{"id":19826,"date":"2020-01-14T17:25:04","date_gmt":"2020-01-14T11:55:04","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19826"},"modified":"2021-07-02T15:28:58","modified_gmt":"2021-07-02T09:58:58","slug":"2nd-puc-basic-maths-question-bank-chapter-18-ex-18-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-18-ex-18-3\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3"},"content":{"rendered":"

Students can Download Basic Maths Exercise 18.3 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 18 Differential Calculus Ex 18.3<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Differential Calculus Ex 18.3 One or Two Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\n3x2<\/sup> + 4y2<\/sup> = 10
\nAnswer:
\nGiven 3x2<\/sup> + 4y2<\/sup> = 10
\nDiff w.r.t x
\n6x + 8y \\(\\frac{d y}{d x}\\) = 0
\n\\(\\Rightarrow \\quad \\frac{d y}{d x}=\\frac{-8 y}{6 x}=\\frac{-4 y}{3 x}\\)<\/p>\n

Question 2.
\n\\(\\sqrt{x}+\\sqrt{y}=3\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\ny2<\/sup> = 4ax.
\nAnswer:
\nGiven y2<\/sup> = 4ax.
\nDifferentiate with respect to x
\n2y \\(\\frac{d y}{d x}\\) = 4a.1 \u21d2 \\(\\frac{d y}{d x}\\) = \\(\\frac{4 a}{2 y}=\\frac{2 a}{y}\\)<\/p>\n

Question 4.
\n\\(x^{\\frac{2}{3}}+y^{\\frac{2}{3}}=a^{\\frac{2}{3}}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 5.
\nx2<\/sup> = 4ay
\nAnswer:
\nGiven x2<\/sup> = 4ay
\nDifferentiate with respect to x, 2x = 4a \\(\\frac{d y}{d x}\\) \u21d2 \\(\\frac{d y}{d x}\\) = \\(\\frac{2 x}{4 a}=\\frac{x}{2 a}\\)<\/p>\n

Question 6.
\n\\(\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 7.
\nx3<\/sup> + y3<\/sup> = 3axy
\nAnswer:
\nGiven x3<\/sup> + y3<\/sup> = 3axy
\n3x2<\/sup> + 3y2<\/sup> \\(\\frac{d y}{d x}\\) = 3a \\(\\left(x \\cdot \\frac{d y}{d x}+y \\cdot 1\\right)\\)
\n\\(\\frac{d y}{d x}\\) (3y2<\/sup> – 3ax) = 3ay – 3x2<\/sup> = \\(\\frac{d y}{d x}=\\frac{a y-x^{2}}{y^{2}-a x}\\)<\/p>\n

Question 8.
\nx – y = 0
\nAnswer:
\nGiven x – y = 0
\nDifferentiate with respect to x,
\n1 – \\(\\frac{d y}{d x}\\) = 0 \u21d2 \\(\\frac{d y}{d x}\\) = 1<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nx2<\/sup> – y2<\/sup> = a2<\/sup>
\nAnswer:
\nGiven x2<\/sup> – y2<\/sup> = a2<\/sup>
\nDifferentiate with respect to x we get,
\n2x – 2y. \\(\\frac{d y}{d x}\\) = 0 \u21d2 \\(\\frac{d y}{d x}\\) = \\(\\frac{2 x}{2 y}=\\frac{x}{y}\\)<\/p>\n

Question 10.
\nx + \\(\\sqrt{x y}\\) = x2<\/sup>.
\nAnswer:
\n\"2nd<\/p>\n

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

2nd PUC Basic Maths Differential Calculus Ex 18.3 Three marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nlog(xy) = x2<\/sup> + y2<\/sup>
\nAnswer:
\nGiven log(xy) = x2<\/sup> + y2<\/sup>
\nlog x + log y = x2<\/sup> + y2<\/sup> differentiate w.r.t x
\n\"2nd<\/p>\n

