{"id":19692,"date":"2020-01-13T17:53:21","date_gmt":"2020-01-13T12:23:21","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19692"},"modified":"2021-07-02T15:29:39","modified_gmt":"2021-07-02T09:59:39","slug":"2nd-puc-basic-maths-question-bank-chapter-19-ex-19-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-19-ex-19-3\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.3"},"content":{"rendered":"

Students can Download Basic Maths Exercise 19.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.3<\/h2>\n

Part- A<\/span><\/p>\n

2nd PUC Basic Maths Application of Derivatives Ex 19.3 Two or Three Marks Questions with Answers.<\/strong><\/p>\n

Question 1.
\nFind the maximum and minimum value of the following function.
\n(i) f(x) = x3<\/sup> – 3x
\n(ii) f(x) = x3<\/sup> – 6x2<\/sup> + 9x + 15(0 \u2264 x \u2264 6)
\n(iii) f(x) = x4<\/sup> – 62x2<\/sup> + 120x + 9
\n(iv) f(x) = 2x3<\/sup> – 3x2<\/sup> – 12x + 12
\n(v) f(x) = 2x3<\/sup> – 3x2<\/sup> – 36x + 10
\n(vi) f(x) = 9x2<\/sup> + 12x + 2
\n(vii) f(x) = 2x3<\/sup> – 15x2<\/sup> + 36x + 10
\n(viii) f(x) = 2x3<\/sup> – 21x2<\/sup> + 36x – 20
\n(ix) f(x) = 2x3<\/sup> – 15x2<\/sup> + 36x + 10
\n(x) f(x) = 12x5<\/sup> – 45x4<\/sup> + 40x3<\/sup> + 6
\nAnswer:
\n(i) Given f(x) = x3<\/sup> – 3x …..(1)
\nf'(x) = 3x2<\/sup> – 3 = 3(x2<\/sup> – 1) = 3(x – 1) (x + 1) …..(2)
\nf'(x) = 0 \u21d2 x = \u00b11
\nf”(x) = 6x – (3)
\nAt x = 1, f”(1) = 6 > 0, f(x) is minimum at x = 1
\n& minimum value is f(1) = 1 – 3 = -2
\nAt x = -1, f”(-1) = -6 < 0 f(x) is maximum at x = -1
\nMaximum value is f(-1) = -1 + 3 = 2<\/p>\n

\"KSEEB<\/p>\n

(ii) f(x) = x3<\/sup> – 6x2 <\/sup>+ 9x + 15 (0 \u2264 x \u2264 6) – (1)
\nf'(x) = 3x2<\/sup> – 12x + 9
\n= 3(x2<\/sup> – 4x + 3) = 3 (x – 3) (x + 1] = 0
\nf “(x) = 6x – 12
\nf'(x) = 0 \u21d2 x = 1 or 3
\nAt x = 1 f”(1) = 6 – 12 = -6 < 0 the function is maximum at x = 1.
\nAnd maximum value is f (1) = 1 – 6 + 9 + 15 = 19
\nAt x = 3, f”(3) = 6 \u00d7 3 – 12 = 18 -12 = 6 > 0, f is minimum at x = 3
\nAnd minimum value is f(3) = 33<\/sup> – 6.32<\/sup> + 9.3 + 15 = 27 – 54 + 27 + 15 = 15.<\/p>\n

(iii) Given f(x) = x4<\/sup> – 62x2<\/sup> + 120x + 9 …..(1)
\nf”(x) = 4x3<\/sup> – 124x + 120 ….(2)
\nf”(x) = 12x2<\/sup> – 124 …..(3)
\nfor a function to be maximum or minimum of f'(x) = 0.
\n\u21d2 x3<\/sup>\u00a0– 31x + 30 = 0 here x = 1 is a root
\n\"2nd
\n\u21d2 x2<\/sup> + x – 30 = 0 \u21d2 (x + 6)(x – 5) = 0 \u21d2 x = 5 – 6
\nPut x = 1 in (3] we get f”(x) = (12 – 124) < 0
\nf(x) att-ains maximum at = 1 & maximum at x = 1 & max value is
\nf(1) = 1 – 62 + 120 + 9 = 68.
\nAt x = 5, f”(5) = 12(5)2<\/sup> 124 = 300 – 124 > 0 f(x) attains
\nminimum at x = 5, & minimum value is f(5) = 625 – 1550 + 600 + 9 = – 316
\nAt x = – 6, f “(-6) = 12 (-6)2<\/sup> – 124 = 432 – 124 > 0
\nf(x) attains minimum at x = -6 & minimum value is
\nf (-6) = (-6)4<\/sup> – 62(-6)2<\/sup> + 120 (-6) + 9
\n= 1296 – 2232 – 720 + 9 = -1647.<\/p>\n

