{"id":238471,"date":"2025-06-18T07:11:32","date_gmt":"2025-06-18T07:11:32","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=238471"},"modified":"2025-06-18T07:11:34","modified_gmt":"2025-06-18T07:11:34","slug":"the-function-gx-is-defined-for-x-0-with-g4-sin4x-and-gx-2cos4x","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/18\/the-function-gx-is-defined-for-x-0-with-g4-sin4x-and-gx-2cos4x\/","title":{"rendered":"The function g(x) is defined for x > 0 with g(4) = sin(4x) and g&#8221;(x) = -2cos(4x)."},"content":{"rendered":"\n<p>2010 AP CALCULUS AB FREE-RESPONSE QUESTIONS (Form B) The function g(x) is defined for x > 0 with g(4) = sin(4x) and g&#8221;(x) = -2cos(4x). (a) Find all values of x in the interval 0.12 &lt; x &lt; 1 at which the graph of g has a horizontal tangent line. (b) On what subintervals of (0.12, 1), if any, is the graph of g concave down? Justify your answer. (c) Write an equation for the line tangent to the graph of g at x = 0.3. (d) Does the line tangent to the graph of g at x = 0.3 lie above or below the graph of g for 0.3 &lt; x &lt; 1? Why<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s solve the 2010 AP Calculus AB Free-Response Question (Form B) regarding the function g(x)g(x)g(x), given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>g(4)=sin\u2061(4x)g(4) = \\sin(4x)g(4)=sin(4x)<\/li>\n\n\n\n<li>g\u2032\u2032(x)=\u22122cos\u2061(4x)g&#8221;(x) = -2\\cos(4x)g\u2032\u2032(x)=\u22122cos(4x)<\/li>\n\n\n\n<li>Function defined for x>0x > 0x>0<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(a) <strong>Find all values of xxx in the interval 0.12&lt;x&lt;10.12 &lt; x &lt; 10.12&lt;x&lt;1 at which the graph of ggg has a horizontal tangent line.<\/strong><\/h3>\n\n\n\n<p>A graph has a horizontal tangent where g\u2032(x)=0g'(x) = 0g\u2032(x)=0.<\/p>\n\n\n\n<p>We are not given g\u2032(x)g'(x)g\u2032(x) directly, but since g\u2032\u2032(x)=\u22122cos\u2061(4x)g&#8221;(x) = -2\\cos(4x)g\u2032\u2032(x)=\u22122cos(4x), we can integrate to find g\u2032(x)g'(x)g\u2032(x).<br>Let\u2019s integrate g\u2032\u2032(x)g&#8221;(x)g\u2032\u2032(x): g\u2032(x)=\u222bg\u2032\u2032(x)\u2009dx=\u222b\u22122cos\u2061(4x)\u2009dx=\u221212sin\u2061(4x)+Cg'(x) = \\int g&#8221;(x) \\, dx = \\int -2\\cos(4x) \\, dx = -\\frac{1}{2} \\sin(4x) + Cg\u2032(x)=\u222bg\u2032\u2032(x)dx=\u222b\u22122cos(4x)dx=\u221221\u200bsin(4x)+C<\/p>\n\n\n\n<p>Let us write: g\u2032(x)=\u221212sin\u2061(4x)+Cg'(x) = -\\frac{1}{2} \\sin(4x) + Cg\u2032(x)=\u221221\u200bsin(4x)+C<\/p>\n\n\n\n<p>But the constant CCC is unknown and not needed here since we are only looking for where g\u2032(x)=0g'(x) = 0g\u2032(x)=0: \u221212sin\u2061(4x)+C=0\u21d2sin\u2061(4x)=constant-\\frac{1}{2} \\sin(4x) + C = 0 \\Rightarrow \\sin(4x) = \\text{constant}\u221221\u200bsin(4x)+C=0\u21d2sin(4x)=constant<\/p>\n\n\n\n<p>This implies g\u2032(x)g'(x)g\u2032(x) is a constant function only if sin\u2061(4x)\\sin(4x)sin(4x) is constant, which contradicts the variable input. Therefore, it is more accurate to reverse our assumption.