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Class 11th Exemplar Solution

Where does $z$ lie, if $\frac{z - 5 i}{z + 5 i} = \frac{1}{5} ?$

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | \frac{z - 5 i}{z + 5 i} | = 1 \\ L e t z = x + y i \\ ∴ | \frac{x + y i - 5 i}{x + y i + 5 i} | = 1 \Rightarrow | \frac{x + (y - 5) i}{x + (y + 5) i} | = 1 \\ \Rightarrow | x + (y - 5) i | = | x + (y + 5) i | \\ \Rightarrow x^{2} + {(y - 5)}^{2} = x^{2} + {(y + 5)}^{2} \\ \Rightarrow {(y - 5)}^{2} = {(y + 5)}^{2} \\ \Rightarrow y^{2} + 25 - 10 y = y^{2} + 25 + 10 y \\ \Rightarrow 20 y = 0 \Rightarrow y = 0 \\ H e n c e, z l i e s o n x - a x i s i . e ., r e a l a x i s . \end{array}$

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a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

Find the principal argument of ${(1 + i \sqrt[]{3})}^{2}$ .

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : {(1 + i \sqrt[]{3})}^{2} = 1 + i^{2} . 3 + 2 \sqrt[]{3} i \\ = 1 - 3 + 2 \sqrt[]{3} i = - 2 + 2 \sqrt[]{3} i \\ t a n α = | \frac{2 \sqrt[]{3}}{- 2} | [? t a n α = | \frac{I m g (z)}{R e (z)} |] \\ \Rightarrow t a n α = | - \sqrt[]{3} | = \sqrt[]{3} \\ \Rightarrow t a n α = t a n \frac{π}{3} ∴ α = \frac{π}{3} \\ N o w R e (z) < 0 a n d i m a g e (z) > 0 \\ ∴ a r g (z) = π - α = π - \frac{π}{3} = \frac{2 π}{3} \\ H e n c e, t h e p r i n c i p l e a r g = \frac{2 π}{3} . \end{array}$

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New answer posted

a year ago

Class 11th Physics

4.22 $\hat{i}$ and $\hat{j}$ are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors $\hat{i}$ + $\hat{j}$ , and $\hat{i}$ - $\hat{j}$ ? What are the components of a vector A= 2 $\hat{i}$ + 3 $\hat{j}$ along the directions of $\hat{i}$ + $\hat{j}$ , and $\hat{i}$ - $\hat{j}$ ? [You may use graphical method]

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Vishal Baghel

Contributor-Level 10

4.22

Let us consider a vector $\overset{?}{P}$ . The equation can be written as

Px = Py = 1 $|\overset{?}{P}|$ = $\sqrt (P_{x}^{2} + P_{y}^{2})$ $|\overset{?}{P}|$ = $\sqrt (1^{2} + 1^{2}) |\overset{?}{P}|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\overset{?}{i}$ + $\overset{?}{j}$ = $\sqrt{2}$

Let $θ$ be the angle made by vector $\overset{?}{P}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(P x / P y) θ$ = $\tan^{- 1} ? (1 / 1)$ , $θ$ = 45 $°$ with the x axis

Let $\overset{?}{Q}$ = $\overset{?}{i}$ - $\overset{?}{j}$

$\overset{?}{Q_{x}} \overset{?}{i}$ – $\overset{?}{Q_{y}}$ $\overset{?}{j}$ = ( $\overset{?}{i}$ – $\overset{?}{j})$

$\overset{?}{Q_{x}}$ = $\overset{?}{Q_{y}}$ = 1

$|\overset{?}{Q}|$ = $\sqrt{{Q_{x}}^{2} + {Q_{y}}^{2}}$ = $\sqrt 2$

Hence $|\overset{?}{Q}|$ = $\sqrt 2$ . Therefore the magnitude of ( $\overset{?}{i}$ + $\overset{?}{j})$ = $\sqrt 2$

