{"id":5507,"date":"2024-01-10T09:04:25","date_gmt":"2024-01-10T03:34:25","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=5507"},"modified":"2024-01-11T17:52:10","modified_gmt":"2024-01-11T12:22:10","slug":"maths-1b-errors-and-approximations-important-questions","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/maths-1b-errors-and-approximations-important-questions\/","title":{"rendered":"TS Inter 1st Year Maths 1B Errors and Approximations Important Questions"},"content":{"rendered":"

Students must practice these Maths 1B Important Questions<\/a> TS Inter 1st Year Maths 1B Errors and Approximations Important Questions to help strengthen their preparations for exams.<\/p>\n

TS Inter 1st Year Maths 1B Errors and Approximations Important Questions<\/h2>\n

Question 1.
\nIf y = x2<\/sup> + 3x + 6, find \u2206y and dy when x = 10, \u2206x = 0.01. [Mar. ’15 (TS), ’14, ’11, ’05; May ’15 (AP)]
\nSolution:
\nLet y = f(x) = x2<\/sup> + 3x + 6, x = 10, \u2206x = 0.01
\n\u2206y = f(x + \u2206x) – f(x)
\n= (x + \u2206x)2<\/sup> + 3(x + \u2206x) + 6 – (x2<\/sup> + 3x + 6)
\n= x2<\/sup> + (\u2206x)2<\/sup> + 2 . x . \u2206x + 3x + 3(\u2206x) + 6 – x2<\/sup> – 3x – 6
\n= (\u2206x)2<\/sup> + 2 . x . \u2206x + 3 . \u2206x
\n= (2x + 3) \u2206x + (\u2206x)2<\/sup>
\n= (2.10 + 3) (0.01) + (0. 01)2<\/sup>
\n= (23)(0.01) + 0.0001
\n= 0.23 + 0.0001
\n= 0.2301
\ndy = \\(\\frac{d y}{d x}\\) \u00d7 \u2206x
\n= (2x + 3) \u2206x
\n= (2.10 + 3)(0.01)
\n= 23(0.01)
\n= 0.23<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind \u2206y and dy for the function y = ex<\/sup> + x when x = 5, \u2206x = 0.02. [May ’13]
\nSolution:
\nLet y = f(x) = ex<\/sup> + x, x = 5, \u2206x = 0.02
\n\u2206y = f(x + \u2206x) – f(x)
\n= e(x+\u2206x)<\/sup> + (x + \u2206x) – (ex<\/sup> + x)
\n= ex+\u2206x<\/sup> + x + \u2206x – ex<\/sup>\u00a0– x
\n= ex+\u2206x<\/sup> + \u2206x – ex<\/sup>
\n= e5+0.02<\/sup> + 0.02 – e5<\/sup>
\ndy = \\(\\frac{d y}{d x}\\) \u00d7 \u2206x
\n= (ex<\/sup> + 1) \u00d7 \u2206x
\n= (e5<\/sup> + 1) 0.02<\/p>\n

Question 3.
\nFind the approximate value of \u221a82. [Mar. ’13; May ’09]
\nSolution:
\nf(x) = \u221ax, x = 81, \u2206x = 1
\n\u2206f = df = \\(\\frac{d f}{d x}\\) . \u2206x
\n= \\(\\frac{1}{2 \\sqrt{81}} \\cdot 1\\)
\n= \\(\\frac{1}{18}\\)
\n= 0.05556
\nNow, \u221a82 = f(x + \u2206x) = \u2206f + f(x)
\n= 0.05556 + 9
\n= 9.05556<\/p>\n

Question 4.
\nFind the approximate value of \\(\\sqrt[3]{65}\\).
\nSolution:
\nLet f(x) = \\(\\sqrt[3]{65}\\), x = 64, \u2206x = 1
\n\"TS
\nf(x) = \\(\\sqrt[3]{x}\\) = \\(\\sqrt[3]{64\\) = 4
\nNow, \\(\\sqrt[3]{65}\\) = f(x + \u2206x) = \u2206f + f(x)
\n= 4 + 0.02083
\n= 4.02083<\/p>\n

