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New answer posted

4 months ago

Class 11th Chemistry

On addition of acetate ions to an acetic acid solution, the concentration of hydrogen ions, [H⁺]

(a) Becomes zero (b) Is unchanged (c) Increases (d) Decreases

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Vishal Baghel

Contributor-Level 10

Answer: (d) decreases

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New answer posted

4 months ago

Class 11th Chemistry

As the size of A increases down the group, H-A bond strength decreases and so the acid strength

(a) Increases (b) Decreases

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Vishal Baghel

Contributor-Level 10

Answer: (a) increases

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New answer posted

4 months ago

Class 11th Limits and Derivatives

42. Find the derivative of the following functions:

(i)sin x cos x (ii) $s e c x$ (iii)5 sec x + 4 cosx

(iv) $c o s e c x$ (v)3cot x + 5 cosec x

(vi) $5 s i n x - 6 c o s x + 7$ (vii) $2 t a n x - 7 s e c x$

0 Follower 7 Views

alok kumar singh

Contributor-Level 10

(i) f(x)=sin x cos x

So, $f^{?} (x) = \underset{h ? 0}{l i m} \frac{f (x + h) ? f (x)}{h}$

$= \underset{h ? 0}{l i m} \frac{s i n (x + h) c o s (x + h) ? s i n x c o s x}{h}$

$= \underset{h ? 0}{l i m} \frac{1}{2 h} * [2 s i n (x + h) c o s (x + h) ? 2 s i n x c o s x]$

$= \underset{h ? 0}{l i m} \frac{1}{2 h} [s i n 2 (x + h) ? s i n 2 x]$

$= \underset{h ? 0}{l i m} \frac{1}{2 h} [2 c o s \frac{2 (x + h) + 2 x}{2} s i n \frac{2 (x + h) ? 2 x}{2}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [c o s (2 x + h) s i n h]$

$= \underset{h ? 0}{l i m} c o s (2 x + h) * \underset{h ? 0}{l i m} \frac{s i n h}{h}$

$= c o s (2 x + 0)$

$= c o s 2 x$

$(ii) f (x) = s e c x$

So, $f^{?} (x) = \underset{h ? 0}{l i m} \frac{f (x + h) ? f (x)}{h}$

$= \underset{h ? 0}{l i m} \frac{1}{h} [s e c (x + h) ? s e c x]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{1}{c o s (x + h)} ? \frac{1}{c o s x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{c o s x ? c o s (x + h)}{c o s (x + h) c o s x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{? 2 s i n (\frac{x + x + h}{2}) s i n (\frac{x ? (x + h)}{2})}{c o s (x + h) c o s x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [? \frac{2 s i n (\frac{2 x + h}{2}) s i n (? h / 2)}{c o s (x + h) c o s x}]$

$= \underset{h ? 0}{l i m} \frac{(? 1 s i n (\frac{2 x + h}{2})}{c o s (x + h) c o s x} * \underset{h ? 0}{l i m} (? 1) \frac{s i n h / 2}{h / 2}$

$= \frac{s i n x}{c o s x ? c o s x} * 1$

$= t a n x ? s e c x .$

(iii) Given f(x)=5 sec x+4 cosx.

So, $f^{?} (x) = \underset{h ? 0}{l i m} \frac{f (x + h) ? f (x)}{h}$

$\begin{array}{l} = \underset{h ? 0}{l i m} \frac{5 s e c (x + h) + 4 c o s (x + h) ? [5 s e c x + 4 c o s x]}{h} . \end{array}$

$= \underset{h ? 0}{l i m} \frac{5}{h} [s e c (x + h) ? s e c x] + \underset{h ? 0}{l i m} \frac{4}{h} [c o s (x + h) ? c o s x]$

$= \underset{h ? 0}{l i m} \frac{5}{h} [\frac{1}{c o s (x + h)} ? \frac{1}{c o s x}] + \underset{h ? 0}{l i m} \frac{4}{h} [? 2 s i n (\frac{x + h + x}{2}) s i n (\frac{x + h ? x}{2})]$

$= \underset{h ? 0}{l i m} \frac{5}{h} [\frac{c o s x ? c o s (x + h)}{c o s (x + h) (c o s x)}] + \underset{h ? 0}{l i m} \frac{4}{h} [? 2 s i n (\frac{2 x + h}{2}) s i n \frac{h}{2}]$

$= \underset{h ? 0}{l i m} \frac{5}{h} [? \frac{2 s i n (\frac{2 x + h}{2}) s i n (? h / 2)}{c o s (x + h) c o s x}] ? 4 \underset{h ? 0}{l i m} s i n (\frac{2 x + h}{2}) \underset{h ? 0}{l i m} \frac{s i n h / 2}{h / 2}$