Let $θ$ be the angle made

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4.22

Let us consider a vector $\overset{?}{P}$ . The equation can be written as

Px = Py = 1 $|\overset{?}{P}|$ = $\sqrt (P_{x}^{2} + P_{y}^{2})$ $|\overset{?}{P}|$ = $\sqrt (1^{2} + 1^{2}) |\overset{?}{P}|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\overset{?}{i}$ + $\overset{?}{j}$ = $\sqrt{2}$

Let $θ$ be the angle made by vector $\overset{?}{P}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(P x / P y) θ$ = $\tan^{- 1} ? (1 / 1)$ , $θ$ = 45 $°$ with the x axis

Let $\overset{?}{Q}$ = $\overset{?}{i}$ - $\overset{?}{j}$

$\overset{?}{Q_{x}} \overset{?}{i}$ – $\overset{?}{Q_{y}}$ $\overset{?}{j}$ = ( $\overset{?}{i}$ – $\overset{?}{j})$

$\overset{?}{Q_{x}}$ = $\overset{?}{Q_{y}}$ = 1

$|\overset{?}{Q}|$ = $\sqrt{{Q_{x}}^{2} + {Q_{y}}^{2}}$ = $\sqrt 2$

Hence $|\overset{?}{Q}|$ = $\sqrt 2$ . Therefore the magnitude of ( $\overset{?}{i}$ + $\overset{?}{j})$ = $\sqrt 2$

Let $θ$ be the angle made by vector $\overset{?}{Q}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(Q x / Q y) θ$ = ${- \tan}^{- 1} ? (1 / 1)$ , $θ$ = - 45 $°$ with the x axis

It is given that $\overset{?}{A}$ = 2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ . We can write Ax $\overset{?}{i}$ + Ay $\overset{?}{j}$ = 2 $\overset{?}{i}$ + 3 $\overset{?}{j}$

Comparing the components on both sides, we get

Ax =2 and Ay = 3

$|\overset{?}{A}|$ = $\sqrt{{(2}^{2}} + 3^{2}$ ) = $\sqrt 13$

Let $\overset{?}{A_{x}}$ make an angle $θ$ with the x axis, then

$\tan ? θ$ = (Ax / Ay), $θ$ = $\tan^{- 1} (? 3 / 2)$ = 56.31 $°$

So the angle between (2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ ) and ( $\overset{?}{i}$ + $\overset{?}{j}$ )

$θ'$ = 56.31 – 45 = 11.31 $°$

Component of vector $\overset{?}{A}$ , along the direction of $\overset{?}{P}$ , making an angle $θ$

= ( A cos $θ'$ ) $\overset{?}{P}$

= ( A cos 11.31) $\frac{(\overset{?}{i} + \overset{?}{j})}{\sqrt{2}}$

= $\sqrt{13}$ $* \frac{0.9806}{\sqrt{2}}$ ( $\overset{?}{i}$ + $\overset{?}{j}$ )

= 2.5 $*$ $\sqrt{2}$

= $\frac{5}{\sqrt{2}}$

Let $θ'$ be the angle between the vectors (2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ ) and ( $\overset{?}{i}$ - $\overset{?}{j})$

$θ " = 45 + 56.31 = 101.31 °$

Component of vector $\overset{?}{A}$ , along the direction of $\overset{?}{Q}$ , making an angle $θ$

= (A cos $θ "$ ) $\frac{(\overset{?}{i} - \overset{?}{j})}{\sqrt{2}}$

= $\sqrt{13}$ cos ( $101.31 °) \frac{(\overset{?}{i} - \overset{?}{j})}{\sqrt{2}}$

