![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-119.png\")
<\/figure>\n\n\n\nThe Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the intervals where the graph of ( y = \\sec(2x) ) is increasing, we need to follow a few steps. First, we should find the derivative of the function, as the derivative tells us the slope of the graph. If the derivative is positive, the function is increasing.<\/p>\n\n\n\n We begin by taking the derivative of ( y = \\sec(2x) ) with respect to ( x ). To differentiate ( \\sec(2x) ), we use the chain rule. The derivative of ( \\sec(u) ) is ( \\sec(u) \\tan(u) ), and then we multiply by the derivative of the inside function, which is ( 2x ), giving us:<\/p>\n\n\n\n [ The graph of the function is increasing where the derivative is positive, which means:<\/p>\n\n\n\n [ Since ( \\sec(2x) ) is always positive when ( \\tan(2x) ) is positive, we need to determine where ( \\tan(2x) ) is positive. The tangent function is positive in the first and third quadrants. Therefore, we need to find the intervals where ( \\tan(2x) > 0 ).<\/p>\n\n\n\n We know that the tangent function has a period of ( \\pi ). Hence, ( \\tan(2x) ) has a period of ( \\frac{\\pi}{2} ). For ( \\tan(2x) ) to be positive, ( 2x ) must be in the first or third quadrant, i.e., in the interval ( (0, \\pi) ) and ( (\\pi, 2\\pi) ), which gives:<\/p>\n\n\n\n [ Dividing by 2, we get the intervals where ( x ) is increasing:<\/p>\n\n\n\n [ The intervals where the graph of ( y = \\sec(2x) ) is increasing are:<\/p>\n\n\n\n [ These are the intervals where the derivative is positive, meaning the function is increasing.<\/p>\n","protected":false},"excerpt":{"rendered":" State the intervals where the graph of y = sec(2x) is increasing in between The Correct Answer and Explanation is : To determine the intervals where the graph of ( y = \\sec(2x) ) is increasing, we need to follow a few steps. First, we should find the derivative of the function, as the derivative […]<\/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-188490","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188490","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=188490"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188490\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188490"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188490"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188490"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. Finding the Derivative:<\/h3>\n\n\n\n
\\frac{d}{dx}[\\sec(2x)] = 2 \\sec(2x) \\tan(2x)
]<\/p>\n\n\n\n2. When is the Derivative Positive?<\/h3>\n\n\n\n
2 \\sec(2x) \\tan(2x) > 0
]<\/p>\n\n\n\n3. Solving ( \\tan(2x) > 0 ):<\/h3>\n\n\n\n
0 < 2x < \\pi \\quad \\text{or} \\quad \\pi < 2x < 2\\pi
]<\/p>\n\n\n\n
0 < x < \\frac{\\pi}{2} \\quad \\text{or} \\quad \\pi < x < \\frac{3\\pi}{2}
]<\/p>\n\n\n\n4. Conclusion:<\/h3>\n\n\n\n
(0, \\frac{\\pi}{2}) \\cup (\\pi, \\frac{3\\pi}{2})
]<\/p>\n\n\n\n