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Tan(xo+45o)

Posted by Ashok Kumar S (Jul 21, 2017 7:03 a.m.) (Question ID: 6808)

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  • Using tan (a+b) = {tex}tan \ a +tan \ b\over 1- tan\ a.tan\ b{/tex}

    we get

    {tex}tan (x+45) = { tan x + tan 45\over 1-tanx\times tan 45}{/tex}

    {tex}{tan\ x + 1 \over 1-tan\ x}\\ [ as \ tan \ 45 = 1]{/tex}

    {tex}1+tan \ x\over 1- tan \ x{/tex}

     

    Answered by Sahdev Sharma (Jul 21, 2017 7:32 a.m.)
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1.defination of function?

Posted by Rohit Sharma (Jul 20, 2017 4:52 a.m.) (Question ID: 6785)

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  • A function is a special relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.

    Answered by Sahdev Sharma (Jul 20, 2017 8:40 a.m.)
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What is trace matrix?

 

Posted by Ashok Kumar S (Jul 19, 2017 3:38 p.m.) (Question ID: 6755)

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How to solve a approximate value of sin 59°

Posted by Harshit Mishra (Jul 18, 2017 6:58 p.m.) (Question ID: 6739)

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Y=Sin-1(2x+1/1+4x)

 

Posted by Ashok Kumar S (Jul 18, 2017 1:41 p.m.) (Question ID: 6733)

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Cosy = xcos(a+y) show that Dy/dx = cos2(a+y)/Sina

Posted by Ashok Kumar S (Jul 20, 2017 12:50 p.m.) (Question ID: 6702)

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  • Given cos y =x cos(a+y)-------(1)

    Differentiating both sides w.r.t. x we get

    -siny{tex}{dy\over dx}{/tex}= x {tex} { d (cos(a+y)) \over dx}{/tex}+1× cos(a+y)     [ by using product rule and chain rule]

    -siny {tex}{dy\over dx}{/tex}=x (-sin(a+y) {tex}{d(a+y) \over dx}{/tex}) +cos(a+y)

    -siny {tex}{dy \over dx}{/tex}= x(-sin(a+y)(0+{tex}{dy\over dx}{/tex}))+cos(a+y)

    -siny {tex}{dy\over dx}{/tex}=-x sin(a+y) {tex}{dy \over dx}{/tex}+cos(a+y)

    -siny {tex}{dy \over dx}{/tex}+x sin(a+y) {tex}{dy \over dx}{/tex}=cos(a+y)

    (-siny +x sin(a+y)) {tex}{dy \over dx}{/tex}= cos(a+y)

    (-siny + {tex}{Cos y \over cos(a+y)}{/tex}sin(a+y)){tex}{dy\over dx}{/tex}=cos(a+y) [ by eq.(1) substitute the value of x]                         

    {tex}{-siny ×cos(a+y)+cosy ×sin(a+y)\over cos(a+y)}{/tex}){tex}{dy\over dx}{/tex}=cos(a+y)

    ({tex}{sin(a+y)×cos y -cos(a+y)×siny \over cos(a+y)}{/tex}){tex}{dy \over dx}{/tex}=cos(a+y)

    ({tex}{sin(a+y-y) \over cos(a+y)}{/tex}{tex}{dy\over dx}{/tex}=cos(a+y)  [ by using identity sin(A-B)=sinAcosB-cosAsinB]

    ({tex}{sin a\over cos(a+y)}{/tex}){tex}{dy\over dx}{/tex}=cos(a+y)

    {tex}{dy \over dx}{/tex}={tex}{cos^2(a+y) \over sin a}{/tex}                Hence Proved.