= -0.5( $\overset{?}{i}$ - $\overset{?}{j})$

= $- \frac{1}{\sqrt{2}}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

Find $| (1 + i) \frac{(2 + i)}{(3 + i)} |$

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | (1 + i) \frac{(2 + i)}{(3 + i)} | \\ | (1 + i) \frac{(2 + i)}{(3 + i)} * \frac{3 - i}{3 - i} | = | (1 + i) . \frac{6 - 2 i + 3 i - i^{2}}{9 - i^{2}} | \\ = | \frac{(1 + i) . (7 + i)}{9 + 1} | = | \frac{7 + i + 7 i + i^{2}}{10} | = | \frac{7 + 8 i - 1}{10} | \\ = | \frac{6 + 8 i}{10} | = | \frac{3}{5} + \frac{4}{5} i | = \sqrt[]{{(\frac{3}{5})}^{2} + {(\frac{4}{5})}^{2}} = \sqrt[]{\frac{9}{25} + \frac{16}{25}} = \sqrt[]{\frac{25}{25}} = 1 \\ H e n c e, | (1 + i) \frac{(2 + i)}{(3 + i)} | = 1 \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

If $? z_{1} ? = ? z_{2} ? = 1$ , is it necessary that $z_{1} = z_{2}$ ?

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z_{1} = x_{1} + y_{1} i a n d z_{2} = x_{2} + y_{2} i \\ ∴ | x_{1} + y_{1} i | = | x_{2} + y_{2} i | \\ \Rightarrow \sqrt[]{x_{1}^{2} + y_{1}^{2}} = \sqrt[]{x_{2}^{2} + y_{2}^{2}} \\ \Rightarrow x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2} \Rightarrow x_{1}^{2} = x_{2}^{2} a n d y_{1}^{2} = y_{2}^{2} \\ \Rightarrow x_{1} = \pm x_{2} a n d y_{1} = \pm y_{2} \\ S o, z_{1} = x_{1} + y_{1} i a n d z_{2} = \pm x_{2} \pm y_{2} i \\ ∴ z_{1} \neq z_{2} \\ H e n c e, i t i s n o t n e c e s s a r y t h a t z_{1} = z_{2} . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

If $? z_{1} ? = ? z_{2} ? = 1$ , is it necessary that $z_{1} = z_{2}$ ?

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

What is the conjugate of $\frac{2 - i}{(1 - 2 i)}$ z

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t z = \frac{2 - i}{{(1 - 2 i)}^{2}} = \frac{2 - i}{1 + 4 i^{2} - 4 i} = \frac{2 - i}{1 - 4 - 4 i} \\ = \frac{2 - i}{- 3 - 4 i} = \frac{2 - i}{- 3 - 4 i} * \frac{- 3 + 4 i}{- 3 + 4 i} \\ = \frac{- 6 + 8 i + 3 i - 4 i^{2}}{{(- 3)}^{2} - {(4 i)}^{2}} = \frac{- 6 + 11 i + 4}{9 - 16 i^{2}} \\ = \frac{- 2 + 11 i}{9 + 16} = \frac{- 2}{25} + \frac{11}{25} i \\ ∴ \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \\ H e n c e, \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find $z$ if $? z ? = 4$ and $(z) = \frac{5 π}{6}$ .

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | z | = 4 a n d a r g (z) = \frac{5 π}{6} \Rightarrow θ = \frac{5 π}{6} \\ | z | = 4 \Rightarrow r = 4 \\ S o P o l a r f o r m o f z = r [c o s θ + i s i n θ] \\ = 4 [c o s \frac{5 π}{6} + i s i n \frac{5 π}{6}] \\ = 4 [c o s (π - \frac{π}{6}) + i s i n (π - \frac{π}{6})] \\ = 4 [- c o s \frac{π}{6} + i s i n \frac{π}{6}] = 4 [\frac{\sqrt[]{3}}{2} + i . \frac{1}{2}] \\ = - 2 \sqrt[]{3} + 2 i \\ H e n c e, z = - 2 \sqrt[]{3} + 2 i . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

What is the conjugate of $\frac{2 - i}{(1 - 2 i)}$ z

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

(i) The order relation is defined on the set of complex numbers.

(ii) Multiplication of a non-zero complex number by $- i$ rotates the point about the origin through a right angle in the anti-clockwise direction.

(iii) For any complex number $z$ , the minimum value of $? z + \frac{1}{z} ?$ is 1.

(iv) The locus represented by $? z - 1 ? = ? z - i ?$ is a line perpendicular to the join of $(1, 0)$ and $(0, 1)$ .

