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The sums of n terms of two A.P.'s are in the ratio 5n + 4 : 9n + 6 . Find the ratio of their 18th terms

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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.
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If f(x)=(ax^{n})^{1/n},a Is greater than 0 and n belongs to N then prove that f(f(x))=x for all x.
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If f is a real function defined by f(x)=x1/x+1,then prove that : f(2x)=3f(x)+1/f(x)+3

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Convert 4 Radian into degree measure

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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
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Prove that cos^2A+cos^2(A+120)+cos^2(A120)=3/2

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{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}
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Q.1 If θ is the positive acute angle , then solve the equation
4 cos2 θ  4 sinθ = 1 .
Q.2 Find the value of sin 2θ , when tan θ = 247 and θ lies in 3rd quadrant.
Q.3 If sinθ = 1213 , then find the quadrants in which θ can lie. Also , find the values of the remaining trigonometric ratios of θ .
Q.4 A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88m, when it has traced out 720 at the centre. Then, find the length of rope.

It is 24 /7 in question no. 2

It is 12/13 in question no. 3
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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}
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If theta is positive acute angle then solve the equation:
4cos^2 theta  4 sin theta =1

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{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}
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convert 3i into polar form.

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Let {tex}3i= 03i=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}
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now days 11th class mathatics follow modern abc book so ur solution ncert not sufficient so plz update with modern book solution

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If a,b,c € R, then find the relation between a,b&c such that (bx)²4(ax)(cx)=0 has real & equal roots.

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Find domain of1/(x4)(x1)

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Ans. {tex}f(x) = {1\over (x4)(x1)}{/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}(x4)(x1) \ne 0 {/tex}
=> {tex}x4\ne 0 \space and\space x1\ne 0 {/tex}
{tex}x\ne 4 \space and\space x\ne 1 {/tex}
So Domain of f(x) = R  {1,4}
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Find domain of(x4)(x1)?

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{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
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evaluate {tex}evaluate sigma superscript 13 subscript n=1 i^n + i^n+1) where n belongs to N{/tex}

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Write the following sets in roster form
1) B = {x : x is a positive factor of a prime number}
2) C={x : x2=x where x belongs to Real numbers}
Answer it quickl
Fast.pls its very urgent..

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Write the following sets in roster form
1) B = {x : x is a positive factor of a prime number}
2) C={x : x^{2}=x where x belongs to Real numbers}
Answer it quickly

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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}
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If AunionB=AintersectionB then proved A=B

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if f is a function satisfying f(x+y)=f(x) f(y) for all x,y belongs to n such that f(1)=3 and sigma x =1 to n f(x)=120 then find n.

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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
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please what is the meaning of general and pricipal solution in trignometry.

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4 cards are drawn at a time from a pack of 52 cards find the probability of getting all the 4 cards of the same suit.

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How to find the ratio in which the line segment forming the points (4,8,10) and (6,10, 8) is divided by yz plane

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How to find the square root of 724i ?

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Ans.
Let {tex}\sqrt {724i} = x+yi{/tex}
Squaring Both Sides,
=> 724i = x^{2}  y^{2 }+ 2xyi
on comparing, we get
=> x^{2}  y^{2 }= 7 ……(1)
and 2xy = 24 ………(2)
We know
(x^{2}+y^{2})^{2 }= (x^{2}y^{2})^{2} + 4x^{2}y^{2}
=> (x^{2}+y^{2})^{2} = 49 + 576
=> (x^{2}+y^{2})^{2 }= 625
=> x^{2}+ y^{2} = 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 {724i} = \pm (34i){/tex}
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Find the term independent of x in the expansion (x/√3)+(√3/2x²)^10

your question is worng when i am solved then i find r=10/3 which is not natural
corrwct question is {tex}{\left( {\sqrt {{x \over 3}} + {3 \over {2{x^2}}}} \right)^{10}}{/tex}

Please answer this question I need it urgently
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Answers:

{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}
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find the derivative of sin^{2x}

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{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}
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Find derivative of function f(x)=sinx/sinxcosx with respect to x.

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Ans. {tex}f(x) = {sinx\over sinxcosx}{/tex}
Using quotient Rule, we get
{tex}f'(x) ={ {(sinxcosx){d(sinx)\over dx} sinx{d(sinxcosx)\over dx}}\over (sinxcosx)^2}{/tex}
{tex}=> { (sinxcosx)cosx  sinx(cosx(sinx)) \over (sinxcosx)^2}{/tex}
{tex}=> {sinx cosx  cos^2x sinxcosx sin^2x\over (sinxcosx)^2}{/tex}
{tex}=> { (cos^2x +sin^2x)\over (sinxcosx)^2}{/tex}
{tex}=> {1\over (sinxcosx)^2}{/tex}
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I have a doubt in chapter 9 ie, sequences and series.
Question. insert 3 nos. b/w 1 & 256 so that the resulting sequence is a GP

anybody to answer this ques.
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{tex}let\,three\,numbers\,b/w\,1\,and\,256\,be\,{G_1},\,{G_2},\,and\,{G_3}{/tex}
{tex}1,{G_1},\,{G_2},\,{G_3},256\,are\,in\,GP{/tex}
{tex}a = 1{/tex}
{tex}l = 256{/tex}
{tex}a{r^4} = 256{/tex}
{tex}{r^4} = 256{/tex}
{tex}r = 4{/tex}
{tex}Now\,{G_1} = ar = 1 \times 4 = 4{/tex}
{tex}{G_2} = a{r^2} = 1 \times {\left( 4 \right)^2} = 16{/tex}
{tex}{G_3} = a{r^3} = 1 \times {\left( 4 \right)^3} = 64{/tex}
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find limits of lim_{x}→infinity (root a+x  root 9/x)

please update your question. please wirte question correctly
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let f:rr be a function given by f(x)=x^{2}+2.find f^{1}(27).

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{tex}let\,{f^{  1}}\left( {27} \right) = x{/tex}
{tex} \Rightarrow f\left( x \right) = 27{/tex}
{tex} \Rightarrow {x^2} + 2 = 27{/tex}
{tex} \Rightarrow {x^2} = 25{/tex}
{tex} \Rightarrow x = \pm 5{/tex}
{tex}Hence{/tex}
{tex}{f^{  1}}\left( {27} \right) = \left\{ {  5,5} \right\}{/tex}
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find two no. such that their arithematic mean is 5 and G.M is 9 without using the identity (a+b)^{2}=(ab)^{2}+4ab

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There are no two numbers such that their arithmatic mean is 5 and G.M. is 9
Because we have know that
AM>GM for between any two numbers a and b.
here AM=5 and GM=9
9>5 or GM>AM which is invalid
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