{"id":8148,"date":"2021-11-14T17:49:55","date_gmt":"2021-11-14T12:19:55","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8148"},"modified":"2021-11-15T16:03:52","modified_gmt":"2021-11-15T10:33:52","slug":"it-is-given-that-for-the-function-fx-x3-6x2-ax-b-on-1-3-rolless-theorem-holds-with-c-2-1over-sqrt3-find-the-values-of-a-and-b-if-f1-f3-0","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/it-is-given-that-for-the-function-fx-x3-6x2-ax-b-on-1-3-rolless-theorem-holds-with-c-2-1over-sqrt3-find-the-values-of-a-and-b-if-f1-f3-0\/","title":{"rendered":"It is given that for the function f(x) = \\(x^3 – 6x^2 + ax + b\\) on [1, 3], Rolles’s theorem holds with c = \\(2 +{1\\over \\sqrt{3}}\\). Find the values of a and b, if f(1) = f(3) = 0."},"content":{"rendered":"

Solution :<\/h2>\n

We are given that f(1) = f(3) = 0.<\/p>\n

\\(\\therefore\\)\u00a0 \\(1^3 – 6 \\times 1 + a + b\\) = \\(3^3 – 6 \\times 3^2 + 3a + b\\) = 0<\/p>\n

\\(\\implies\\)\u00a0 a + b = 5 and 3a + b = 27<\/p>\n

Solving these two equations for a and b, f'(c) is zero or not.<\/p>\n

We have,<\/p>\n

f(x) = \\(x^3 – 6x^2 + ax + b\\)<\/p>\n

\\(\\implies\\)\u00a0 f(x) = \\(x^3 – 6x^2 + 11x – 6\\)<\/p>\n

\\(\\implies\\)\u00a0 f'(x)\u00a0 = \\(3x^2 – 12x + 11\\)<\/p>\n

\\(\\therefore\\)\u00a0 f'(c) = \\(3c^2 – 12c + 11\\) = 3\\((2 +{1\\over \\sqrt{3}})^2\\) – 12(\\(2 +{1\\over \\sqrt{3}}\\)) + 11<\/p>\n

= 12 + \\(12\\over \\sqrt{3}\\) + 1 – 24 – \\(12\\over \\sqrt{3}\\) + 11 = 0<\/p>\n

Hence, a = 11 and b = -6.<\/p>\n

\n

Similar Questions<\/h3>\n

Find the approximate value of f(3.02), where f(x) = \\(3x^2 + 5x + 3\\).<\/a><\/p>\n

Verify Rolle\u2019s theorem for the function f(x) = \\(x^2\\) \u2013 5x + 6 on the interval [2, 3].<\/a><\/p>\n

If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.<\/a><\/p>\n

Find the point on the curve y = cos x \u2013 1, x \\(\\in\\) \\([{\\pi\\over 2}, {3\\pi\\over 2}]\\) at which tangent is parallel to the x-axis.<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : We are given that f(1) = f(3) = 0. \\(\\therefore\\)\u00a0 \\(1^3 – 6 \\times 1 + a + b\\) = \\(3^3 – 6 \\times 3^2 + 3a + b\\) = 0 \\(\\implies\\)\u00a0 a + b = 5 and 3a + b = 27 Solving these two equations for a and b, f'(c) is …<\/p>\n

It is given that for the function f(x) = \\(x^3 – 6x^2 + ax + b\\) on [1, 3], Rolles’s theorem holds with c = \\(2 +{1\\over \\sqrt{3}}\\). Find the values of a and b, if f(1) = f(3) = 0.<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","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":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[66,43],"tags":[],"yoast_head":"\nIt is given that for the function f(x) = \\(x^3 - 6x^2 + ax + b\\) on [1, 3], Rolles's theorem holds with c = \\(2 +{1\\over \\sqrt{3}}\\). Find the values of a and b, if f(1) = f(3) = 0.<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/it-is-given-that-for-the-function-fx-x3-6x2-ax-b-on-1-3-rolless-theorem-holds-with-c-2-1over-sqrt3-find-the-values-of-a-and-b-if-f1-f3-0\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"It is given that for the function f(x) = \\(x^3 - 6x^2 + ax + b\\) on [1, 3], Rolles's theorem holds with c = \\(2 +{1\\over \\sqrt{3}}\\). Find the values of a and b, if f(1) = f(3) = 0.\" \/>\n<meta property=\"og:description\" content=\"Solution : We are given that f(1) = f(3) = 0. (therefore)\u00a0 (1^3 – 6 times 1 + a + b) = (3^3 – 6 times 3^2 + 3a + b) = 0 (implies)\u00a0 a + b = 5 and 3a + b = 27 Solving these two equations for a and b, f'(c) is … It is given that for the function f(x) = (x^3 – 6x^2 + ax + b) on [1, 3], Rolles’s theorem holds with c = (2 +{1over sqrt{3}}). Find the values of a and b, if f(1) = f(3) = 0. 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