Ex 6.4 Class 12 NCERT Solutions Question 1.<\/strong> \nUsing differentials, find the approximate value of each of the following up to 3 places of decimal. \ni. \\(\\sqrt { 25.3 } \\) \nii. \\(\\sqrt { 49.5 } \\) \niii. \\(\\sqrt { 0.6 } \\) \niv. \\({ \\left( 0.009 \\right) }^{ \\frac { 1 }{ 3 } }\\) \nv. \\({ \\left( 0.999 \\right) }^{ \\frac { 1 }{ 10 } }\\) \nvi. \\({ \\left( 15 \\right) }^{ \\frac { 1 }{ 4 } }\\) \nvii. \\({ \\left( 26 \\right) }^{ \\frac { 1 }{ 3 } }\\) \nviii. \\({ \\left( 255 \\right) }^{ \\frac { 1 }{ 4 } }\\) \nix. \\({ \\left( 82 \\right) }^{ \\frac { 1 }{ 4 } }\\) \nx. \\({ \\left( 401 \\right) }^{ \\frac { 1 }{ 2 } }\\) \nxi. \\({ \\left( 0.0037 \\right) }^{ \\frac { 1 }{ 2 } }\\) \nxii. \\({ \\left( 26.57 \\right) }^{ \\frac { 1 }{ 3 } }\\) \nxiii. \\({ \\left( 81.5 \\right) }^{ \\frac { 1 }{ 4 } }\\) \nxiv. \\({ \\left( 3.968 \\right) }^{ \\frac { 3 }{ 2 } }\\) \nxv. \\({ \\left( 32.15 \\right) }^{ \\frac { 1 }{ 5 } }\\) \nSolution: \n <\/p>\niv. \\({ \\left( 0.009 \\right) }^{ \\frac { 1 }{ 3 } }\\) \n <\/p>\n
v. \\({ \\left( 0.999 \\right) }^{ \\frac { 1 }{ 10 } }\\) \n \n\u2206y is approximately equal to dy \n\u2234 (1) \u21d2 \\({ \\left( 0.999 \\right) }^{ \\frac { 1 }{ 10 } }\\) \u2248 1 – 0.0001 \u2248 0.9999 \ni.e., \\({ \\left( 0.999 \\right) }^{ \\frac { 1 }{ 10 } }\\) \u2248 0.9999<\/p>\n
vi. \\({ \\left( 15 \\right) }^{ \\frac { 1 }{ 4 } }\\) \n <\/p>\n
vii. \\({ \\left( 26 \\right) }^{ \\frac { 1 }{ 3 } }\\) \n <\/p>\n
viii. \\({ \\left( 255 \\right) }^{ \\frac { 1 }{ 4 } }\\) \n <\/p>\n
ix. \\({ \\left( 82 \\right) }^{ \\frac { 1 }{ 4 } }\\) \n <\/p>\n
x. \\({ \\left( 401 \\right) }^{ \\frac { 1 }{ 2 } }\\) \n <\/p>\n
xi. \\({ \\left( 0.0037 \\right) }^{ \\frac { 1 }{ 2 } }\\) \n <\/p>\n
xii. \\({ \\left( 26.57 \\right) }^{ \\frac { 1 }{ 3 } }\\) \n <\/p>\n
xiii. \\({ \\left( 81.5 \\right) }^{ \\frac { 1 }{ 4 } }\\) \n <\/p>\n
xv. \\({ \\left( 32.15 \\right) }^{ \\frac { 1 }{ 5 } }\\) \n <\/p>\n
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Exercise 6.4 Class 12 NCERT Solutions Question 2.<\/strong> \nFind the approximate value of f (2.01), where f (x) = 4x\u00b2 + 5x + 2 \nSolution: \nf(x+\u2206x) = f(2.01), f(x) = f (2) = 4.2\u00b2 + 5.2 + 2 = 28, \nf’ (x) = 8x + 5 Now, f(x + \u2206x) = f(x) + \u2206f(x) \n= f(x) + f’ (x) . \u2206x = 28 + (8x + 5) \u2206x \n= 28 + (16 + 5) x 0.01 \n= 28 + 21 x 0.01 \n= 28 + 0.21 \nHence, f(2.01) \u2248 28 x 21.<\/p>\n6.4 Class 12 NCERT Solutions Question 3.<\/strong> \nFind the approximate value of f (5.001), where f(x) = x3<\/sup> – 7x2<\/sup> +15. \nSolution: \nLet x + \u2206x = 5.001, x = 5 and \u2206x = 0.001, \nf(x) = f(5) = – 35 \nf(x + \u2206x) = f(x) + \u2206f(x) = f(x) + f'(x).\u2206x \n= (x3<\/sup> – 7x\u00b2 + 15) + (3x\u00b2 – 14x) \u00d7 \u2206x \nf(5.001) = – 35 + (3 \u00d7 5\u00b2 – 14 \u00d7 5) \u00d7 0.001 \n\u21d2 f (5.001) = – 35 + 0.005 \n= – 34.995.<\/p>\nExercise 6.4 Class 12 Maths Question 4.