\"TS<\/p>\n

Question 5.
\nIf the increase in the side of a square is 2% then find the approximate percentage of increase in its area. [Mar. ’18 (TS); Mar. ’12, ’08; May ’05]
\nSolution:
\n\"TS
\nLet x be the side and A be the area of a square.
\nGiven, that % error in x = 2
\n\\(\\frac{\\Delta \\mathrm{x}}{\\mathrm{x}}\\) \u00d7 100 = 2
\nThe area of a square, A = x2<\/sup>
\n\"TS
\n\u2234 % error in area = 4.<\/p>\n

Question 6.
\nThe diameter of a sphere is measured to be 40 cm. If an error of 0.02 cm is made in it, then find approximate errors in the volume and surface area of the sphere. [Mar. ’13(old), ’09; May ’03]
\nSolution:
\n\"TS
\nLet r, d, s, v be the radius, diameter, surface area, and volume of a sphere.
\nGiven, diameter, d = 40 cm,
\nradius, r = 20 cm
\nerror in diameter, \u2206d = 0.02 cm
\nerror in radius, \u2206r = 0.01 cm
\n(i) Volume of a sphere, V = \\(\\frac{4}{3}\\)\u03c0r3<\/sup>
\nerror in V = \u2206V = \\(\\frac{\\mathrm{dV}}{\\mathrm{dr}}\\) . \u2206r
\n= \\(\\frac{4}{3}\\)\u03c0(3r2<\/sup>) . \u2206r
\n= 4\u03c0r2<\/sup> . \u2206r
\n= 4\u03c0(20)2<\/sup> (0.01)
\n= 16\u03c0 cu. cm
\n(ii) Surface area of a sphere, s = 4\u03c0r2<\/sup>
\nerror in surface area of a sphere \u2206s = \\(\\frac{\\mathrm{ds}}{\\mathrm{dr}}\\) \u2206dr
\n= 4\u03c0(2r) . \u2206r
\n= 8\u03c0(20)(0.01)
\n= 1.6\u03c0 sq. cm
\n\u2234 Approximate error in the volume of sphere = 16\u03c0 cu. cm.
\n\u2234 Approximate error in the area of sphere = 1.6\u03c0 sq. cm.<\/p>\n

Some More Maths 1B Errors and Approximations Important Questions<\/h3>\n

Question 7.
\nIf y = 5x2<\/sup> + 6x + 6, find \u2206y and dy when x = 2, \u2206x = 0.001.
\nSolution:
\nLet y = f(x) = 5x2<\/sup> + 6x + 6, x = 2, \u2206x = 0.001
\n\u2206y = f(x + \u2206x) – f(x)
\n= 5(x + \u2206x)2<\/sup> + 6(x + \u2206x) + 6 – (5x2<\/sup> + 6x + 6)
\n= 5(x2<\/sup> + (\u2206x)2<\/sup> + 2 . x . \u2206x) + 6x + 6(\u2206x) + 6 – 5x2<\/sup> – 6x – 6
\n= 5x2<\/sup> + 5(\u2206x)2<\/sup> + 10 . x . \u2206x + 6(\u2206x) – 5x2<\/sup>
\n= [5(\u2206x) + 10x + 6)] \u2206x
\n= [5(0.001) + 10(2) + 6] (0.001)
\n= [0.005 + 20 + 6] 0.001
\n= 0.026005
\ndy = \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) \u00d7 \u2206x
\n= (5(2x) + 6) \u2206x
\n= (10x + 6) \u2206x
\n= (10 . 2 + 6) 0.001
\n= (26) 0.001
\n= 0.026<\/p>\n

\"TS<\/p>\n

Question 8.
\nFind the approximate value of \\(\\sqrt{\\mathbf{25.001}}\\).
\nSolution:
\nLet f(x) = \u221ax, x = 25, \u2206x = 0.001
\n\"TS
\nf(x) = \u221ax = \u221a25 = 5
\nNow, \\(\\sqrt{\\mathbf{25.001}}\\) = f(x + \u2206x)
\n= \u2206f + f(x)
\n= 0.0001 + 5
\n= 5.0001<\/p>\n