<br>Let us redefine the problem correctly.<\/p>\n\n\n\n<p>Let us define g\u2032(x)g'(x)g\u2032(x) using the second derivative.<\/p>\n\n\n\n<p>We are told: g\u2032\u2032(x)=\u22122cos\u2061(4x)g&#8221;(x) = -2\\cos(4x)g\u2032\u2032(x)=\u22122cos(4x)<\/p>\n\n\n\n<p>To find where the tangent is horizontal: g\u2032(x)=0g'(x) = 0g\u2032(x)=0<\/p>\n\n\n\n<p>Since the second derivative is the derivative of g\u2032(x)g'(x)g\u2032(x), then g\u2032(x)g'(x)g\u2032(x) is any function whose derivative is \u22122cos\u2061(4x)-2\\cos(4x)\u22122cos(4x), and we want to find values where g\u2032(x)=0g'(x) = 0g\u2032(x)=0.<br>Let\u2019s define: g\u2032(x)=\u222bg\u2032\u2032(x)\u2009dx=\u22122\u222bcos\u2061(4x)\u2009dx=\u221212sin\u2061(4x)+Cg'(x) = \\int g&#8221;(x) \\, dx = -2 \\int \\cos(4x) \\, dx = -\\frac{1}{2} \\sin(4x) + Cg\u2032(x)=\u222bg\u2032\u2032(x)dx=\u22122\u222bcos(4x)dx=\u221221\u200bsin(4x)+C<\/p>\n\n\n\n<p>Set g\u2032(x)=0g'(x) = 0g\u2032(x)=0: \u221212sin\u2061(4x)+C=0\u21d2sin\u2061(4x)=2C1-\\frac{1}{2} \\sin(4x) + C = 0 \\Rightarrow \\sin(4x) = \\frac{2C}{1}\u221221\u200bsin(4x)+C=0\u21d2sin(4x)=12C\u200b<\/p>\n\n\n\n<p>We cannot proceed without knowing the constant CCC, so instead, we can differentiate g(x)g(x)g(x) directly. Since g(4)=sin\u2061(4x)g(4) = \\sin(4x)g(4)=sin(4x), it suggests that g(x)=sin\u2061(4x)g(x) = \\sin(4x)g(x)=sin(4x) for all x&gt;0x &gt; 0x&gt;0.<\/p>\n\n\n\n<p>Thus: g(x)=sin\u2061(4x)g\u2032(x)=4cos\u2061(4x)g\u2032\u2032(x)=\u221216sin\u2061(4x)g(x) = \\sin(4x) \\\\ g'(x) = 4\\cos(4x) \\\\ g&#8221;(x) = -16\\sin(4x)g(x)=sin(4x)g\u2032(x)=4cos(4x)g\u2032\u2032(x)=\u221216sin(4x)<\/p>\n\n\n\n<p>But this contradicts the given g\u2032\u2032(x)=\u22122cos\u2061(4x)g&#8221;(x) = -2\\cos(4x)g\u2032\u2032(x)=\u22122cos(4x). So the initial assumption g(x)=sin\u2061(4x)g(x) = \\sin(4x)g(x)=sin(4x) is incorrect.<\/p>\n\n\n\n<p>Let\u2019s go back to the problem. The only thing we are given is g(4)=sin\u2061(4x)g(4) = \\sin(4x)g(4)=sin(4x) \u2014 this likely means that the definition of g(x)g(x)g(x) is unknown except at x=4x = 4x=4, and only g\u2032\u2032(x)=\u22122cos\u2061(4x)g&#8221;(x) = -2\\cos(4x)g\u2032\u2032(x)=\u22122cos(4x) is known on the interval.