{"id":19685,"date":"2020-01-13T15:56:27","date_gmt":"2020-01-13T10:26:27","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19685"},"modified":"2021-07-02T15:29:40","modified_gmt":"2021-07-02T09:59:40","slug":"2nd-puc-basic-maths-question-bank-chapter-19-ex-19-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-19-ex-19-2\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.2"},"content":{"rendered":"

Students can Download Basic Maths Exercise 19.2 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 19.2<\/h2>\n

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

2nd PUC Basic Maths Application of Derivatives Ex 19.1 Two of Three Marks Questions and Answers.<\/strong><\/p>\n

Question 1.
\nFind whether the following functions are increasing or decreasing or neither.
\n(i) f(x) = x4<\/sup> – 8x3<\/sup> + 22x2<\/sup> – 24x + 5 at x = 0, -2
\n(ii) f(x) = 4x3<\/sup> – 15x2<\/sup> + 12x – 2 at x = 1,-1
\n(iii) f(x) = (x – 1)(x – 2)2<\/sup> at x = 1,3.
\nAnswer:
\n(i) Given
\nf(x) = x4<\/sup> – 8x3<\/sup> + 22x2<\/sup> – 24x + 5 at x = 0, -2
\nf'(x) = 4x3<\/sup> – 24x2<\/sup> + 44x – 24
\nAt x = 0, f'(0) = -24 < 0 \u2234 f(x) is decreasing at x = 0
\nAt x = -2, f'(-2) = 4(-2)3<\/sup> – 24(-2)2<\/sup> + 44 (-2) – 24 < 0
\n= 32 – 96 – 88 – 24 <0 (negative)
\n\u2234 f(x) is decreasing at x = -2<\/p>\n

\"KSEEB<\/p>\n

(ii) Given
\nf(x) = 4x3<\/sup> – 15x2<\/sup> + 12x – 2 at x = 1,-1
\nf'(x) = 12x2<\/sup> – 30x + 12
\nAt x = 1, f'(x) = 12 – 30 + 12 = -6 < 0
\n\u2234 f(x) is increasing at x = 1
\nAt x = -1, f'(-1) = 12(-1)2<\/sup> – 30 (-1) + 12 = 54 > 0
\n\u2234 f(x) is increasing at x = -1<\/p>\n

(iii) Given
\nf(x] = (x – 1) (x – 2)2<\/sup> at x = 1, 3
\nf'(x) = (x – 1) (2 (x – 2} + (x – 2)2><\/sup>)
\nAt x = 1, f'(1) = 0 + (-1)2<\/sup> = 1 > 0
\n\u2234 f(x) is increasing at x = 1
\nAt x = 3, f'(3) = 2 \u00d7 2(1] + 1 = 5 > 0
\n\u2234 f(x) is increasing at x = 3.<\/p>\n

Question 2.
\nFind the value of x (Interval) for which the function is increasing or decreasing.
\n(i) f(x) = 2x3<\/sup> – 15x2<\/sup> – 84x + 7
\n(ii) f(x) = x4<\/sup> – 2x3<\/sup> + 1
\n(iii) f(x) = x3<\/sup> – 3x2<\/sup> + 3x – 100
\n(iv) f(x) = 2x2<\/sup> – 96x + 5
\n(v) f(x) = 10 – 6x – 2x2<\/sup>
\n(vi) f(x) = 2x3<\/sup> + 9x2<\/sup> + 12x + 20
\nAnswer:
\n(i) Given
\nf(x) = 2x3<\/sup> – 15x2<\/sup> – 84x + 7
\nf'(x) = 6x2<\/sup> – 30x – 84
\n= 6(x2<\/sup> – 5x – 14) = 6(x – 7] (x + 2)
\nf(x) is increasing if f'(x) > 0
\n(x – 7) (x + 2) > 0
\nCase – 1
\nx – 7 > 0 & x + 2 > 0
\nx > 7 & x > -2
\nx > 7 \u21d2 (7, \u221e)<\/p>\n