(iv) f(x) = 2x3<\/sup> – 3x2<\/sup> – 12x + 12 ….(1)
\nf'(x) = 6x2<\/sup> – 6x – 12 = 6 (x2<\/sup> – x – 2) …(2)
\nf'(x) = 12x – 6 ….(3)
\nfor a function to be maximum or minimum f ‘(x) = 0 \u21d2 (x – 2) (x + 1) = 0
\n\u21d2 x = 2 or – 1
\nPut x = (-1) in (3) we get f “(-1) = -12 – 6 = -18 < 0
\nf(x) att-ains maximum at x = -1 & maximum value is
\nf(-1) = 2 (-1)3 – 3 (-1)2 – 12(-1) + 12 = 19
\nPut x = 2 in(3) f”(2) = 24 – 6 = 18 > 0
\nf(x) attains minimum at x = 2 & minimum value is
\nf(2) = 2 (2)3<\/sup> – 3(2)2<\/sup> – 12(2) + 12 = 6- ^-24+ >^=8
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(v) Given f(x) = 2x3<\/sup> – 3x2<\/sup> – 36x + 10 ….(1)
\nf ‘(x) = 6x2<\/sup> – 6x – 36 = 6 (x2<\/sup> – x – 6) …..(2)
\nf “(x) = 12x – 6 ……(3)
\nFor a function to be maximum of minimum f ‘(x) = 0
\n\u21d2 (x2<\/sup> – x – 6) = 0 => (x – 3) (x + 2) = 0
\n\u21d2 x = 3 or – 2
\nPut x = 3 in equation (3) we get
\nf “(3) = 36 – 6 = 30 >0
\n\u21d2 f(x) attains minimum at x = 3
\nMinimum value is f(3) = 2(3)3<\/sup> – 3 (32<\/sup>) – 36 (3) + 10
\nf(3) = 54 – 27 – 108 + 10 = -71
\nPut x = -2 in equation (3) we get
\nf”(-2) = -24 – 6 = – 30 < 0 \u21d2 f (x) attains maximum at x = -2
\nMaximum value is f(-2) = 2(-2)3<\/sup> -3(-2)2<\/sup> – 36 (-2] + 10
\nf(-2) -16-12 + 72 + 10 = 54<\/p>\n

(vi) Given f(x) = 9x2<\/sup> + 12x + 2 ….(1)
\nf'(x) = 18x + 12 …(2)
\nf”(x) = 18 > 0 ……(3)
\n\u21d2 f(x) attains minimum
\nf'(x) = 0 \u21d2 18x+ 12 = 0 \u21d2 x = \\(-\\frac{2}{3}\\)
\n& f” \\(\\left(-\\frac{2}{3}\\right)\\) 18 > 0 \u21d2 f(x) is minimum & the minimum value is
\nf\\(\\left(-\\frac{2}{3}\\right)\\) = 9\\(\\left(\\frac{4}{9}\\right)\\) + 12 \\(\\left(-\\frac{2}{3}\\right)\\) + 2
\n= 4 – 8 + 2 = -2<\/p>\n

(vii) f(x) = 2x3<\/sup> – 15x2<\/sup> + 36x + 10 …….(1)
\nf ‘(x) = 6x2<\/sup> – 30x + 36 = 6 (x2<\/sup> – 5x + 6) ……..(2)
\nf”(x) = 12x – 30 …..(3)
\nf'(x) = 0 \u21d2 x2<\/sup> -5x + 6 = 0 \u21d2 (x – 3)(x – 2) = 0 \u21d2 x = 3 or 2
\nwhen x = 3 f “(x) = 12x – 30
\nf “(3) = 36 – 30 = 6 > 0 \u21d2 f(x) has minimum
\nMinimum value is f(3) = 2(3)3<\/sup> – 15(3)2<\/sup> + 36(3) + 10
\nf(3) = 54 – 135 + 108 + 10 = 37
\nwhen x = 2, f”(2) = 24 – 30 = -6 < 0 \u21d2 f(x) has maximum
\nmaximum value is f(2) = 2(2)3<\/sup> – 15(2) + 36(2) + 10
\nf(2) = 16 – 60 + 72 + 10 = 38.<\/p>\n

\"KSEEB<\/p>\n

(viii) Given f(x) = 2x3<\/sup> – 21x2<\/sup> + 36x – 20 ….. (1)
\nf'(x) = 6x2<\/sup> – 42x + 36 …… (2)
\n= 6(x2<\/sup> – 7x + 6)
\nf'(x) = 6 (x – 1) (x – 6) = 0 \u21d2 x = 1 or 6
\nf”(x) = 12x – 42 … (3)
\nwhen x = l,f “(1) = 12 – 42 = -30 < 0 \u21d2 f(x) is maximum
\nmaximum value is f(1) = 2 – 21 + 36 – 20 = -3
\nwhen x = 6, f “(6) = 72 – 42 = 30 > 0 \u21d2 f(x) is minimum
\nminimum value is f(6) = 2(6)33<\/sup> – 21 (6)2<\/sup> + 36(6) – 20
\nf(6) = 432 – 756 + 216 – 20 = -128<\/p>\n