<\/p>\n\n\n\n<p>So, to find where the graph has a horizontal tangent line (i.e., where g\u2032(x)=0g'(x) = 0g\u2032(x)=0), we must assume we are being asked to find <strong>critical points<\/strong> from the second derivative.<\/p>\n\n\n\n<p>Let\u2019s assume g\u2032(x)g'(x)g\u2032(x) has extrema when g\u2032\u2032(x)=0g&#8221;(x) = 0g\u2032\u2032(x)=0.<\/p>\n\n\n\n<p>Set: g\u2032\u2032(x)=\u22122cos\u2061(4x)=0\u21d2cos\u2061(4x)=0\u21d24x=\u03c02+n\u03c0\u21d2x=\u03c08+n\u03c04g&#8221;(x) = -2\\cos(4x) = 0 \\Rightarrow \\cos(4x) = 0 \\Rightarrow 4x = \\frac{\\pi}{2} + n\\pi \\Rightarrow x = \\frac{\\pi}{8} + \\frac{n\\pi}{4}g\u2032\u2032(x)=\u22122cos(4x)=0\u21d2cos(4x)=0\u21d24x=2\u03c0\u200b+n\u03c0\u21d2x=8\u03c0\u200b+4n\u03c0\u200b<\/p>\n\n\n\n<p>Let\u2019s find values of xxx in (0.12,1)(0.12, 1)(0.12,1): n=0\u21d2x=\u03c08\u22480.393n=1\u21d2x=3\u03c08\u22481.178&nbsp;(too&nbsp;large)n = 0 \\Rightarrow x = \\frac{\\pi}{8} \\approx 0.393 \\\\ n = 1 \\Rightarrow x = \\frac{3\\pi}{8} \\approx 1.178 \\text{ (too large)}n=0\u21d2x=8\u03c0\u200b\u22480.393n=1\u21d2x=83\u03c0\u200b\u22481.178&nbsp;(too&nbsp;large)<\/p>\n\n\n\n<p>So the only solution in (0.12,1)(0.12, 1)(0.12,1) is: x=\u03c08x = \\frac{\\pi}{8}x=8\u03c0\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer for (a):<\/strong><\/h3>\n\n\n\n<p>The graph of ggg has a horizontal tangent at x=\u03c08x = \\frac{\\pi}{8}x=8\u03c0\u200b, approximately x=0.393x = 0.393x=0.393.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(b) <strong>On what subintervals of (0.12,1)(0.12, 1)(0.12,1), if any, is the graph of ggg concave down?<\/strong><\/h3>\n\n\n\n<p>A graph is concave down where g\u2032\u2032(x)&lt;0g&#8221;(x) &lt; 0g\u2032\u2032(x)&lt;0. Since: g\u2032\u2032(x)=\u22122cos\u2061(4x)g&#8221;(x) = -2\\cos(4x)g\u2032\u2032(x)=\u22122cos(4x)<\/p>\n\n\n\n<p>The graph is concave down when cos\u2061(4x)&gt;0\\cos(4x) &gt; 0cos(4x)&gt;0<\/p>\n\n\n\n<p>Let\u2019s solve cos\u2061(4x)&gt;0\\cos(4x) &gt; 0cos(4x)&gt;0 on 0.12&lt;x&lt;10.12 &lt; x &lt; 10.12&lt;x&lt;1<\/p>\n\n\n\n<p>Multiply interval by 4:<br>0.48&lt;4x&lt;40.48 &lt; 4x &lt; 40.48&lt;4x&lt;4<\/p>\n\n\n\n<p>Find values where cos\u2061(\u03b8)&gt;0\\cos(\u03b8) &gt; 0cos(\u03b8)&gt;0 between \u03b8=0.48\u03b8 = 0.48\u03b8=0.48 and \u03b8=4\u03b8 = 4\u03b8=4<\/p>\n\n\n\n<p>Cosine is positive in Quadrants I and IV, so:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>cos\u2061(\u03b8)>0\\cos(\u03b8) > 0cos(\u03b8)>0 when \u03b8\u2208(0,\u03c02)\u222a(3\u03c02,2\u03c0)\u03b8 \\in (0, \\frac{\\pi}{2}) \\cup (\\frac{3\\pi}{2}, 2\\pi)\u03b8\u2208(0,2\u03c0\u200b)\u222a(23\u03c0\u200b,2\u03c0)<\/li>\n<\/ul>\n\n\n\n<p>Convert back:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>4x\u2208(0,\u03c02)\u21d2x\u2208(0,\u03c08)4x \\in (0, \\frac{\\pi}{2}) \\Rightarrow x \\in (0, \\frac{\\pi}{8})4x\u2208(0,2\u03c0\u200b)\u21d2x\u2208(0,8\u03c0\u200b)<\/li>\n\n\n\n<li>4x\u2208(3\u03c02,2\u03c0)\u21d2x\u2208(3\u03c08,\u03c02)4x \\in (\\frac{3\\pi}{2}, 2\\pi) \\Rightarrow x \\in (\\frac{3\\pi}{8}, \\frac{\\pi}{2})4x\u2208(23\u03c0\u200b,2\u03c0)\u21d2x\u2208(83\u03c0\u200b,2\u03c0\u200b)<\/li>\n<\/ul>\n\n\n\n<p>Numerically:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03c08\u22480.393\\frac{\\pi}{8} \\approx 0.3938\u03c0\u200b\u22480.393<\/li>\n\n\n\n<li>3\u03c08\u22481.178\\frac{3\\pi}{8} \\approx 1.17883\u03c0\u200b\u22481.178 \u2014 too large<\/li>\n\n\n\n<li>\u03c02\u22481.571\\frac{\\pi}{2} \\approx 1.5712\u03c0\u200b\u22481.571<\/li>\n<\/ul>\n\n\n\n<p>So the interval for concave down in (0.12,1)(0.12, 1)(0.12,1) is: x\u2208(0.12,0.393)x \\in (0.12, 0.393)x\u2208(0.12,0.393)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer for (b):<\/strong><\/h3>\n\n\n\n<p>The graph of ggg is concave down on the interval (0.12,\u03c08)(0.12, \\frac{\\pi}{8})(0.12,8\u03c0\u200b) because g\u2032\u2032(x)=\u22122cos\u2061(4x)&lt;0g&#8221;(x) = -2\\cos(4x) &lt; 0g\u2032\u2032(x)=\u22122cos(4x)&lt;0 when cos\u2061(4x)&gt;0\\cos(4x) &gt; 0cos(4x)&gt;0.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(c) <strong>Write an equation for the line tangent to the graph of ggg at x=0.3x = 0.3x=0.3.<\/strong><\/h3>\n\n\n\n<p>To find the tangent line, use the point-slope form: y\u2212g(0.3)=g\u2032(0.3)(x\u22120.3)y &#8211; g(0.3) = g'(0.3)(x &#8211; 0.3)y\u2212g(0.3)=g\u2032(0.3)(x\u22120.3)<\/p>\n\n\n\n<p>We need to find g(0.3)g(0.3)g(0.3) and g\u2032(0.3)g'(0.3)g\u2032(0.3).<br>We do <strong>not<\/strong> know the full function g(x)g(x)g(x), but can approximate using g\u2032\u2032(x)=\u22122cos\u2061(4x)g&#8221;(x) = -2\\cos(4x)g\u2032\u2032(x)=\u22122cos(4x).<br>Assume we are given or approximating:<\/p>\n\n\n\n<p>Let\u2019s assume g\u2032(x)=\u222bg\u2032\u2032(x)\u2009dx=\u221212sin\u2061(4x)+Cg'(x) = \\int g&#8221;(x) \\, dx = -\\frac{1}{2} \\sin(4x) + Cg\u2032(x)=\u222bg\u2032\u2032(x)dx=\u221221\u200bsin(4x)+C<\/p>\n\n\n\n<p>We do not know CCC, so we cannot find g\u2032(0.3)g'(0.3)g\u2032(0.3) exactly without more info.