$= \frac{s i n (\frac{2 x + 0}{2})}{c o s (x + 0) c o s x} * 1 ? 4 s i n (\frac{2 x}{2})$

$= 5 \frac{s i n x}{c o s x} ? \frac{1}{c o s x} ? 4 s i n x$

$= 5 t a n x ? s e c x ? 4 s i n x$

$(iv) Given f (x) = c o s e c x$

$f^{?} (x) = \underset{h ? 0}{l i m} \frac{f (x + h) ? f (x)}{h}$

$= \underset{h ? 0}{l i m} \frac{1}{h} [c o s e c (x + h) ? c o s e c x]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{1}{s i n (x + h)} ? \frac{1}{s i n x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{s i n x ? s i n (x + h)}{s i n (x + h) s i n x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{2 c o s (\frac{x + x + h}{2}) s i n (\frac{x ? (x + h)}{2})]}{s i n (x + h) s i n x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{2 c o s (\frac{x + x + h}{2}) s i n (\frac{x ? (x + h)}{2})}{s i n (x + h) s i n x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{2 c o s (\frac{2 x + h}{2}) s i n (? h / 2)]}{s i n (x + h) s i n x}]$

$= \underset{h ? 0}{l i m} \frac{c o s (\frac{2 x + h}{2})}{s i n (x + h) s i n x} * \frac{(? 1) s i n (2)}{h / 2})$

$= \frac{c o s (\frac{2 x + 0}{2})}{s i n (x + 0) s i n x} * (? 1)$

$= ? \frac{c o s x}{s i n x} * \frac{1}{s i n x}$

$= ? c o t x ? c o s e c x$

(v) Given,f(x)=3 cot x+5cosecx.

So, $f^{?} (x) = \underset{h ? 0}{l i m} \frac{f (x + h) ? f (x)}{h}$ $= \frac{2}{c o s (x + 0) c o s x} * 1 + 7 s i n (\frac{2 x + 0}{2}) * (? 1)$

$= \underset{h ? 0}{l i m} 3 c o t (x + h) + 5 c o s e c (x + h) ? [3 c o t x + 5 c o s x)$

$_{h ? 0} \frac{3}{h} [c o t (x + h) ? c o t x] + \underset{h ? 0}{l i m}$

$= ? \frac{5}{h} [c o s e c (x + h) ? c o s e c x]$

$= \underset{h ? 0}{l i m} \frac{3}{h} [\frac{c o s (x + h)}{s i n (x + h)} ? \frac{c o s x}{s i n x}] + \underset{h ? 0}{l i m} \frac{5}{h} [\frac{1}{s i n (x + h)} ? \frac{1}{s i n x}]$

$= \underset{h ? 0}{l i m} \frac{3}{h} [\frac{c o s (x + h) s i n x ? c o s x s i n (x + h)}{s i n (x + h) s i n x}] + \underset{h ? 0}{l i m} \frac{5}{h} [\frac{s i n x ? s i n (x + h)}{s i n (x + h) s i n x}]$

$= \underset{h ? 0}{l i m} \frac{3}{h} [\frac{s i n (x ? (x + h))}{s i n (x + h) s i n x} + \underset{h ? 0}{l i m} \frac{5}{h} [\frac{2 c o s (\frac{x + x + h}{2}) s i n (\frac{x ? (x + h)}{2})}{s i n (x + h) s i n x}$

$= \underset{h ? 0}{l i m} \frac{3}{s i n (x + h) s i n x} * \underset{h ? 0}{l i m} (? 1) \frac{s i n h}{h} + \underset{h ? 0}{l i m} \frac{5}{h} [\frac{2 c o s (\frac{2 x + h}{2}) s i n (? h / 2)}{s i n (x + h) s i n x}$