    Answered by Dharmendra Kumar (Jul 20, 2017 12:50 p.m.)
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If y= 1/√(b​​​​​​2​​​​​- c​​​​​​2) log [√(b+a)+√(b-a)tan(x/2)/√(b+a)-√(b-a)tan(x/2)]

Prove that dy/dx=1/(a+b cos x)

Please solve this problem 

Posted by Harshit Mishra (Jul 16, 2017 11:03 a.m.) (Question ID: 6680)

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If     f(x)=[(c+x)/(d+x)]c+d+2x

Show that f'(0)=[c+d]^c+d  [2 log c/d+(d^2-c^2)/cd]

please solve this problem

 

Posted by Harshit Mishra (Jul 16, 2017 10:49 a.m.) (Question ID: 6678)

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I =cosec^3 x dx by integration by parts 

 

Posted by Riya Jain (Jul 15, 2017 9:08 p.m.) (Question ID: 6667)

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Cos-1 (cos (680)

Posted by Deepanshu Saini (Jul 14, 2017 7:02 a.m.) (Question ID: 6628)

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  • {tex}cos^{-1}[cos(680^°)]{/tex}

    {tex}= cos^{-1}[cos(4\pi -40^°)]{/tex}

    {tex}= cos^{-1}[cos(40^°)]{/tex}

    ={tex}= 40^°{/tex}

    Answered by Sahdev Sharma (Jul 14, 2017 7:22 a.m.)
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  • {tex}cos^{-1}[cos(680^°)]{/tex}

    {tex}cos^{-1}[cos(4\pi -40^°)]{/tex}

    {tex}= cos^{-1}[cos(40^°)]{/tex}

    {tex}40^°{/tex}

    Answered by Payal Singh (Jul 14, 2017 7:21 a.m.)
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Sir I failed in 12th cbse in 2017 with subjects English,Physics,Chemistry,Biology and Math.Now I want to reappear in 2018.Can I change Math with any other subject as I want to leave math.Kindly give reply immediately. Thankyou

Posted by Syed Masroor Husain (Jul 06, 2017 7:05 p.m.) (Question ID: 6429)

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  • No.

    Answered by Dr Pathikrt Banerjee (Jul 07, 2017 2:20 p.m.)
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Intrigate √tanx

Posted by Priyanshu Mehta (Jul 06, 2017 5:36 p.m.) (Question ID: 6426)

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  • {tex}\int tan x. dx{/tex}

    {tex}\int{ sin x\over cos x} dx{/tex}

    {tex}take \ cos\ x = t{/tex}

    {tex}so\ sin \ x\ dx = -dt {/tex}

    {tex}\int-{1\over t} dt{/tex}

    {tex}= - log\ t + c {/tex}

    {tex}= - log (cos \ x)+ c {/tex}

    {tex}log |sec\ x| + c{/tex}

    Answered by Payal Singh (Jul 07, 2017 5:54 a.m.)
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  • {tex}\int tan x. dx{/tex}

    {tex}\int{ sin x\over cos x} dx{/tex}

    Put cos x = t

    Then

    sin x dx = -dt

    {tex}-\int{1\over t} dt{/tex}

    = - log t + c

    = - log (cos x)+ c

    {tex}log |sec\ x| + c{/tex}

    Answered by Sahdev Sharma (Jul 07, 2017 5:48 a.m.)
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(Sin-1x)2+(cos-1x)2=5π2/8

 

Posted by Nachiket Sudewad (Jul 06, 2017 3:49 p.m.) (Question ID: 6423)

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differentiate w.r.t x:

xsin¯¹x÷ (1-x²)

 

​​

Posted by Yatin Jindal (Jul 05, 2017 8:23 p.m.) (Question ID: 6399)

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  • Let {tex}y = {{x{{\sin }^{ - 1}}x} \over {1 - {x^2}}}{/tex}

    =>     {tex}{{dy} \over {dx}} = {{\left( {1 - {x^2}} \right){d \over {dx}}\left( {x{{\sin }^{ - 1}}x} \right) - x{{\sin }^{ - 1}}x{d \over {dx}}\left( {1 - {x^2}} \right)} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}

    =>     {tex}{{dy} \over {dx}} = {{\left( {1 - {x^2}} \right)\left[ {x.{1 \over {\sqrt {1 - {x^2}} }} + {{\sin }^{ - 1}}x.1} \right] + 2{x^2}{{\sin }^{ - 1}}x} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}

    =>     {tex}{{dy} \over {dx}} = {{x\sqrt {1 - {x^2}} + \left( {1 - {x^2}} \right){{\sin }^{ - 1}}x + 2{x^2}{{\sin }^{ - 1}}x} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}

