)} \over 2} + {{1 + \cos \left( {{\rm{2A}} - {{240}^ \circ }} \right)} \over 2}{/tex}

= {tex}{{3 + 1 + \cos 2{\rm{A + }}1 + \cos \left( {{\rm{2A + }}{{240}^ \circ }} \right) + 1 + \cos \left( {{\rm{2A}} - {{240}^ \circ }} \right)} \over 2}{/tex}

= {tex}{{3 + 3 + \cos 2{\rm{A}} + \cos \left( {{\rm{2A + }}{{240}^ \circ }} \right) + \cos \left( {{\rm{2A}} - {{240}^ \circ }} \right)} \over 2}{/tex}

= {tex}{{3 + 3 + \cos 2{\rm{A}} + 2\cos 2{\rm{A}}.\cos {{240}^ \circ }} \over 2}{/tex}

= {tex}{{3 + 3 + \cos 2{\rm{A}} - \cos 2{\rm{A}}} \over 2}{/tex}

= {tex}{6 \over 2}{/tex}

{tex}\Rightarrow{/tex} {tex}{\cos ^2}2{\rm{A}} + {\cos ^2}2\left( {{\rm{A}} + {{120}^ \circ }} \right) + {\cos ^2}2\left( {{\rm{A}} - {{120}^ \circ }} \right) = {3 \over 2}{/tex}

Q.2     {tex}\sin 2\theta = {{2\tan \theta } \over {1 + {{\tan }^2}\theta }}{/tex}

Since {tex}\theta {/tex} is in third quadrant, therefore value of tangent is positive.

{tex}\Rightarrow{/tex}     {tex}\sin 2\theta = {{2 \times {{24} \over 7}} \over {1 + {{\left( {{{24} \over 7}} \right)}^2}}}{/tex}

{tex}\Rightarrow{/tex}     {tex}\sin 2\theta = {{{{48} \over 7}} \over {1 + {{576} \over {49}}}}{/tex}

{tex}\Rightarrow{/tex}     {tex}\sin 2\theta = {{48} \over 7} \times {{49} \over {625}} = {{336} \over {625}}{/tex}

{tex}4{\cos ^2}\theta - 4\sin \theta = 1{/tex}

=>          {tex}4\left( {1 - {{\sin }^2}\theta } \right) - 4\sin \theta = 1{/tex}

=>          {tex}4 - 4{\sin ^2}\theta - 4\sin \theta = 1{/tex}

=>          {tex} - 4{\sin ^2}\theta - 4\sin \theta + 4 - 1 = 0{/tex}

=>          {tex}4{\sin ^2}\theta + 4\sin \theta - 3 = 0{/tex}

=>          {tex}4{\sin ^2}\theta + 6\sin \theta - 2\sin \theta - 3 = 0{/tex}

=>          {tex}2\sin \theta \left( {2\sin \theta + 3} \right) - 1\left( {2\sin \theta + 3} \right) = 0{/tex}

=>          {tex}\left( {2\sin \theta + 3} \right)\left( {2\sin \theta - 1} \right) = 0{/tex}

=>          {tex}2\sin \theta + 3 = 0{/tex}  or   {tex}2\sin \theta - 1 = 0{/tex}

=>          {tex}\sin \theta = {{ - 3} \over 2}{/tex}  or  {tex}\sin \theta = {1 \over 2}{/tex}

But {tex}\theta {/tex} is positive, therefore, taking

{tex}\sin \theta = {1 \over 2}{/tex}

=>        {tex}\sin \theta = \sin {30^ \circ }{/tex}

=>        {tex}\theta = {30^ \circ }{/tex}

Let {tex}-3i= 0-3i=r\left( cos\theta +isin\theta \right) {/tex}

Comparing real and imaginary parts

{tex}rcos\theta =0.......................\left( i \right) \\ rsin\theta =-3....................\left( ii \right) \\ {/tex}

Squaring and adding equations (i) and (ii)

{tex}{ r }^{ 2 }{ cos }^{ 2 }\theta +{ { r }^{ 2 } }sin^{ 2 }\theta =9\\ \Rightarrow { r }^{ 2 }=9\quad \quad \quad \quad \quad \quad \quad \left( { \because cos }^{ 2 }\theta +sin^{ 2 }\theta =1 \right) {/tex}

{tex}\Rightarrow r=3{/tex}

Substituting {tex}r=3{/tex} in equations (i) and (ii), we get 

{tex}cos\theta =0\quad and\quad sin\theta =-1\\ {/tex}

So {tex}\theta {/tex} lies in fourth quadrant 

{tex}\therefore \quad \theta =\frac { -\Pi }{ 2 } {/tex}     [ in the fourth quarant format of amplitude or principal argument is {tex}-\theta {/tex} (where {tex}\theta =\frac { \Pi }{ 2 } {/tex} ) ]

{tex}-3i=3\left( cos\left( \frac { -\Pi }{ 2 } \right) +isin\left( \frac { -\Pi }{ 2 } \right) \right) \\ \Rightarrow -3i=3\left( cos\left( \frac { \Pi }{ 2 } \right) -isin\left( \frac { \Pi }{ 2 } \right) \right) \quad \quad \left[ \because cos\left( -\theta \right) =cos\theta \quad ,sin\left( -\theta \right) =-sin\theta \right] {/tex}

Ans. {tex}f(x) = {1\over (x-4)(x-1)}{/tex}

 We need to find where the expression is undefined. These values are not part of the domain.