$= \frac{3}{s i n x ? s i n x} * (? 1) + 5 \underset{h00New answer posted 4 months ago Class 11th Limits and Derivatives 29. Kindly consider the following 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 29. Given, f (x) = (x - a 1) (x - a 2)… (x - a n)So, \underset{x \to a_{1}}{l i m} f (x) = \underset{x \to a_{1}}{l i m} (x - a_{1}) \underset{x \to a_{1}}{l i m} (x - a_{2}) ? \underset{x \to a_{1}}{l i m} (x - a^{n}) = (a 1 - a 1) (a 1 - a 2) … (a 1 - a n) = 0 (a 1 - a 2) … (a 1 - a n) = 0 And \underset{x \to a}{l i m} f (x) = \underset{x \to a}{l i m} (x - a_{1}) \underset{x \to a}{l i m} (x - a_{2}) \dots \underset{x \to a}{l i m} (x - a n) = (a - a 1) (a - a 2) … (a - a n) 00New answer posted 4 months ago Class 11th Limits and Derivatives 17. Kindly consider the following 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 17. Kindly go through the solution = = \underset{x \to 0}{l i m} \frac{2 s i n^{2} x}{2 s i n^{2} \frac{x}{2}} = \frac{\underset{x \to 0}{l i m} {(\frac{s i n x}{x})}^{2} * x^{2}}{\underset{x \to 0}{l i m} {(\frac{s i n^{} \frac{x}{2}}{\frac{x}{2}})}^{2} {\frac{x}{2}}^{2}} = \frac{{(1)}^{2} * x^{2} * 4}{{(1)}^{2} * x^{2}} = 4 00New question posted 4 months ago Class 11th Ncert Solutions Maths class 11th Using binomial theorem, evaluate each of the following: 3. (102) 5 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 4 months ago Class 11th Ncert Solutions Maths class 11th 44. Kindly consider the following Sec x = 2 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 44. Sec x = 2. . We have, Sec x = 2, | Sec x is (+)ve the principal solution lies in I st and IV th quadrent. = \frac{π}{3} and \frac{6 π - π}{3} = \frac{π}{3} and \frac{5π}{3} . As secx = sec \frac{π}{3} . cosx = cos \frac{π}{3} [∴ s e c x = \frac{1}{c o s x}] The general solution is x = 2nπ ± \frac{π}{3}, n ∈ z. 00New answer posted 4 months ago Class 11th Straight Lines 28. Find equation of the line through the point (0, 2) making an angle \frac{2π}{3} with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin. 0 Follower 9 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 28. Kindly consider the following 00New answer posted 4 months ago Class 11th Ncert Solutions Maths class 11th 30. Kindly consider the following y + 8 ≥ 2x 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 30. 30. For inequality y+8 ≤ 2x, the equation of the line is y+8=2x. We consider the table below to plot of y+8=2x. \begin{array}{l} x \\ y \end{array} | \begin{array}{l} 0 \\ - 8 \end{array} | \begin{array}{l} 4 \\ 0 \end{array} | The line devides the xy-plane into half planer I and II. We select a point (0,0) and check the correctness of the inequality. i.e., 0+8 ≤ 2 * 0 0 ≤ 0 which is true. So, the solution region is I which includes the rigin (0,0). The continuous line also indicates that any points on the line also satisfy the given inequality. 00New answer posted 4 months ago Class 11th Ncert Solutions Maths class 11th 29. Kindly consider the following 3x + 4y ≤ 12 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 29. For inequality 3x+4y≥ 12 the equation of the line is 3x+4y=12 We consider the table below to plot 3x+4y=12. \begin{array}{l} x \\ y \end{array} | \begin{array}{l} 0 \\ 3 \end{array} | \begin{array}{l} 4 \\ 0 \end{array} | This line devides the xy-plane into half planer I and II. We select point 0 (0,0) and check the correctness of the inequality. i.e., 3 * 0+4 * 0 ≤ 12. 0+0 ≤ 12 0 ≤12 which is true. So, the solution region is I which includes the origin (0,0). The continuous line also indicates that any points in the line also satisfy the given inequality. 00❮ 476477478479480481482483484 ❯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 679k Reviews 1800k Answers .write-a-review-widget h2.tbSec2 {
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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
$j(document).ready(function(){
window.contentMapping = {"Filters":"Filter rankings by publisher, exams, location, specialization etc.","DEB":"Download details of eligibility, admissions, fees, infra etc.","Shortlist":"Save your shortlist, get updates, college recommendations etc.","Compare":"Compare colleges on ranking, placements, reviews, fees etc.","Reviews":"Get honest reviews from students who studied in this college","Questions":"Get answers from the students currently studying in this college","Institute":"View college details, courses offered & recommendations","Course":"View course details, admission process & insights","Exam":"View Exam Detail","QuestionSearch":"Now you can also search from 5 lakh+ answered questions","DownloadGuide":"Get all details offline, receive important updates & more.","QuestionPapers":"Get prep tips & previous years question papers."};
initializeWebsiteTour('cta','anaDesktopV2',contentMapping);
});
lazyLoadCss('//css.shiksha.ws/public/css/build/quesDiscPosting.min.99ce37.css');}{l i m}$

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