(v) If $z$ is a complex number such that $z \neq 0$ and Re $(z) = 0$ , then Im $(z^{2}) = 0$ .

(vi) The inequality $? z - 4 ? < ? z - 2 ?$ represents the region given by $x > 3$ .

(vii) Let $z_{1}$ and $z_{2}$ be two complex numbers such that $? z_{1} + z_{2} ? = ? z_{1} ? + ? z_{2} ?$ , then ar

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(i) The order relation is defined on the set of complex numbers.

(ii) Multiplication of a non-zero complex number by $- i$ rotates the point about the origin through a right angle in the anti-clockwise direction.

(iii) For any complex number $z$ , the minimum value of $? z + \frac{1}{z} ?$ is 1.

(iv) The locus represented by $? z - 1 ? = ? z - i ?$ is a line perpendicular to the join of $(1, 0)$ and $(0, 1)$ .

(v) If $z$ is a complex number such that $z \neq 0$ and Re $(z) = 0$ , then Im $(z^{2}) = 0$ .

(vi) The inequality $? z - 4 ? < ? z - 2 ?$ represents the region given by $x > 3$ .

(vii) Let $z_{1}$ and $z_{2}$ be two complex numbers such that $? z_{1} + z_{2} ? = ? z_{1} ? + ? z_{2} ?$ , then arg $(z_{1} - z_{2}) = 0$ .

(viii) $2$ is not a complex number.

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Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} (i) C o m p a r i s o n o f t w o p u r e l y i m a g i n a r y c o m p l e x n u m b e r s i s n o t p o s s i b l e . H o w e v e r, t h e t w o \\ p u r e l y i m a g i n a r y r e a l c o m p l e x n u m b e r s c a n b e c o m p a r e d . \\ S o, i t i s' F a l s e' . \\ (i i) L e t z = x + y i \\ z . i = (x + y i) i = x i - y w h i c h r o t a t e s a t a n g l e o f 18 0^{0} \\ S o, i t i s' F a l s e' . \\ (i i i) L e t z = x + y i \\ ∴ | z | + | z + 1 | = \sqrt[]{x^{2} + y^{2}} + \sqrt[]{{(x - 1)}^{2} + y^{2}} \\ T h e v a l u e o f | z | + | z + 1 | i s m i n i m u m w h e n x = 0, y = 0 i . e ., 1 . \\ H e n c e, i t i s' T r u e' . \\ (i v) L e t z = x + y i \\ G i v e n t h a t : | z - 1 | = | z - i | \\ t h e n | x + y i - 1 | = | x + y i - i | \\ \Rightarrow | (x - 1) + y i | = | x - (1 - y) i | \\ \Rightarrow \sqrt[]{{(x - 1)}^{2} + y^{2}} = \sqrt[]{x^{2} + {(1 - y)}^{2}} \\ \Rightarrow {(x - 1)}^{2} + y^{2} = x^{2} + {(1 - y)}^{2} \\ \Rightarrow x^{2} - 2 x + 1 + y^{2} = x^{2} + 1 + y^{2} - 2 y \\ \Rightarrow - 2 x + 2 y = 0 \\ \Rightarrow x - y = 0 w h i c h i s a s t r a i g h t l i n e . \\ S l o p e = 1 \\ N o w e q u a t i o n o f a l i n e t h r o u g h t h e point (1, 0) a n d (0, 1) \\ y - 0 = \frac{1 - 0}{0 - 1} (x - 1) \\ \Rightarrow y = - x + 1 w h o s e s l o p e = - 1 . \\ N o w t h e m u l t i p l i c a t i o n o f t h e s l o p e s o f t w o l i n e s = - 1 * 1 = - 1, \\ S o t h e y a r e p e r p e n d i c u l a r . \\ H e n c e, i t i s' T r u e' . \end{array}$