    =>     {tex}{{dy} \over {dx}} = {{x\sqrt {1 - {x^2}} + {{\sin }^{ - 1}}x - {x^2}{{\sin }^{ - 1}}x + 2{x^2}{{\sin }^{ - 1}}x} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}

    =>     {tex}{{dy} \over {dx}} = {{x\sqrt {1 - {x^2}} + {{\sin }^{ - 1}}x + {x^2}{{\sin }^{ - 1}}x} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}

    =>     {tex}{{dy} \over {dx}} = {{x\sqrt {1 - {x^2}} + {{\sin }^{ - 1}}x\left( {1 + {x^2}} \right)} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}

     

    Answered by Rashmi Bajpayee (Jul 08, 2017 11:59 a.m.)
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Intrigate  √tanx

Posted by Priyanshu Mehta (Jul 05, 2017 4:07 p.m.) (Question ID: 6395)

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  • {tex}\int tan x\ dx{/tex}

    {tex}= \int {sin x\over cos x} dx{/tex}

    Put cos x = t

    Then, -sinx dx = dt

    And sin x dx = -dt

    {tex}=- \int {1\over t} dt{/tex}

    = -log t + c

    = -log (cos x) + c

    = log |sec x| + c

    Answered by Sahdev Sharma (Jul 05, 2017 10:06 p.m.)
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Primitive of sin^-1x

Posted by Priyanshu Mehta (Jul 02, 2017 11:43 a.m.) (Question ID: 6308)

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  • {tex}\int sin^{-1}x \ dx{/tex}

    {tex}\int sin^{-1}x.1 \ dx{/tex}

    Now Integrate by parts, Using ILATE first Function is sin​​​​-1 x and 2nd Function is 1.

    {tex}sin^{-1}x\int 1\ dx- \int {1\over \sqrt {1-x^2}}xdx{/tex}

    {tex}= sin^{-1}x.x- I_1{/tex} .....(1)

    Where {tex}I_1 = \int {1\over \sqrt {1-x^2}}xdx {/tex}

    Put {tex}{1-x^2} = t{/tex}

    -2xdx = dt

    {tex}=> I_1 = -{1\over 2}\int {1\over \sqrt t}dt{/tex}

    {tex}=> I_1 = - {\sqrt t}{/tex}

    {tex}=> I_1 = -{\sqrt {1-x^2}}{/tex}

    Put in (1), we get

    {tex}sin^{-1}x.x+\sqrt{1-x^2} + c{/tex}

     

    Answered by Payal Singh (Jul 02, 2017 4:22 p.m.)
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Using differentials, find the approximate value of sin (22/14).

Posted by Somnath Roy (Jul 01, 2017 9:51 p.m.) (Question ID: 6301)

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Sin-1x + sin-1(1-x) = cot-1

Find x.

Pls answer fast

Posted by Akash Panwar (Jun 29, 2017 5:24 p.m.) (Question ID: 6254)

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  • {tex}\eqalign{ & {\sin ^{ - 1}}x + {\sin ^{ - 1}}(1 - x) = {\cot ^{ - 1}}x \cr & {\sin ^{ - 1}}(x + 1 - x) = {\cot ^{ - 1}}x \cr & {\sin ^{ - 1}}(1) = {\cot ^{ - 1}}x \cr & {90^ \circ } = {\cot ^{ - 1}}x \cr & x = \cot {90^ \circ } \cr & x = 0 \cr} {/tex}

    Answered by Arun Soni (Jul 01, 2017 3:27 p.m.)
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Is there is any period of function f(x) = x ..if yes then what is it ? And if not why ???? ..plz tell me

 

Posted by Mintu Rawat (Jun 29, 2017 2:16 p.m.) (Question ID: 6249)

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  •  A periodic function is a function that repeats its values in regular intervals or periods

    For a periodic function {tex}f\left( x+p \right) =f\left( x \right) ,\quad p{/tex} is the period, so that  for a periodic function its value always oscillates between a fixed range

    Now consider the function {tex}f\left( x \right) =x{/tex}

    We have for every unique value of x we always have a unique value of y, which means it cannot repeats itself after every fixed interval .