For the above function, if it is defined then

{tex}(x-4)(x-1) \ne 0 {/tex}

=> {tex}x-4\ne 0 \space and\space x-1\ne 0 {/tex}

{tex}x\ne 4 \space and\space x\ne 1 {/tex}

So Domain of f(x) = R - {1,4}

{tex}\left( {x - 4} \right)\left( {x - 1} \right) = {x^2} - 5x + 4{/tex} is a polynomial.

we know that Polynomial is defined for every real x.

hence Domain of {tex}\left( {x - 4} \right)\left( {x - 1} \right){/tex}=R

1) B={2,3,5,7,11,...} since prime or positive factors of prime numbers are itself.

2)C={0,1} since C={x: x²=x where x belongs to R}

                               ={x: x²-x=0 where x belongs to R}

                               ={x: x(x-1)=0 where x belongs to R}

                               ={x: x=0 and x-1=0 where x belongs to R}

                               ={x: x=0 and x=1where x belongs to R}

                               ={0,1}

Prime numbers have only two factors i.e. 1 and itself

{tex}B = \left\{ {1,x} \right\}{/tex}

As we know

 {tex}\eqalign{ & {\left( 0 \right)^2} = 0 \cr & {\left( 1 \right)^2} = 1 \cr} {/tex}

{tex}C = \left\{ {0,1} \right\}{/tex}

X÷1+x*

2x^

Ans. Given: f(x + y) = f(x)*f (y) for all x, y {tex}\in {/tex} N ......(1)

{tex}\displaystyle\sum_{i=1}^{n} f(x) = 120 \space \space ............(2){/tex}

f(1) = 3

Taking x = y = 1 in (1), we obtain

f(1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

Similarly, f(3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27

f(4) = f(1 + 3) = f (1) f (3) = 3 × 27 = 81

f(1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P.

First term is 3 and common ratio is also 3. 

Also we know 

{tex}=> S_n = {a(r^n -1)\over r - 1}{/tex}       

{tex}=> S_n = {3(3^n -1)\over 3 - 1}{/tex}  .................... (3)

From (2) and (3), We get 

{tex}=> 120 = {3(3^n -1)\over 3 - 1}{/tex}

{tex}=> 120 = {3(3^n -1)\over 2}{/tex}

{tex}=> 80 = 3^n -1{/tex}

{tex}=> 81 = 3^n {/tex}

{tex}=> 3^4 = 3^n {/tex}

n = 4,

So Value of n = 4

Four cards can be selected from 52 cards in ⁵²C₄ ways. 
Now, there are four suits, e.g. club, spade, heart and diamond each of 13 cards. 
So total number of ways of getting all the four cards of the same suit:
⇒ ¹³C₄+¹³C₄+¹³C₄+¹³C₄=4×¹³C₄

{tex}\therefore{/tex} the probability of getting 4 cards from the same suit={tex}\frac{{4{ \times ^{13}}{C_4}}}{{^{52}{C_4}}}{/tex}

= {tex}\frac{{198}}{{20825}}{/tex}

Ans. 

Let {tex}\sqrt {-7-24i} = x+yi{/tex}

Squaring Both Sides,

=> -7-24i = x² - y² + 2xyi

on comparing, we get

=> x² - y² = -7   ……(1)

and 2xy = -24 ………(2)

We know

(x²+y²)² = (x²-y²)² + 4x²y²

=> (x²+y²)² = 49 + 576

=> (x²+y²)² = 625

=> x²+ y² = 25  …………(3)

solving (2) and (3), We get

{tex}x = \pm 3 \space \space \space and \space y =\pm4{/tex}

from (2) as product of xy is -ve, so it means x and y are of opposite sign.

So, when x = 3, y = -4

when x = -3, y = 4

Therefore, 

{tex}\sqrt {-7-24i} = \pm (3-4i){/tex}

{tex}{\left( {\sqrt {{x \over 3}} + {3 \over {2{x^2}}}} \right)^{10}}{/tex}

{tex}{T_{r + 1}}{ = ^{10}}{C_r}{\left( {\sqrt {{x \over 3}} } \right)^{10 - r}}{\left( {{3 \over {2{x^2}}}} \right)^r}{/tex}

{tex}{ = ^{10}}{C_r}{\left( x \right)^{5 - {r \over 2} - 2r}}{\left( 3 \right)^{ - 5 + {r \over 2} + r}}{\left( 2 \right)^{ - r}}{/tex}

{tex}{ = ^{10}}{C_r}{\left( x \right)^{{{10 - 5r} \over 2}}}{\left( 3 \right)^{{{3r - 10} \over 2}}}{\left( 2 \right)^{ - r}}{/tex}

{tex}for\,independent\,of\,x{/tex}

{tex}{\left( x \right)^{{{10 - 5r} \over 2}}} = {x^0}{/tex}

{tex}{{10 - 5r} \over 2} = 0{/tex}

{tex}10 - 5r = 0{/tex}

{tex}r = 2{/tex}

{tex}independent\,term\,{T_3}{ = ^{10}}{C_2}{\left( 3 \right)^{ - 2}}{\left( 2 \right)^{ - 2}}{/tex}

{tex} = {{10!} \over {2!8!}} \times {1 \over {36}}{/tex}

{tex} = {{10 \times 9} \over {2 \times 36}}{/tex}

{tex} = {5 \over 4}{/tex}

{tex}let\,y = \sin 2x{/tex}

{tex}differentiating\,with\,respect\,to\,x{/tex}

{tex}{{dy} \over {dx}} = {d \over {dx}}\left( {\sin 2x} \right){/tex}

{tex} = \cos 2x{d \over {dx}}\left( {2x} \right){/tex}

{tex} = 2\cos 2x{/tex}