$\begin{array}{l} (v) L e t z = x + y i, z \neq 0 a n d R e (z) = 0 \\ Since r e a l p a r t i s 0 \Rightarrow x = 0 \\ ∴ z = 0 + y i = y i \\ ∴ I m (z^{2}) = y^{2} i^{2} = - y^{2} w h i c h i s r e a l . \\ H e n c e, i t i s' F a l s e' . \\ (v i) G i v e n t h a t : | z - 4 | < | z - 2 | \\ L e t z = x + y i \\ \Rightarrow | x + y i - 4 | < | x + y i - 2 | \\ \Rightarrow | (x - 4) + y i | < | (x - 2) + y i | \\ \Rightarrow \sqrt[]{{(x - 4)}^{2} + y^{2}} < \sqrt[]{{(x - 2)}^{2} + y^{2}} \\ \Rightarrow {(x - 4)}^{2} + y^{2} < {(x - 2)}^{2} + y^{2} \\ \Rightarrow {(x - 4)}^{2} < {(x - 2)}^{2} \\ \Rightarrow x^{2} - 8 x + 16 < x^{2} - 4 x + 4 \\ \Rightarrow - 8 x + 4 x < - 16 + 4 \\ \Rightarrow - 4 x < - 12 \\ \Rightarrow x > 3 \\ H e n c e, i t i s' T r u e' . \\ (v i i) L e t z_{1} = x_{1} + y_{1} i a n d z_{2} = x_{2} + y_{2} i \\ \Rightarrow | z_{1} + z_{2} | = | z_{1} | + | z00❮ 394395396397398399400401402 ❯Taking an Exam? Selecting a College? Get authentic answers from experts, students and alumni that you won't find anywhere else Sign Up on Shiksha On Shiksha, get access to 67k Colleges 1.2k Exams 717k Reviews 1850k Answers .write-a-review-widget h2.tbSec2 {
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function jsb9onUnloadTracking() { jsb9eraseCookie("jsb9Track"); var date = new Date(); var presentTime = date.getTime(); var presentUrl = window.location.href; jsb9createCookie("jsb9Track",presentTime+"|"+presentUrl,5); } var date = new Date(); jsb9TrackEndTime = date.getTime(); function jsb9TrackTime() { var jsb9date = new Date(); jsb9TrackFinalLoad=jsb9date.getTime(); if(typeof(jsb9TrackVal) == "string") { var cookieArr = jsb9TrackVal.split('|'); var prevTime = cookieArr[0]; var refererUrl = cookieArr[1]; var jsb9Iframe = document.createElement('div');jsb9Iframe.style.cssText='height:0px;overflow:hidden;'; jsb9Iframe.id = 'jsb9Div'; var style = 'border:0;width:0;height:0;display:none'; var jsb9ServerTime = 195.183992386; var presentUrl = window.location.href; var customTrack = ""; for(var i in jsb9recordTimes) { customTrack += "|"+i+":"+jsb9recordTimes[i]; } var data = presentUrl+"|"+refererUrl+"|"+prevTime+"|"+jsb9TrackStartTime+"|"+jsb9TrackEndTime+"|"+jsb9TrackFinalLoad+"|"+jsb9ServerTime+customTrack; jsb9Iframe.innerHTML = '<iframe border="0" height=0 widht=0 style="visibility: hidden" src="https://track.99acres.com/images/zero.gif?data='+data+'"></iframe>';
document.body.appendChild(jsb9Iframe); } } }catch(e){ }var shikshaProduct = 'anaDesktopV2';/**
* Method loads all required scripts for Unified Registration
* callback function is loadRequiredDataForUnifiedRegistrationProcess
* callback API will be alwys called either file is already loaded or not
*/
function initForUnifiedRegistration()
{
LazyLoad.loadOnce([
'//js.shiksha.ws/public/js/build/ajax-api.1e2339.js'
],loadRequiredDataForUnifiedRegistrationProcess,null,null,true);
}