    Hence {tex}f\left( x \right) =x{/tex} is not a periodic function.

     

     

    Answered by Rasmi Rv (Jul 03, 2017 2:59 p.m.)
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chapter 9 "continuity",
R.D.SHARMA
sums no. 26,first exercise

 

Posted by Abhay Paul (Jun 29, 2017 12:09 a.m.) (Question ID: 6240)

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sin-1(x)+cos-1(1/2)=pi/2 find x

Posted by Sakthivel R (Jun 27, 2017 7:24 p.m.) (Question ID: 6214)

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  • We have {tex}\sin ^{ -1 }{ x } +\cos ^{ -1 }{ x } =\frac { \Pi }{ 2 } ,\quad x\epsilon \left[ -1,1 \right] {/tex}

    Using this result if {tex}\sin ^{ -1 }{ x } +\cos ^{ -1 }{ 1/2 } =\frac { \Pi }{ 2 } {/tex} we can say x= {tex}\frac { 1 }{ 2 } {/tex}

    Answered by Rasmi Rv (Jul 03, 2017 2:27 p.m.)
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Find dy/dx::

sin xy + cos xy =1

Posted by Rohit Kumar (Jun 26, 2017 9:16 p.m.) (Question ID: 6194)

  • Good question

    Posted by Onkar Yadav (Jun 26, 2017 11:55 p.m.)
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  • sin xy + cos xy = 1

    Differentiate with respect to x, we get

    cos xy ( y + x{tex}dy\over dx{/tex}) - sin xy (y+x{tex}dy\over dx{/tex}) = 0

    => cos xy ( y + x{tex}dy\over dx{/tex}) = sin xy (y+x{tex}dy\over dx{/tex})

    => tan xy = 1

    Differentiate with respect to x,

    => sec​​​​2 xy (y+ x{tex}dy\over dx{/tex}) = 0

    => y+ x {tex}dy\over dx{/tex}= 0

    =>  {tex}dy\over dx{/tex} = {tex}-y\over x{/tex}

    Answered by Payal Singh (Jun 27, 2017 6:15 a.m.)
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differentiate xx+x1/x

Posted by Santhosh Pk (Jun 26, 2017 7:54 p.m.) (Question ID: 6190)

  • Good question

    Posted by Onkar Yadav (Jun 26, 2017 11:55 p.m.)
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  • Ans. 

    y = {tex}x^x + x^{1\over x}{/tex}

    Let {tex}y = u + v {/tex}

    where {tex}u = x^x \ \ \ and \ \ v = x^{1\over x}{/tex}

    then 

    {tex}{dy\over dx} = {du\over dx} + {dv\over dx}{/tex}  ............... (1)

    Now, 
    {tex}u = x^x {/tex}
    Taking log both sides, 

    {tex}log \ u = log\ x^x{/tex}

    {tex}=> log \ u = xlog\ x{/tex}

    Differentiate w.r.to x, we get 

    {tex}=> {1\over u}{du\over dx} = \left (x \times {1\over x} \right ) + log \ x{/tex}

    {tex}=> {1\over u} {du\over dx}= 1 + log \ x{/tex}

    {tex}=> {du\over dx}= x^x(1 + log \ x){/tex} ........... (2)

    Now, {tex}v= x^{1\over x}{/tex}

    Taking log both sides, 

    {tex}=> log \ v = log \ x^{1\over x}{/tex}
    {tex}=> log \ v = {1\over x}log \ x{/tex}

    Differentiate w.r.to x 

    {tex}=> {1\over v} {dv\over dx} = \left ( {1\over x}\times {1\over x} \right ) + log \ x\times {-1\over x^2}{/tex}

    {tex} => {dv\over dx} = x^{1\over x} {1\over x^2}\left [1 - log \ x \right ]{/tex}   ........... (3)

    From (2) and (3) we get 

    {tex}=> {dy\over dx}= x^x(1 + log \ x) + x^{1\over x} {1\over x^2}\left [1 - log \ x \right ]{/tex}