Question 9.
\nIf the increase in the side of a square is 4% then find the approximate percentage of increase in the area of the square.
\nSolution:
\nLet x be the side and A be the area of a square.
\nGiven, that % error in the side = 4
\n\u2234 \\(\\frac{\\Delta x}{x}\\) \u00d7 100 = 4
\nArea of a square, A = x2<\/sup>
\n\u2206A = 2x . \u2206x (\u2235 \u2206A = \\(\\frac{\\mathrm{dA}}{\\mathrm{dx}}\\) . \u2206x)
\nNow, % error in area = \\(\\frac{\\Delta \\mathrm{A}}{\\mathrm{A}}\\) \u00d7 100
\n= \\(\\frac{2 \\mathrm{x} \\cdot \\Delta \\mathrm{x}}{\\mathrm{x}^2}\\) \u00d7 100
\n= \\(\\frac{2 \\cdot \\Delta x}{x}\\) \u00d7 100
\n= 2 \u00d7 4
\n= 8
\n\u2234 % error in area = 8<\/p>\n

Question 10.
\nFind dy and \u2206y of y = f(x) = x2<\/sup> + x, at x = 10 when \u2206x = 0.1. [Mar. ’16 (TS), ’15 (AP); May ’15 (TS); Mar. ’17 (AP & TS)]
\nSolution:
\nGiven, y = f(x) = x2<\/sup> + x, x = 10, \u2206x = 0.1
\n\u2206y = f(x + \u2206x) – f(x)
\n= (x + \u2206x)2<\/sup> + x + \u2206x – (x2<\/sup> + x)
\n= x2<\/sup> + (\u2206x)2<\/sup> + 2 . x . \u2206x + x + \u2206x – x2<\/sup> – x
\n= (\u2206x)2<\/sup> + 2 . x . \u2206x + \u2206x
\n= (2x + 1) \u2206x + (\u2206x)2<\/sup>
\n= (2 . 10 + 1) 0.1 + (0.1)2<\/sup>
\n= (21) (0.1) + 0.01
\n= 2.1 + 0.01
\n= 2.11
\ndy = \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) \u00d7 \u2206x
\n= (2x + 1) \u2206x
\n= (2 . 10 + 1) 0.1
\n= (21) (0.1)
\n= 2.1<\/p>\n

\"TS<\/p>\n

Question 11.
\nFind \u2206y and dy of y = \\(\\frac{1}{x+2}\\), x = 8 and \u2206x = 0.02.
\nSolution:
\nLet y = f(x) = \\(\\frac{1}{x+2}\\), x = 8, \u2206x = 0.02
\n\u2206y = f(x + \u2206x) – f(x)
\n\"TS
\n\"TS<\/p>\n

Question 12.
\nFind \u2206y and dy of y = cos x, x = 60\u00b0, and \u2206x = 1\u00b0. [Mar. ’19 (TS)]
\nSolution:
\nLet y = f(x) = cos x, x = 60\u00b0,
\n\u2206x = 1\u00b0 = 1(0.01745) = 0.01745
\n\u2206y = f(x + \u2206x) – f(x)
\n= cos(x + \u2206x) – cos x
\n= cos (60\u00b0 + 1\u00b0) – cos 60\u00b0
\n= cos 61\u00b0 – cos 60\u00b0
\n= 0.4848 – \\(\\frac{1}{2}\\)
\n= 0.4848 – 0.5
\n= -0.0152
\ndy = \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) . \u2206x
\n= -sin x . \u2206x
\n= -sin 60\u00b0 . (0.01745)
\n= \\(\\frac{-\\sqrt{3}}{2}\\) . (0.01745)
\n= -(0.8660)(0.01745)
\n= -0.015<\/p>\n

Question 13.
\nFind the approximate value of \\(\\sqrt[3]{999}\\). [Mar. ’19 (AP)]
\nSolution:
\n\"TS<\/p>\n

Question 14.
\nFind the approximate value of \\(\\sqrt[3]{7.8}\\).
\nSolution:
\n\"TS<\/p>\n

Question 15.
\nFind the approximate value of sin 62\u00b0.
\nSolution:
\nLet y = f(x) = sin x, x = 60\u00b0,
\n\u2206x = 2\u00b0 = 2(0.01745) = 0.03490
\n\u2206f = df = \\(\\frac{d f}{d x}\\) . \u2206x
\n= cos x . \u2206x
\n= cos 60\u00b0 (0.03490)
\n= \\(\\frac{0.03490}{2}\\)
\n= 0.01745
\nf(x) = sin x = sin 60\u00b0 = 0.8660
\nNow, sin 62\u00b0 = f(x + \u2206x) = f(x) + \u2206f
\n= 0.8660 + 0.01745
\n= 0.88345<\/p>\n