/*
Function which put values into global vars. like is_ldb user or cross btn click
*/

function loadRequiredDataForUnifiedRegistrationProcess()
{
/* ajax to set if user register or not START */
//checkLdbUser();
/* ajax to set if user register or not END */
/* set variable to check whether user has clicked unified overlay or not*/
unified_form_overlay1_cancel_clicked = getCookie('is_unified_overlay1_clicked');
unified_form_overlay2_cancel_clicked = getCookie('is_unified_overlay2_clicked');
unified_form_overlay3_cancel_clicked = getCookie('is_unified_overlay3_clicked');
/* set Form submit url for diff types of overlays */
if(typeof(arr_unified) !== 'undefined') {
ShikshaUnifiedRegistarion.url_unified = ShikshaUnifiedRegistarion.ajaxUrlHelper(arr_unified);
}
if($('homepagePromotionCarousel')) {
//if($('questionText')) { $('questionText').value = 'Make an informed career choice, ask the expert now!';}
}
// showResetPasswordLayer();
}
/* UNIFIED REGISTRATION APIs END */$j = $.noConflict();var lazyLoadJsFiles = []
function lazyLoadJsFilesInParallel() {
if(lazyLoadJsFiles.length > 0) {
LazyLoad.loadOnce(lazyLoadJsFiles,function(){},null,null,true);
}
}

function LazyLoadAnAContactUsCallback(){
$LAB.script('//js.shiksha.ws/public/js/build/quesDiscPosting.8361d6.js').wait(function(){
LazyLoadAnAContactUs();
});
}

function loadJsFilesInParallel(){
$j(document).trigger('labLoadedJs');
$LAB
.script("//js.shiksha.ws/public/js/build/AutoSuggestor.512dd5.js")
.wait(function(){
if(SHOW_AUTOSUGGESTOR_JS){
autosuggestorInstanceCheck = setInterval(function(){
var fileLoaded = false;
try{
var aso = new AutoSuggestor();
fileLoaded = true;
} catch(e) {
fileLoaded = false;
}
if(fileLoaded){
clearInterval(autosuggestorInstanceCheck);
if(typeof(initializeAutoSuggestorInstance) == 'function') {
initializeAutoSuggestorInstance();
}
if(typeof(initializeAutoSuggestorInstanceSearchV2) == 'function') {
try {
initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","tabPressedHandler":"handleAutoSuggestorTabPressed","enterPressedHandler":"handleAutoSuggestorEnterPressedSearch","mouseClickedHandler":"handleAutoSuggestorMouseClickedSearch","navigationKeysHandler":"handleNavigationKeysHandler","objectName":"autoSuggestorInstanceSearch","searchBoxId":"searchby-college","suggestionContainerId":"search-college-layer","typeAhead":0,"suggestionType":"course_and_institute","suggestionCount":8,"autoSuggestorUrl":"\/search\/AutoSuggestorV2\/getSuggestionsFromSolr\/","isMobileSearch":false,"showGrouping":1,"getTrendingSuggestions":1}');
initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","tabPressedHandler":"handleAutoSuggestorTabPressedCareers","enterPressedHandler":"handleAutoSuggestorEnterPressedCareers","mouseClickedHandler":"handleAutoSuggestorEnterPressedCareers","objectName":"autoSuggestorInstanceCareer","searchBoxId":"searchby-career","suggestionContainerId":"search-career-layer","getTrendingSuggestions":1,"typeAhead":0,"disableCache":1,"suggestionType":"career","suggestionCount":8}');
initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","tabPressedHandler":"handleAutoSuggestorTabPressedQuestion","enterPressedHandler":"handleAutoSuggestorEnterPressedQuestion","mouseClickedHandler":"handleAutoSuggestorEnterPressedQuestion","mouseClickedHandlerLastElement":"handleAutoSuggestorEnterPressedAskQuestion","askNewQuestionURL":"https:\/\/ask.shiksha.com","mouseClickedHandlerCTA":"handleAutoSuggestorEnterPressedCTA","backKeyHandler":"handleAutoSuggestorBackKeyPressedQuestion","objectName":"autoSuggestorInstanceQuestion","searchBoxId":"searchby-question","suggestionContainerId":"search-question-layer","removeExactMatch":1,"suggestionType":"question","suggestionCount":8,"isMobileSearch":false,"autoSuggestorUrl":"\/search\/AutoSuggestorV2\/getSuggestionsFromSolr\/","showGrouping":1,"getTrendingSuggestions":1}');
initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","tabPressedHandler":"handleAutoSuggestorTabPressedExams","enterPressedHandler":"handleAutoSuggestorEnterPressedExams","mouseClickedHandler":"handleAutoSuggestorEnterPressedExams","backKeyHandler":"handleAutoSuggestorBackKeyPressedExams","objectName":"autoSuggestorInstanceExam","searchBoxId":"searchby-exam","suggestionContainerId":"search-exam-layer","suggestionType":"exam","suggestionCount":8,"autoSuggestorUrl":"\/search\/AutoSuggestorV2\/getSuggestionsFromSolr\/","showGrouping":1,"getTrendingSuggestions":1}');
tagsASConfig = '{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","enterPressedHandler":"handleAutoSuggestorEnterPressedTags","mouseClickedHandler":"handleAutoSuggestorEnterPressedTags","objectName":"autoSuggestorInstanceForTags","searchBoxId":"tagSearch","suggestionContainerId":"tagSearch_container","typeAhead":0,"suggestionType":"tag","suggestionCount":10,"autoSuggestorUrl":"\/Tagging\/TaggingDesktop\/getAutoSuggestor","showGrouping":1,"disableCache":1,"removePartialMatch":0}';
initializeAutoSuggestorInstanceSearchV2(tagsASConfig);
initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","searchVersion":"2","enterPressedHandler":"handleAutoSuggestorEnterPressedTags","mouseClickedHandler":"handleAutoSuggestorEnterPressedTags","objectName":"autoSuggestorInstanceForTagsQDP","searchBoxId":"tagSearchQDP","suggestionContainerId":"tagSearchQDP_container","typeAhead":0,"suggestionType":"tag","suggestionCount":6,"autoSuggestorUrl":"\/Tagging\/TaggingDesktop\/getAutoSuggestor","showGrouping":1,"disableCache":1,"callBackFunctionAfterBuildingSuggestionContainer":"showSuggestedTags","callBackFunctionOnInputKeysPressed":"showSuggestedTags"}');
}
catch(e) {
console.log(e);
}
}
if(typeof(initializeAutoSuggestorInstancesForCollgeReviews) == 'function') {
initializeAutoSuggestorInstancesForCollgeReviews();
}
if(typeof(initializeAutoSuggestorInstancesForCampusConnect) == 'function') {
initializeAutoSuggestorInstancesForCampusConnect();
}