     

    Answered by Naveen Sharma (Jun 30, 2017 12:14 p.m.)
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Tan(π/4+1/2cos-¹a/b)+tan(π/4-1/2cos-¹a/b)=2b/a

Prove that

Posted by Amit Pandey (Jun 30, 2017 12:42 p.m.) (Question ID: 6080)

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  • L.H.S. {tex}\tan \left[ {{\pi \over 4} + {1 \over 2}{{\cos }^{ - 1}}{a \over b}} \right] + \tan \left[ {{\pi \over 4} - {1 \over 2}{{\cos }^{ - 1}}{a \over b}} \right]{/tex}

    Let {tex}{\cos ^{ - 1}}{a \over b} = \theta ,{/tex} then {tex}\cos \theta = {a \over b}{/tex} and {tex}{1 \over 2}{\cos ^{ - 1}}{a \over b} = {\theta \over 2}{/tex}

    = {tex}\tan \left[ {{\pi \over 4} + {\theta \over 2}} \right] + \tan \left[ {{\pi \over 4} - {\theta \over 2}} \right]{/tex}

    = {tex}{{\tan {\pi \over 4} + \tan {\theta \over 2}} \over {1 - \tan {\pi \over 4}\tan {\theta \over 2}}} + {{\tan {\pi \over 4} - tan{\theta \over 2}} \over {1 + \tan {\pi \over 4}\tan {\theta \over 2}}}{/tex}

    = {tex}{{1 + \tan {\theta \over 2}} \over {1 - \tan {\theta \over 2}}} + {{1 - tan{\theta \over 2}} \over {1 + \tan {\theta \over 2}}}{/tex}

    = {tex}{{{{\left( {1 + \tan {\theta \over 2}} \right)}^2} + {{\left( {1 - \tan {\theta \over 2}} \right)}^2}} \over {1 - {{\tan }^2}{\theta \over 2}}}{/tex}

    = {tex}{{2\left( {1 + {{\tan }^2}{\theta \over 2}} \right)} \over {1 - {{\tan }^2}{\theta \over 2}}}{/tex}

    = {tex}2\left( {{1 \over {\cos \theta }}} \right){/tex}

    = {tex}{{2b} \over a}{/tex}

    = R.H.S.

    Answered by Rashmi Bajpayee (Jun 30, 2017 12:42 p.m.)
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When to use product rule in differentiation and when to use chain rule??

Posted by Siddhant Gangly (Jun 14, 2017 8:39 p.m.) (Question ID: 5909)

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  • When two or more functions given in product, we have to use product rule.

    Example : xsinx, sinxcosx, logxsinx

    When one function comes with in the bracket of another function ( called function of function), we have use chain rule.

    Example: (i)sin(cosx) here sin and cos not in product form either cosx is with in the fiction of sin.

                    (ii) log(sinx) here sin within the function of log.

    Answered by Justin Santhiagu (Jun 15, 2017 6:41 a.m.)
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Tell the values of sin(π) and cos(π).

Posted by Priyam Chouhan (Jun 11, 2017 7:49 p.m.) (Question ID: 5860)

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  • Sin {tex}\pi = 0{/tex}

    Cos {tex}\pi = -1{/tex}

    Answered by Payal Singh (Jun 11, 2017 10:43 p.m.)
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the angles of a triangle are in a.p. if the greatest angle is twice the smallest angle, find the angles using matrics method?

 

Posted by Sruthi K (Jun 11, 2017 12:53 p.m.) (Question ID: 5852)

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  • Given the angles of the triangle are in A.P

    So let them be a-d,a,a+d.