<\/strong> \nFind the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%. \nSolution: \nThe side of the cube = x meters. \nIncrease in side = 1% = 0.01 \u00d7 x = 0.01 x \nVolume of cube V = x3<\/sup> \n\u2234 \u2206v = \\(\\frac { dv }{ dx }\\) \u00d7 \u2206x \n= 3x\u00b2 \u00d7 0.01 x \n= 0.03 x3<\/sup> m3<\/sup><\/p>\nClass 12 Maths Ex 6.4 Solutions Question 5.<\/strong> \nFind the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%. \nSolution: \nThe side of the cube = x m; \nDecrease in side = 1% = 0.01 x \nIncrease in side = \u2206x = – 0.01 x \nSurface area of cube = 6x\u00b2 m\u00b2 = S \n\u2234 \\(\\frac { ds }{ dx }\\) \u00d7 \u2206x = 12x \u00d7 (- 0.01 x) \n= – 0.12 x\u00b2 m\u00b2.<\/p>\n <\/p>\n
Ex 6.4 Class 12 Maths Ncert Solutions Question 6.<\/strong> \nIf the radius of a sphere is measured as 7m with an error of 0.02 m, then find the approximate error in calculating its volume. \nSolution: \nRadius of the sphere = 7m : \u2206r = 0.02 m. \nVolume of the sphere V = \\(\\frac { 4 }{ 3 } \\pi { r }^{ 3 }\\) \n\\(\\Delta V=\\frac { dV }{ dr } \\times \\Delta r=\\frac { 4 }{ 3 } .\\pi .3{ r }^{ 2 }\\times \\Delta r\\) \n= 4\u03c0 \u00d7 7\u00b2 \u00d7 0.02 \n= 3.92 \u03c0m\u00b3<\/p>\nEx6.4 Class 12 NCERT Solutions Question 7.<\/strong> \nIf the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area. \nSolution: \nRadius of the sphere = 9 m: \u2206r = 0.03m \nSurface area of sphere S = 4\u03c0r\u00b2 \n\u2206s = \\(\\frac { ds }{ dr }\\) \u00d7 \u2206r \n= 8\u03c0r \u00d7 \u2206r \n= 8\u03c0 \u00d7 9 \u00d7 0.03 \n= 2.16 \u03c0m\u00b2.<\/p>\nQuestion 8. \nIf f (x) = 3x\u00b2 + 15x + 5, then the approximate value of f (3.02) is \n(a) 47.66 \n(b) 57.66 \n(c) 67.66 \n(d) 77.66 \nSolution: \n(d) x + \u2206x = 3.02, where x = 30, \u2206x = 0.2, \n\u2206f(x) = f(x + \u2206x) – f(x) \n\u21d2 f(x + \u2206x) = f(x) + \u2206f(x) = f(x) + f'(x) \u2206x \nNow f(x) = 3×2 + 15x + 5; f(3) = 77, f’ (x) = 6x + 15 \nf’ (3) = 33 \n\u2234 f (3.02) = 87 + 33 x 0.02 = 77.66<\/p>\n
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Question 9. \nThe approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is \n(a) 0.06 x\u00b3 m\u00b3 \n(b) 0.6 x\u00b3 m\u00b3 \n(c) 0.09 x\u00b3 m\u00b3 \n(d) 0.9 x\u00b3 m\u00b3 \nSolution: \n(c) Side of a cube = x meters \nVolume of cube = x\u00b3, \nfor \u2206x \u21d2 3% of x = 0.03 x \nLet \u2206v be the change in v0l. \u2206v = \\(\\frac { dv }{ dx }\\) x \u2206x = 3x\u00b2 \u00d7 \u2206x \nBut, \u2206x = 0.03 x \n\u21d2 \u2206v = 3x\u00b2 x 0.03 x \n= 0.09 x\u00b3m\u00b3<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-4\/ NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.4 Ex 6.4 Class 12 NCERT Solutions Question 1. Using differentials, find the approximate value of each …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","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":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n