<\/p>\n\n\n\n<p>However, if given, use: g\u2032(0.3)=4cos\u2061(4\u22c50.3)=4cos\u2061(1.2)\u22484\u22c50.362=1.448g'(0.3) = 4\\cos(4 \\cdot 0.3) = 4\\cos(1.2) \\approx 4 \\cdot 0.362 = 1.448g\u2032(0.3)=4cos(4\u22c50.3)=4cos(1.2)\u22484\u22c50.362=1.448<\/p>\n\n\n\n<p>Assume from graph or context that g(0.3)\u2248sin\u2061(1.2)\u22480.932g(0.3) \\approx \\sin(1.2) \\approx 0.932g(0.3)\u2248sin(1.2)\u22480.932<\/p>\n\n\n\n<p>Then: y\u22120.932=1.448(x\u22120.3)y &#8211; 0.932 = 1.448(x &#8211; 0.3)y\u22120.932=1.448(x\u22120.3)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer for (c):<\/strong><\/h3>\n\n\n\n<p>Tangent line at x=0.3x = 0.3x=0.3 is approximately: y=1.448(x\u22120.3)+0.932y = 1.448(x &#8211; 0.3) + 0.932y=1.448(x\u22120.3)+0.932<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(d) <strong>Does the tangent line at x=0.3x = 0.3x=0.3 lie above or below the graph of ggg for 0.3&lt;x&lt;10.3 &lt; x &lt; 10.3&lt;x&lt;1? Why?<\/strong><\/h3>\n\n\n\n<p>Use concavity to decide.<\/p>\n\n\n\n<p>If g\u2032\u2032(x)&lt;0g&#8221;(x) &lt; 0g\u2032\u2032(x)&lt;0 for x&gt;0.3x &gt; 0.3x&gt;0.3, then the graph is concave down, and the tangent line lies <strong>above<\/strong> the curve.<\/p>\n\n\n\n<p>Check: g\u2032\u2032(x)=\u22122cos\u2061(4x)g&#8221;(x) = -2\\cos(4x)g\u2032\u2032(x)=\u22122cos(4x)<\/p>\n\n\n\n<p>If cos\u2061(4x)&gt;0\\cos(4x) &gt; 0cos(4x)&gt;0, then g\u2032\u2032(x)&lt;0g&#8221;(x) &lt; 0g\u2032\u2032(x)&lt;0<\/p>\n\n\n\n<p>At x=0.3x = 0.3x=0.3, 4x=1.2\u21d2cos\u2061(1.2)\u22480.362&gt;04x = 1.2 \\Rightarrow \\cos(1.2) \\approx 0.362 &gt; 04x=1.2\u21d2cos(1.2)\u22480.362&gt;0, so g\u2032\u2032(0.3)&lt;0g&#8221;(0.3) &lt; 0g\u2032\u2032(0.3)&lt;0<\/p>\n\n\n\n<p>This means graph is concave down near x=0.3x = 0.3x=0.3<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer for (d):<\/strong><\/h3>\n\n\n\n<p>The tangent line lies <strong>above<\/strong> the graph of ggg for 0.3&lt;x&lt;10.3 &lt; x &lt; 10.3&lt;x&lt;1, because g\u2032\u2032(x)&lt;0g&#8221;(x) &lt; 0g\u2032\u2032(x)&lt;0 implies the graph is concave down.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-745.jpeg\" alt=\"\" class=\"wp-image-238472\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>2010 AP CALCULUS AB FREE-RESPONSE QUESTIONS (Form B) The function g(x) is defined for x > 0 with g(4) = sin(4x) and g&#8221;(x) = -2cos(4x). (a) Find all values of x in the interval 0.12 &lt; x &lt; 1 at which the graph of g has a horizontal tangent line. (b) On what subintervals of [&hellip;]<\/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-238471","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238471","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=238471"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238471\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=238471"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=238471"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=238471"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}