\"TS<\/p>\n

Question 16.
\nFind the approximate value of cos(60\u00b05′).
\nSolution:
\nLet y = f(x) = cos x, x = 60\u00b0,
\n\u2206x = 5′ = \\(\\left(\\frac{5}{60}\\right)^{\\circ}=\\left(\\frac{1}{12}\\right)^{\\circ}\\)
\n= \\(\\frac{1}{12}\\)(0.01745)
\n= 0.001454
\n\u2206f = df = \\(\\frac{\\mathrm{df}}{\\mathrm{dx}}\\) . \u2206x
\n= -sin x . \u2206x
\n= -sin 60\u00b0 . (0.001454)
\n= -0.8660 . (0.001454)
\n= -0.001259
\nf(x) = cos x = cos 60\u00b0 = 0.5
\nNow, cos(60\u00b0 5′) = f(x + \u2206x) = f(x) + \u2206x
\n= 0.5 – 0.001259
\n= 0.498741<\/p>\n

Question 17.
\nFind the approximate value of \\(\\sqrt[4]{17}\\).
\nSolution:
\n\"TS<\/p>\n

Question 18.
\nThe radius of a sphere is measured as 14 cm. Later it was found that there is an error of 0.02 cm in measuring the radius. Find the approximate error in the surface area of the sphere.
\nSolution:
\nLet r be the radius and s be the surface area of the sphere.
\nGiven, radius, r = 14 cm
\nerror in radius, \u2206r = 0.02 cm
\nthe surface area of a sphere, s = 4\u03c0r2<\/sup>
\nerror in surface area of sphere = \u2206s
\n= \\(\\frac{\\mathrm{ds}}{\\mathrm{dr}}\\) . \u2206r
\n= 4\u03c0(2r) . \u2206r
\n= 8\u03c0(14)(0.02)
\n= 2.24\u03c0
\n= 2.24(3.14)
\n= 7.03356 or 7.04 sq.cm.<\/p>\n

Question 19.
\nThe side of a square is increased from 3 cm to 3.01 cm. Find the approximate increase in the area of the square.
\nSolution:
\nLet x be the side and A be the area of a square.
\nGiven that x = 3, \u2206x = 0.01
\nArea of a square, A = x2<\/sup>
\nerror in area = \u2206A = \\(\\frac{\\mathrm{dA}}{\\mathrm{dx}}\\) . \u2206x
\n= 2x . \u2206x
\n= 2.3(0.01)
\n= 6(0.01)
\n= 0.06 sq. cm<\/p>\n

\"TS<\/p>\n

Question 20.
\nIf the radius of a sphere is increased from 7 cm to 7.02 cm then find the approximate increase in the volume of the sphere.
\nSolution:
\nLer r be the radius of a sphere and V be its volume.
\nGiven that, radius r = 7 cm, \u2206r = 0.02 cm
\nVolume of the sphere, V = \\(\\frac{4}{3} \\pi r^3\\)
\nerror in volume = \u2206V = \\(\\frac{\\mathrm{dV}}{\\mathrm{dr}}\\) . \u2206r
\n\"TS<\/p>\n

Question 21.
\nIf y = f(x) = kxn<\/sup> then show that the approximate relative error (or increase) in y is n times the relative error (or increase) in x where n and k are constants.
\nSolution:
\nGiven that y = f(x) = k . xn<\/sup>
\nerror in y = \u2206y = \\(\\frac{d y}{d x}\\) . \u2206x
\n= k . n . xn-1<\/sup> . \u2206x
\nrelative error in y = \\(\\frac{\\Delta \\mathrm{y}}{\\mathrm{y}}\\)
\n\"TS
\n\u2234 The relative error in y is n times the relative error in x.<\/p>\n

Question 22.
\nThe time t, of a complete oscillation of a simple pendulum of length l, is given by t = \\(2 \\pi \\sqrt{\\frac{l}{g}}\\), where g is gravitational constant. Find the approximate percentage of error in ‘t’ when the percentage of error in l is 1%.
\nSolution:
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Errors and Approximations Important Questions to help strengthen their preparations for exams. TS Inter 1st Year Maths 1B Errors and Approximations Important Questions Question 1. If y = x2 + 3x + 6, find \u2206y and dy when x = 10, … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5507"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=5507"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5507\/revisions"}],"predecessor-version":[{"id":5715,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5507\/revisions\/5715"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=5507"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=5507"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=5507"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}