if(typeof(initializeAutoSuggestorInstanceAlt) == 'function') {
initializeAutoSuggestorInstanceAlt();
}
}
},1000);
}
});
LazyLoadAnADesktopCallback();
}
(function(g,b,d){var c=b.head||b.getElementsByTagName("head"),D="readyState",E="onreadystatechange",F="DOMContentLoaded",G="addEventListener",H=setTimeout;
function f(){ loadJsFilesInParallel(); /*if(typeof CTACallBackSearch == 'function') CTACallBackSearch();*/} H(function(){if("item"in c){if(!c[0]){H(arguments.callee,25);return}c=c[0]}var a=b.createElement("script"),e=false;a.onload=a[E]=function(){if((a[D]&&a[D]!=="complete"&&a[D]!=="loaded")||e){return false}a.onload=a[E]=null;e=true;f()};
a.src="//js.shiksha.ws/public/js/LAB.min.js";c.insertBefore(a,c.firstChild)},0);if(b[D]==null&&b[G]){b[D]="loading";b[G](F,d=function(){b.removeEventListener(F,d,false);b[D]="complete"},false)}})(this,document);lazyLoadCss('//css.shiksha.ws/public/css/build/socialSharingInner.min.5de709.css');var reset_password_token = '';
var reset_usremail = '';
var reset_password = '';
var usergroup = '';