(ix) Given f(x) = 12x5<\/sup> – 45x4<\/sup> + 40 x3<\/sup> + 6 ….(1)
\nf'(x) = 60x4<\/sup> – 180x3<\/sup> + 120x2<\/sup> ….(2)
\n= 60x2<\/sup> (x2<\/sup> – 3x + 2)
\n= 60x2<\/sup> (x – 1) (x – 2)
\nf ‘(x) = 0 \u21d2 60x2<\/sup> (x – 1) (x – 2) = 0 \u21d2 x = 0, 1, 2
\nf “(x) = 60 (4x3<\/sup> – 6x + 4x) ……. (3)
\nwhen x = 0, f”(x) = 0 \u21d2 f(x) has neither maximum nor minimum
\nwhen x = 1, f”(x) = -1 < 0 \u21d2 f(x) has a maximum & maximum value is
\nf(1) = 12 – 45 + 40 + 6 = 13
\nWhen x =2 f”(x) = 4 > 0 \u2234 f(x) has a minimum
\nminimum value is f(2) = 12(32) – 45(16) + 40(8) + 6
\nf(2) = 384 – 720 + 320 + 6 = -10.<\/p>\n

Question 2.
\nThe sum of two natural numbers is 48. Find the numbers when their product is maximum.
\nAnswer:
\nLet the two numbers be x & y.
\nGiven x + y = 48 & product: = xy where y 48 – x.
\nLet p = xy = x (48 – x) = 48x – x2<\/sup>.
\n\\(\\frac{d p}{d x}\\) = 48 – 2x
\n\\(\\frac{d p}{d x}\\) = 0 \u21d2 48 – 2x = 0 x = 24
\n\\(\\frac{\\mathrm{d}^{2} \\mathrm{p}}{\\mathrm{d} \\mathrm{x}^{2}}\\) = -2 < 0 \u21d2 product is maximum
\nx = 24 \u21d2 y = 48 – 24 = 24
\n\u21d2 the two numbers are 24, 24.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nFind two positive numbers whose sum is 14 and the sum of whose square is minimum.
\nAnswer:
\nLet the two numbers be x and y
\nGiven x + y = 14 & S = x2<\/sup> + y2<\/sup> where y = 14 – x
\n\u2234 S = x2<\/sup> + (14 – x)2<\/sup> = x2<\/sup> + 142<\/sup> + x2<\/sup> – 28x = 2x2<\/sup> – 28x + 142<\/sup>
\n\\(\\frac{d s}{d x}\\) = 4x – 28 \u2192 (1) \\(\\frac{d s}{d x}\\) = 0 \u21d2 4x – 28 = 0 \u21d2 x = 7
\n\\(\\frac{\\mathrm{d}^{2} \\mathrm{s}}{\\mathrm{d} \\mathrm{x}^{2}}\\) = 4 > 0, sum is minimum.
\n\u2234 y = 14 – x = 14 – 7 = 7
\n\u2234 the two positive number are 7 & 7.<\/p>\n

Question 4.
\nFind two positive numbers whose sum is 30 and the sum of their cubes is minimum.
\nAnswer:
\nLet the two numbers be x & y.
\nGiven x + y = 30 & S = x3<\/sup> + y3<\/sup> where y = 30 – x
\nS = x3<\/sup> + (30 – x)3<\/sup> = x3<\/sup> + (30)3<\/sup> – x3<\/sup> – 2700x + 90x2<\/sup>
\n\\(\\frac{d s}{d x}\\) = – 2700 + 180x
\n\\(\\frac{d s}{d x}\\) = 0 \u21d2 x = \\(\\frac{2700}{180}\\) = 15
\n\\(\\frac{\\mathrm{d}^{2} \\mathrm{s}}{\\mathrm{d} \\mathrm{x}^{2}}\\) = 180 > 0 \u21d2 sum of cubes is minimum & y = 30 – 15 = 15
\n\u2234 two positive number are 15 & 15.<\/p>\n

Question 5.
\nThe product of two natural numbers is 64. Find the numbers is their sum is minimum
\nAnswer:
\nLet the two numbers be x & y
\nGiven xy = 64 \u21d2 y = \\(\\frac{64}{x}\\)
\nLet s = x + y = x + \\(\\frac{64}{x}\\).
\n\\(\\frac{d s}{d x}\\) = 1 – \\(\\frac{64}{x^{2}}, \\frac{d s}{d x}\\) = 0 \u21d2 x2<\/sup> = 64 \u21d2 x = \u00b18
\n\\(\\frac{d^{2} s}{d x^{2}}=+\\frac{128}{x^{3}}\\)
\nWhen x = 8, \\(\\frac{d^{2} s}{d x^{2}}=\\frac{128}{8^{3}}\\) > 0 \u21d2 s is minimum
\nWhen x = -8, \\(\\frac{d^{2} s}{d x^{2}}=\\frac{128}{8^{3}}\\) < 0 \u21d2 s is maximum
\nThe two numbers are 8 & 8.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 19.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.3 Part- A 2nd PUC …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19692"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=19692"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19692\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19692"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19692"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19692"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}