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Maths NCERT Exemplar Solutions Class 11th Chapter Five

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2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

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

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Five 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

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

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

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Maths NCERT Exemplar Solutions Class 11th Chapter Five Complex Numbers and Quadratic Equations

(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} | + | z00New question posted 2 months ago Maths NCERT Exemplar Solutions Class 11th Chapter Five Class 11th aBSCJhbascjhbASJCbjhASC 0 Follower 2 Views Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel PostNew answer posted 2 months ago Maths NCERT Exemplar Solutions Class 11th Chapter Five Class 11th fEfkjbDSBDCkjbdcjknSAKJCnAJKSC 0 Follower 1 View Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel Post A alok kumar singh Contributor-Level 10 \begin{array}{l} (i) {| a z_{1} ? b z_{2} |}^{2} + {| b z_{1} + a z_{2} |}^{2} \\ ? ? ? ? ? ? = {| a z_{1} |}^{2} + {| b z_{2} |}^{2} ? 2 R e (a z_{1} . b {\bar{z}}_{2}) + {| b z_{1} |}^{2} + {| a z_{2} |}^{2} + 2 R e (a z_{1} . b {\bar{z}}_{2}) \\ ? ? ? ? ? ? = {| a z_{1} |}^{2} + {| b z_{2} |}^{2} + {| b z_{1} |}^{2} + {| a z_{2} |}^{2} \\ ? ? ? ? ? ? = (a^{2} + b^{2}) ({| z_{1} |}^{2} + {| z_{2} |}^{2}) \\ ? ? ? ? ? ? ? ? H e n c e, ? ? t h e ? ? v a l u e ? ? o f ? ? t h e ? ? f i l l e r ? ? i s ? ? (a^{2} + b^{2}) ({| z_{1} |}^{2} + {| z_{2} |}^{2}) . \\ (i i) \sqrt[]{? 25} * \sqrt[]{? 9} = \sqrt[]{? 1} . \sqrt[]{25} * \sqrt[]{? 1} \sqrt[]{9} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = 5 i * 3 i = 15 i^{2} = ? 15 \\ ? ? ? ? ? ? ? H e n c e, ? ? t h e ? ? v a l u e ? ? o f ? ? t h e ? ? f i l l e r ? ? i s ? ? ? 15 . \end{array} 00New question posted 2 months ago Maths NCERT Exemplar Solutions Class 11th Chapter Five Class 11th jcasbcacbksajcbkasbckjsabckjasbckjASC 0 Follower 2 Views Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel PostNew question posted 2 months ago Maths NCERT Exemplar Solutions Class 11th Chapter Five Complex Numbers and Quadratic Equations (i) For any two complex numbers z_{1}, z_{2} and any real numbers a, b, ? a^{2} z_{1} + b^{2} z_{2} ? - ? b^{2} z_{1} + a^{2} z_{2} ? = . . . . . (ii) The value of \sqrt[]{- 25} \times \sqrt[]{- 9} is . (iii) The number \frac{{(1 - i)}^{3}}{1 - i} is equal to . (iv) The sum of the series i + i^{2} + i^{3} + \dots up to 1000 terms is . (v) Multiplicative inverse of 1 + i is . (vi) If z_{1} and z_{2} are complex numbers such that z_{1} + z_{2} is a real number, then z_{2} = . . . . (vii) (z) + z (z \neq 0) is . (viii) If ? z + 4 ? \leq 3, then the greatest and least values of ? z + 1 ? are . and . (ix) If \frac{z - 2}{z + 6} = \frac{π}{6}, then the locus of z is . (x) If ? z ? = 4 and (z) = \frac{5 π}{6}, then z = \dots \dots \dots \dots 0 Follower 4 Views Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel PostNew question posted 2 months ago Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution (i) For any two complex numbers z_{1}, z_{2} and any real numbers a, b, ? a^{2} z_{1} + b^{2} z_{2} ? - ? b^{2} z_{1} + a^{2} z_{2} ? = . . . . . (ii) The value of \sqrt[]{- 25} \times \sqrt[]{- 9} is . (iii) The number \frac{{(1 - i)}^{3}}{1 - i} is equal to . (iv) The sum of the series i + i^{2} + i^{3} + \dots up to 1000 terms is . (v) Multiplicative inverse of 1 + i is . (vi) If z_{1} and z_{2} are complex numbers such that z_{1} + z_{2} is a real number, then z_{2} = . . . . (vii) arg (z) + arg z (z \neq 0) is . (viii) If ? z + 4 ? \leq 3, then the greatest and least values of ? z + 1 ? are . and . (ix) If \frac{z - 2}{z + 6} = \frac{π}{6}, then the locus of z is . (x) If ? z ? = 4 and (z) = \frac{5 π}{6}, then z = \dots \dots \dots \dots 0 Follower 2 Views Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel PostNew answer posted 3 months ago Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution If z and w are two complex numbers such that ? z w ? = 1 and arg (z) - arg (w) = \frac{π}{2}, then show that z^{} w = - i . 0 Follower 2 Views Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel Post P Payal Gupta Contributor-Level 10 This is a long answer type question as classified in NCERT Exemplar \begin{array}{l} L e t z = r_{1} (c o s θ_{1} + i s i n θ_{1}) a n d w = r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ z w = r_{1} r_{2} [(c o s θ_{1} - i s i n θ_{1})] [(c o s θ_{2} + i s i n θ_{2})] \\ | z w | = r_{1} r_{2} = 1 [G i v e n] \\ N o w a r g (z) - a r g (w) = \frac{π}{2} \\ \Rightarrow θ_{1} - θ_{2} = \frac{π}{2} \Rightarrow a r g (\frac{z}{w}) = \frac{π}{2} \\ \bar{z} w = r_{1} (c o s θ_{1} - i s i n θ_{1}) r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ = r_{1} r_{2} [c o s θ_{1} c o s θ_{2} + i c o s θ_{1} s i n θ_{2} - i s i n θ_{1} c o s θ_{2} - i^{2} s i n θ_{1} s i n θ_{2}] \\ = r_{1} r_{2} [(c o s θ_{1} c o s θ_{2} + s i n θ_{1} s i n θ_{2}) + i (c o s θ_{1} s i n θ_{2} - s i n θ_{1} c o s θ_{2})] \\ = r_{1} r_{2} [c o s (θ_{2} - θ_{1}) + i s i n (θ_{2} - θ_{1})] \\ = r_{1} r_{2} [c o s (- \frac{π}{2}) + i s i n (- \frac{π}{2})] \\ = r_{1} r_{2} [c o s \frac{π}{2} - i s i n \frac{π}{2}] = 1 . [0 - i] = - i \\ H e r e \bar{z} w = - i . H e n c e p r o v e d . \end{array} 00❮ 3456789101112 ❯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 65k Colleges 1.2k Exams 688k Reviews 1800k Answers .write-a-review-widget h2.tbSec2 {
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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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