//var registration_context = '';
var user_logged_in_pref_data = "";
var userInfo = null;
var firstTimePageLoad = true;var serviceWorker = "service-worker.js";
if ('serviceWorker' in navigator && typeof serviceWorker == 'string' && serviceWorker.length > 0 && typeof registerServiceWorker == 'function') {
registerServiceWorker(serviceWorker);
}
/* function to bind autosuggestor */
if(typeof(bindAutoSuggestorEvents) == 'function'){
bindAutoSuggestorEvents();
}
/* responsible for showing GNB sub-menu on mouse hover */
if(typeof(activateGnbMouseOver) == 'function'){
activateGnbMouseOver();
}
var homepageMiddlePanel, instituteDetailPageClass, cookieDomain = '.shiksha.com';
$j(document).ready(function(){
shikshaOnreadyEventPC();
getSocialSharingLayer();
});
$j(window).on('load',function(){
shikshaOnloadEventPC();
if(typeof isUserLoggedIn !='undefined' && isUserLoggedIn && typeof getPredictorList != 'undefined' && typeof getPredictorList == 'function'){
getPredictorList();
}
});window.pushNotiConf = {
deviceType : 'desktop'
};
if(document.getElementById("pushNotificationContainer") != null){
var d = new Date();
var dStr = d.getDate()+''+d.getMonth()+''+d.getFullYear()+'v10';
var randomId = Math.floor(Math.random() * 100);

var loadNotificationSDK = function() {
var obj = document.createElement('script');
/* obj.src = 'https://localshiksha.com/public/js/shikshaPushNotification.js?'+dStr+randomId;*/
obj.src = '//js.shiksha.ws/public/js/build/shikshaPushNotification.b0fa7c.js?'+dStr+randomId;
obj.async = true;
document.head.appendChild(obj);
};

var helperObj = document.createElement('script');
/* helperObj.src = 'https://localshiksha.com/pwa/public/js/push-notification-helpers.js?'+dStr;*/
helperObj.src = '//js.shiksha.ws/pwa/public/push-notification-helpers.js?'+dStr+randomId;
helperObj.async = true;
document.head.appendChild(helperObj);

if (typeof helperObj.onload == 'object') {
helperObj.onload = function(e) {
console.log('notification helper sdk loaded');
loadNotificationSDK();
};
} else {
loadNotificationSDK();
}
if(typeof gaTrackEventCustom == 'function'){
window.phpGATrack = gaTrackEventCustom;
}
}setTimeout(function(){if(typeof loadViewsOnPageLoad == 'function'){loadViewsOnPageLoad();} },2000);var multiLocationCourses = [0];if(shikshaProduct=='CMSExam'){
var elem = document.getElementById("cookieAdSlot-d");
elem.parentNode.removeChild(elem);
}facebook Twitter Google +if (document.addEventListener){
document.addEventListener('mouseup', handleShareLayerHide, false);
} else if (document.attachEvent){
document.attachEvent('onmouseup', handleShareLayerHide);
}

function handleShareLayerHide(e)
{
var target = (e && e.target) || (event && event.srcElement);
var obj = $('socialLayer');
if(target!=obj){
obj.style.display='none';
}
}×//Load website tour
$j(document).ready(function(){
window.contentMapping = {"Filters":"Filter rankings by publisher, exams, location, specialization etc.","DEB":"Download details of eligibility, admissions, fees, infra etc.","Shortlist":"Save your shortlist, get updates, college recommendations etc.","Compare":"Compare colleges on ranking, placements, reviews, fees etc.","Reviews":"Get honest reviews from students who studied in this college","Questions":"Get answers from the students currently studying in this college","Institute":"View college details, courses offered & recommendations","Course":"View course details, admission process & insights","Exam":"View Exam Detail","QuestionSearch":"Now you can also search from 5 lakh+ answered questions","DownloadGuide":"Get all details offline, receive important updates & more.","QuestionPapers":"Get prep tips & previous years question papers."};
initializeWebsiteTour('cta','anaDesktopV2',contentMapping);
});
lazyLoadCss('//css.shiksha.ws/public/css/build/quesDiscPosting.min.99ce37.css'); \end{array}$

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