Question 2.
\n2x<\/sup> + 2y<\/sup> = 2x+y<\/sup>
\nAnswer:
\nGiven 2x<\/sup> + 2y<\/sup> = 2x+y<\/sup>
\nDifferentiate w.r.t. x we get
\n2x<\/sup> log 2 + 2y<\/sup> log 2 \\(\\frac{d y}{d x}\\)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nxy<\/sup> = yx<\/sup>.
\nAnswer:
\nGiven xy<\/sup> = yx<\/sup>., taking logm<\/sup> both sides
\ny log x = x log y differentiate
\nBoth sides w.r.t x
\n\"2nd<\/p>\n

Question 4.
\nsin xy = cos(x + y).
\nAnswer:
\nGiven sin xy = cos(x + y), diff w.r.t x.
\ncos(xy) \\(\\left[\\mathrm{x} \\frac{\\mathrm{dy}}{\\mathrm{dx}}+\\mathrm{y}\\right]\\)
\n\\(\\frac{d y}{d x}\\) [sin(x + y) + xcos (xy)]
\n= -sin(x + y) – y cos (xy)
\n\\(\\frac{d y}{d x}=\\frac{-[\\sin (x+y)+\\cos x y]}{(\\sin (x+y)+x \\cos x y)}\\)<\/p>\n

Question 5.
\ny = 4x+y<\/sup>
\nAnswer:
\nGiven y = 4x+y<\/sup>, diff. w r.t. x
\n\\(\\frac{d y}{d x}\\) = 4x+y<\/sup> log 4(1 + \\(\\frac{d y}{d x}\\)) = 4x+y<\/sup>
\n\\(\\frac{d y}{d x}\\)(1 – 4x+y<\/sup> log 4) = 4x+y<\/sup> log 4
\n\u2234 \\(\\frac{d y}{d x}=\\frac{4^{x+y} \\cdot \\log 4}{1-4^{x+y} \\cdot \\log 4}\\)<\/p>\n

\"KSEEB<\/p>\n

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

2nd PUC Basic Maths Differential Calculus Ex 18.3 Five Marks Questions and Answers.<\/strong><\/p>\n

Question 1.
\nIf \\(\\sqrt{\\frac{x}{y}}+\\sqrt{\\frac{y}{x}}\\) = a, Prove that x . \\(\\frac{d y}{d x}\\) = y.
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\nIf xy<\/sup> = ey – x<\/sup>, show that \\(\\frac{d y}{d x}\\) = \\(\\frac{2-\\log x}{(1-\\log x)^{2}}\\)
\nAnswer:
\nGiven xy<\/sup> = ey – x <\/sup>. Taking log both sides
\ny log x = (y – x)log ee<\/sup>
\nx = y (1 – log x) \u2235 log ee<\/sup> = 1
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nIf cos y = x cos(a + y). show that \\(\\frac{d y}{d x}\\) = \\(\\frac{\\cos ^{2}(a+y)}{\\sin a}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 4.
\nIf ex<\/sup> = yx<\/sup> show that \\(\\frac{d y}{d x}\\) = \\(\\frac{(\\log y)^{2}}{\\log y-1}\\)
\nAnswer:
\nGiven ex<\/sup> = yx<\/sup> Taking logm<\/sup> both sides
\ny log ee<\/sup> = x log y
\ny = x log y differentiate w.r.t x
\n\"2nd<\/p>\n

Question 5.
\nIf ex+y<\/sup> = xy show that \\(\\frac{d y}{d x}\\) = \\(\\frac{y(1-x)}{x(y-1)}\\)
\nAnswer:
\nGiven yex+y<\/sup> = xy
\nTaking log m both sides
\n(x + y) loge = log(xy)
\nx+y = log x + log y diff w.r.t x
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nIf yx<\/sup> = xy<\/sup> show that \\(\\frac{d y}{d x}\\) = \\(\\frac{y(y=x \\log y)}{x(x-y \\log x)}\\)
\nAnswer:
\nGiven yx<\/sup> = xy<\/sup>, Taking logm<\/sup> both sides
\nx log y = y log x, diff w.r.t x
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 18.3 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 18 Differential Calculus Ex 18.3 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\/19826"}],"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=19826"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19826\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19826"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19826"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19826"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}