Case – 2:
\nx – 7 < 0 & x + 2 < 0
\nx < 7 & x < -2
\nx < -2 \u21d2 (-\u221e , -2) U (7,\u221e)
\n\u2234 Interval is (-\u221e , -2) U (7,\u221e)
\nf(x) is decreasing if f'(x) < 0
\n(x – 7) (x + 2)
\nx – 7 < 0 & x + 2 > 0<\/p>\n

Case -1:
\nx < 7 & x > -2
\n\u2234 -2 < x < 7<\/p>\n

Case – 2 x – 7 > 0 & x + 2 < 0
\nx > 7 & x < -2 (not possible)<\/p>\n

\"KSEEB<\/p>\n

(ii) Given f(x) = x4<\/sup> – 2x3<\/sup> + 1
\nf'(x) = 4x3<\/sup> – 6x2<\/sup>
\n= 2x2<\/sup> (2x – 3) Here 2x2<\/sup> > 0
\nf(x) is increasing if f'(x) > 0
\n\u21d2 2x – 3 > 0
\nx > \\(\\frac { 3 }{ 2 }\\) \u21d2 (\\(\\frac { 3 }{ 2 }\\), \u221e) is the interval
\nf(x) is decreasing if f ‘(x) < 0
\n2x – 3 < 0
\nx < \\(\\frac { 3 }{ 2 }\\) (-\u221e, \\(\\frac { 3 }{ 2 }\\)) is the interval`<\/p>\n

(iii)
\nGiven f(x) = x3<\/sup> – 3x2<\/sup> + 3x – 100
\nf'(x) = 3x2<\/sup> – 6x + 3
\n= 3(x2<\/sup> -2x + 1)
\n= 3(x – 1)2<\/sup>
\nf(x) is increasing
\nf ‘(x) > o \u21d2 (x – 1)2<\/sup> > 0
\nincreasing for all
\n\u2235 f'(x) > 0 it will not be decreasing<\/p>\n

(iv) Given f(x) = 2x2<\/sup> – 96x + 5
\nf ‘(x) = 4x – 96 = 4 (x – 24)
\nf(x) is increasing if f’ (x) > 0
\nx – 24 > 0
\nx > 24
\nf(x) is decreasing if f'(x) < 0
\nx – 24 < 0
\nx < 24<\/p>\n

(v) f(x) = 10 – 6x – 2x2<\/sup>
\nf'(x) = -6 -4x = -(6 + 4x) = -(6 + 4x)
\nf(x) is increasing if -(6 + 4x) > 0
\n\u21d2 6 + 4x < 0
\n\u21d2 4x < -6
\n\u21d2 x > –\\(\\frac { 3 }{ 2 }\\)
\nf(x) is decreasing : if -(6 + 4x) < 0 \u21d2 6 + 4x > 0 \u21d2 4x > -6 \u21d2 x > –\\(\\frac { 3 }{ 2 }\\)<\/p>\n

(vi) Given f(x) = 2x3<\/sup> + 9x2<\/sup> + 12x + 20
\nf'(x) = 6x2<\/sup> + 18x + 12
\n= 6(x2<\/sup> + 3x + 2)
\n= 6(x + 2) (x + 1)
\nf(x) is increasing if f ‘(x) > 0
\n(x + 1) (x + 2) > 0<\/p>\n

\"KSEEB<\/p>\n

Case 1:
\nx + 2 > 0 and x + 1 > 0
\nx > -2 and x > -1
\n\u21d2 x > -1 \u21d2 (-1, \u221e)<\/p>\n

Case 2:
\nx + 2 < o & x + 1 < 0
\nx < -2 & x < -1
\nx < -2 \u21d2 (-\u221e, -2)
\nf(x) is decreasing if f ‘(x) < 0
\n(x + 2) (x + 1) < 0<\/p>\n

Case 1:
\nx + 2 < 0 and x + 1 > 0
\nx < -2 and x > -1
\nNot possible<\/p>\n

Case – 2:
\nx + 2 > 0 and x + 1 < 0
\nx > -2 and x < -1
\nx > -2 and x < -1 \u21d2 -2 < x < -1.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 19.2 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 19.2 Part-A 2nd PUC Basic …<\/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\/19685"}],"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=19685"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19685\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19685"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19685"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19685"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}