<\/p>\n\n\n\n
For, 1+6x – 2×3 and, e-x + 2xex – x2ex<\/p>\n\n\n\n
The Correct Answer And Eplanation is:<\/mark><\/strong><\/p>\n\n\n\n To solve this problem, we are asked to find a linear differential operator<\/strong> that annihilates<\/strong> the given functions. An annihilator<\/strong> is a differential operator L(D)L(D) such that L(D)[f(x)]=0L(D)[f(x)] = 0, where DD is the derivative operator ddx\\frac{d}{dx}.<\/p>\n\n\n\n This is a polynomial of degree 3<\/strong>. In general:<\/p>\n\n\n\n Here:<\/p>\n\n\n\n \u2705 Annihilator for<\/strong> f1(x)f_1(x) is: This is a linear combination of exponential functions and polynomials multiplied by exponentials. Let’s break it down:<\/p>\n\n\n\n We treat each term:<\/p>\n\n\n\n So:<\/p>\n\n\n\n But since we want an operator that annihilates all three terms<\/strong>, we take the least common multiple (LCM)<\/strong> of the annihilators.<\/p>\n\n\n\n \u2705 So, the annihilator is: For, 1+6x – 2×3 and, e-x + 2xex – x2ex The Correct Answer And Eplanation is: To solve this problem, we are asked to find a linear differential operator that annihilates the given functions. An annihilator is a differential operator L(D)L(D) such that L(D)[f(x)]=0L(D)[f(x)] = 0, where DD is the derivative operator ddx\\frac{d}{dx}. \ud83d\udd37 Function […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-212418","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212418","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=212418"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212418\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=212418"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=212418"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=212418"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n\ud83d\udd37 Function 1: f1(x)=1+6x\u22122x3f_1(x) = 1 + 6x – 2x^3<\/h3>\n\n\n\n
\n
\n
D4\\boxed{D^4}<\/strong><\/p>\n\n\n\n
\n\n\n\n\ud83d\udd37 Function 2: f2(x)=e\u2212x+2xex\u2212x2exf_2(x) = e^{-x} + 2x e^x – x^2 e^x<\/h3>\n\n\n\n
Terms:<\/h4>\n\n\n\n
\n
\n
\n
Annihilators:<\/h4>\n\n\n\n
\n
(D+1)(D\u22121)3\\boxed{(D + 1)(D – 1)^3}<\/strong><\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answers:<\/h3>\n\n\n\n
\n