     We have sum of the three angles of a triangle is {tex}{ 180 }^{ \circ }{/tex}

    {tex}\Rightarrow \left( a-d \right) +a+\left( a+d \right) ={ 180 }^{ \circ }\\ \Rightarrow 3a={ 180 }^{ \circ }\\ \Rightarrow a=\frac { { 180 }^{ \circ } }{ 3 } ={ 60 }^{ \circ }{/tex} 

    According to the question the greatest angle is twice the smallest angle

    {tex}\Rightarrow a+d=2\left( a-d \right) \\ \Rightarrow a+d=2a-2d\\ \Rightarrow 3d=a\\ \Rightarrow d=\frac { a }{ 3 } =\frac { 60 }{ 3 } ={ 20 }^{ \circ }{/tex}

    {tex}\therefore a-d={ 60 }^{ \circ }-{ 20 }^{ \circ }={ 40 }^{ \circ },a={ 60 }^{ \circ },a+d={ 60 }^{ \circ }+{ 20 }^{ \circ }={ 80 }^{ \circ }{/tex}

    So the angles of the triangle which are in A.P are {tex}{ 40 }^{ \circ },{ 60 }^{ \circ }{/tex}and {tex}{ 80 }^{ \circ }{/tex}

     

    Answered by Rasmi Rv (Jun 11, 2017 1:26 p.m.)
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NCERT ch.-2 example-7, in easy

 

Posted by Tanul Rustagi (Jun 08, 2017 7:23 a.m.) (Question ID: 5743)

  • please write the question instead of the question number

    Posted by Rasmi Rv (Jun 09, 2017 12:32 p.m.)
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Integrate (2+3x)/(3-2x)  kindly show the answer step by step. 

Posted by Srinivas Chandra (Jun 05, 2017 11:57 p.m.) (Question ID: 5706)

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  • Ans. {tex}\int{ 2+3x\over 3-2x}dx{/tex}

    As this is not a proper fraction. then first we have to conert it in proper fraction. 

    {tex}=> \ \int ({ -{3\over 2} + {13\over 2}{1\over (-2x+3)} }) dx {/tex}

    {tex}=> \ \int { -{3\over 2} }dx + \int {{13\over 2}{1\over (-2x+3)}} dx {/tex}

    {tex}=> \ -{3\over 2} \int dx + {13\over 2 }\int {1\over -2x+3} dx {/tex} 
    Multiply and divide integral by -2, we get 

    {tex}=> \ -{3\over 2} \int dx - {13\over 4 }\int {-2\over -2x+3} dx {/tex}

    {tex}=> \ -{3\over 2} x - {13\over 4 }I_2 \ \ \ ........... (1){/tex}

    {tex}I_2 = \int {-2\over -2x+3} dx {/tex}

    Put -2x + 3 = t then 

    -2 dx = dt 

    So,

    {tex}I_2 = \int {1\over t}dt{/tex}

    => I2 = log t = log (-2x+3)

    Put in (1). we get 

    {tex}=> \ -{3\over 2} x - {13\over 4 }log (-2x+3){/tex}

    Answered by Naveen Sharma (Jun 06, 2017 9:20 a.m.)
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Integrate 5/root(x+3)-root(x-2) 

Kindly provide the answer for the above question

Posted by Srinivas Chandra (Jun 05, 2017 11:23 p.m.) (Question ID: 5705)

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  • Ans. 

    {tex}\int {5\over {\sqrt {x+3} - \sqrt {x-2}}} dx{/tex} Multiply and divide by {tex}{\sqrt {x+3} + \sqrt {x-2}}{/tex}

    => {tex}\int {5\over {\sqrt {x+3}\ - \sqrt {x-2}} } \times {{\sqrt {x+3} \ + \sqrt {x-2}}\over {\sqrt {x+3} \ + \sqrt {x-2}}} dx{/tex}

    => {tex}\int{ {5( \sqrt {x+3} \ + \sqrt {x-2} )} \over x+3 - x + 2} dx{/tex}

    => {tex}\int{ {5( \sqrt {x+3} \ + \sqrt {x-2} )} \over 5} dx{/tex}

    => {tex}\int{ ({ \sqrt {x+3} \ + \sqrt {x-2} } )dx}{/tex}

    => {tex}\int{ { \sqrt {x+3} }\ dx + \int {\sqrt {x-2} } \ dx}{/tex}

    => {tex}{2\over 3}(x+3)^{3\over 2} + {2\over 3}(x-2)^{3\over 2}{/tex}

    Answered by Naveen Sharma (Jun 06, 2017 9:06 a.m.)
    Thanks (1)
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