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a year ago

Maths 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

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Four

If the system of equations

$x + y + z = 2$

$2 x + 4 y - z = 6$

$3 x + 2 y + λ z = μ$

has infinitely many solutions, then:

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alok kumar singh

Contributor-Level 10

$\overline{x} = 10$

$\Rightarrow \overset{?}{x} = \frac{63 + a + b}{8} = 10$

$\Rightarrow a + b = 17$

Since, variance is independent of origin.

So, we subtract 10 from each observation.

So, $σ^{2} = 13.5 = \frac{79 + (a - 10)^{2} + (b - 10)^{2}}{8}$

$\Rightarrow a^{2} + b^{2} - 20 (a + b) = - 171$

$\Rightarrow a^{2} + b^{2} = 169$

From (1) and (2) ; $a = 12$ and $b = 5$

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New answer posted

a year ago

Maths 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} | + | z00New answer posted a year ago Maths Maths NCERT Exemplar Solutions Class 12th Chapter Twelve What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin? 0 Follower 3 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} G i v e n t h a t e a c h e d g e o f t h e c u b o i d i s 2 u n i t s . \\ ∴ C o o r d i n a t e s o f t h e v e r t i c e s a r e \\ A (2, 0, 0), B (2, 2, 0), C (0, 2, 0), D (0, 2, 2), E (0, 0, 2), F (2, 0, 2), G (2, 2, 2) a n d O (0, 0, 0) . \end{array} 00New answer posted a year ago Maths Maths NCERT Exemplar Solutions Class 12th Chapter Five The function f (x) = \{\begin{array}{l} \frac{π}{4} + {t a n}^{- 1} x, & | x | \leq 1 \\ \frac{1}{2} (| x | - 1), & | x | > 1 \end{array} is: 0 Follower 14 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 A alok kumar singh Contributor-Level 10 \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a b) \frac{2 b^{2}}{a} = 10 \Rightarrow b^{2} = 5 a Now, ? (t) = \frac{5}{12} + t - t^{2} = \frac{8}{12} - {(t - \frac{1}{2})}^{2} ? (t)_{m a x} = \frac{8}{12} = \frac{2}{3} = e \Rightarrow e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{4}{9} \Rightarrow a^{2} = 81 (From (i) and (ii) So, a^{2} + b^{2} = 81 + 45 = 126 00New answer posted a year ago Maths Maths NCERT Exemplar Solutions Class 12th Chapter Eight If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1 + x)^{N + 5} are in the ratio 5 : 10 : 14, then the largest coefficient in the expansion is: 0 Follower 3 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 A alok kumar singh Contributor-Level 10 f (2) = 8, f^{'} (2) = 5, f^{'} (x) \geq 1, f^{''} (x) \geq 4, \forall x \in (1,6) Using LMVT f^{''} (x) = \frac{f^{'} (5) - f^{'} (2)}{5 - 2} \geq 4 \Rightarrow f^{'} (5) \geq 17 f^{'} (x) = \frac{f (5) - f (2)}{5 - 2} \geq 1 \Rightarrow f (5) \geq 11 Therefore f^{'} (5) + f (5) \geq 28 00New answer posted a year ago Maths Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Prove that the points (0, 1, 7), (2, 1, 9), and (6, 5, 13) are collinear. Find the ratio in which the first point divides the join of the other two. 0 Follower 10 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 t h e g i v e n points a r e A (0, - 1, - 7), B (2, 1, - 9), C (6, 5, - 13) \\ A B = \sqrt[]{{(2 - 0)}^{2} + {(1 + 1)}^{2} + {(- 9 + 7)}^{2}} = \sqrt[]{4 + 4 + 4} = \sqrt[]{12} = 2 \sqrt[]{3} \\ B C = \sqrt[]{{(6 - 2)}^{2} + {(5 - 1)}^{2} + {(- 13 + 9)}^{2}} = \sqrt[]{16 + 16 + 16} = \sqrt[]{48} = 4 \sqrt[]{3} \\ A C = \sqrt[]{{(6 - 0)}^{2} + {(5 + 1)}^{2} + {(- 13 + 7)}^{2}} = \sqrt[]{36 + 36 + 36} = \sqrt[]{108} = 6 \sqrt[]{3} \\ 2 \sqrt[]{3} + 4 \sqrt[]{3} = 6 \sqrt[]{3} \\ i . e ., A B + B C = A C \\ ∴ A B : A C = 2 \sqrt[]{3} : 6 \sqrt[]{3} = 1 : 3 \\ H e n c e, point A d i v i d e s B a n d C i n 1 : 3 e x t e r n a l l y . \end{array} 00New answer posted a year ago Maths Maths NCERT Exemplar Solutions Class 12th Chapter Two Contrapositive of the statement: If a function f is differentiable at a, then it is also continuous at a', is: 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 A alok kumar singh Contributor-Level 10 \int_{0}^{3} ? g (x) - f (x) = \int_{0}^{3} ? | | x - 2 | - 2 | d x - \int_{0}^{3} ? | x - 2 | d x = (\frac{1}{2} * 2 * 2 + 1 + \frac{1}{2} * 1 * 1) - (\frac{1}{2} * 2 * 2 + \frac{1}{2} * 1 * 1) = (2 + 1 + \frac{1}{2}) - (2 + \frac{1}{2}) = 1 00New answer posted a year ago Maths Probability Let S = {E_{1}, E_{2}, . . . . . ., E_{8}} be a sample space of a random experiment such that P(E n) = \frac{n}{36} for every n = 1, 2,……, 8. then the number of elements in the set {A \subseteq S : P (A) \geq \frac{4}{5}} is………. 0 Follower 7 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 P (E n) = n/36 for n = 1, 2, 3, …., 8 P (A) = \frac{A n y p o s s i b l e s u m o f (1, 2, 3, . . . . . . . . ., 8) (= α s a y)}{36} \frac{α}{36} \geq \frac{4}{5} ∴ a \geq 29 If one of the number from {1, 2, ….8} is left then total \geq 29 by 3 ways Similarly by leaving terms more 2 or 3 we get 16 more combinations ∴ Total number of different set a possible is 16 + 3 = 19 00New answer posted a year ago Maths Class 12th If the lines \frac{x - 3}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} & \frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5} are perpendicular, find the value of k 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 V Vishal Baghel Contributor-Level 10 The direction of ratios of the lines, \frac{x - 3}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} & \frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5}, are - 3, 2 k, 2 a n d 3 k, 1, - 5 respectively. It is known that two lines with direction ratios, a_{1}, b_{1}, c_{1} a n d a_{2}, b_{2}, c_{2}, are perpendicular, if a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \begin{array}{l} ∴ - 3 (3 k) + 2 k * 1 + 2 (- 5) = 0 \\ \Rightarrow - 9 k + 2 k - 10 = 0 \\ \Rightarrow 7 k = - 10 \\ \Rightarrow k = \frac{- 10}{7} \end{array} Therefore, for k= -10/7, the given lines are perpendicular to each other. 00❮ 293294295296297298299300301 ❯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 718k Reviews 1850k 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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