{"id":20100,"date":"2020-01-17T15:06:20","date_gmt":"2020-01-17T09:36:20","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20100"},"modified":"2021-07-02T15:28:52","modified_gmt":"2021-07-02T09:58:52","slug":"2nd-puc-basic-maths-question-bank-chapter-18-ex-18-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-18-ex-18-5\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.5"},"content":{"rendered":"

Students can Download Basic Maths Exercise 18.5 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 18.5<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Differential Calculus Ex 18.5 Two or Three Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\n\\(x^{\\sqrt{x}}\\)
\nAnswer:
\nLet y = \\(x^{\\sqrt{x}}\\) taking logm<\/sup> both sides
\nlog y = \\(\\sqrt{x} \\cdot \\log x\\)
\n\"2nd<\/p>\n

Question 2.
\nxsin x<\/sup>
\nAnswer:
\nLet y = xsin x<\/sup>, taking logm<\/sup> both sudes
\nlog y = sinx log x, Diff w.r.t x
\n\\(\\frac{1}{y} \\cdot \\frac{d y}{d x}\\) = sin x . \\(\\frac{1}{x}\\) + log x . cos x
\n\\(\\frac{d y}{d x}\\) = y[\\(\\frac{\\sin x}{x}\\) + log x . cos x]<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\n(sin x)x<\/sup>.
\nAnswer:
\nLet y = (sin x)x<\/sup>, Taking logm<\/sup>
\nlog y = x log sin x
\n\\(\\frac{1}{y} \\frac{d y}{d x}\\) = x . \\(\\frac{1}{\\sin x}\\) . cos x + logsin x.1
\n\\(\\frac{d y}{d x}\\) = y[x cot x + logsin x]<\/p>\n

Question 4.
\nx5+log x<\/sup>
\nAnswer :
\nLet y = x5+log x<\/sup>, Taking logm<\/sup> both sides
\nlog y = (5 + log x). log x,
\nDifferentiate W.r.t x
\n\\(\\frac{1}{y} \\cdot \\frac{d y}{d x}\\) = (5 + log x) . \\(\\frac{1}{x}\\) + log(x) . \\(\\frac{1}{x}\\)
\n\\(\\frac{d y}{d x}\\) = y \\(\\left[\\frac{5+\\log x+\\log x}{x}\\right]\\)
\n= y\\(\\left[\\frac{5+2 \\log x}{x}\\right]\\)<\/p>\n

Question 5.
\nx(sinx – cosx)<\/sup>.
\nAnswer:
\nLet y = x(sinx – cosx)<\/sup> ; log y = (sin x – cos x) log x diff w.r.t x.
\n\\(\\frac{1}{y} \\frac{d y}{d x}\\) = (sinx – cosx) . \\(\\frac{1}{x}\\) + log x (cos x + sin x);
\n\\(\\frac{d y}{d x}\\) = y [\\(\\frac{\\sin x-\\cos x}{x}\\) + log x.(cos x + sin x)]<\/p>\n

\"KSEEB<\/p>\n

Part – B<\/span><\/p>\n

2nd PUC Basic Maths Differential Calculus Ex 18.5 Five Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nxlogx<\/sup> + (log x)x<\/sup>
\nAnswer:
\nLet y = xlogx<\/sup> + (log x)x<\/sup>
\ny = u + v
\n\\(\\frac{d y}{d x}=\\frac{d u}{d x}+\\frac{d v}{d x}\\) …(1)
\nWhere u = x and v = (log x)x<\/sup>
\nTaking logm<\/sup> both sides
\nlog u = log x . log x log v = x log (log x)
\nDifferentiate both w.r.t. x
\n\"2nd<\/p>\n

Question 2.
\nx2<\/sup> . ex2<\/sup><\/sup> . log x.
\nAnswer:
\nLet y = x2<\/sup> . ex2<\/sup><\/sup> . log x. taking logm<\/sup> both sides
\nlog y = log(x2<\/sup> . ex2<\/sup><\/sup> . log x) log y = log (x2<\/sup> . ex2<\/sup><\/sup> . log x) differentiate w.r.t x
\nlog y = log x2<\/sup> + log ex2<\/sup><\/sup> + log(log x)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\n(x + 1)2<\/sup> (x + 2)3<\/sup> (x + 3)4<\/sup>.
\nAnswer:
\nLet y = (x + 1)2<\/sup> (x + 2)3<\/sup> (x + 3)4<\/sup>, taking logm<\/sup>
\nlog y = log(x + 1)2<\/sup> + (x + 2)3<\/sup> + (x + 3)4<\/sup>
\nlog y = 2 log (x + 1) + 3 log (x + 2) + 4 log (x + 3)
\nDifferentiate w.r.t x
\n\"2nd<\/p>\n

Question 4.
\nx2x<\/sup> + xx2<\/sup><\/sup>.
\nAnswer:
\nLet y = x2x<\/sup> + xx2<\/sup><\/sup>
\ny = u + v
\n\\(\\frac{d y}{d x}=\\frac{d u}{d x}+\\frac{d v}{d x}\\) ….(1)
\nWhere u = x2x<\/sup> and v = xx2<\/sup><\/sup>
\nTaking logm<\/sup> both sides
\nlog u = 2x logx logv = x2<\/sup> log x
\nDifferentiate both w.r.t x
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\n\\(=x^{\\left(1+\\frac{1}{x}\\right)}+\\left(1+\\frac{1}{x}\\right)^{x}\\).
\nAnswer:
\nLet y = \\(=x^{\\left(1+\\frac{1}{x}\\right)}+\\left(1+\\frac{1}{x}\\right)^{x}\\) y = u + v \\(\\frac{d y}{d x}=\\frac{d u}{d x}+\\frac{d v}{d x}\\)
\nWhere u = \\(x^{1+\\frac{1}{x}}\\) v = \\(\\left(1+\\frac{1}{x}\\right)^{x}\\)
\nTaking logm<\/sup> both sides
\n\"2nd<\/p>\n

Question 6.
\n\\(\\sqrt{\\frac{\\sqrt{(x-1)(x-2)}}{(x-3)(x-4)(x-5)}}\\)
\nAnswer:
\nLet y = \\(\\sqrt{\\frac{\\sqrt{(x-1)(x-2)}}{(x-3)(x-4)(x-5)}}\\)
\nTaking logm<\/sup> both sides
\nlog y = \\(\\frac { 1 }{ 2 }\\) [log (x – 1) + log (x – 2) – log (x – 3) – log (x – 4) – log (x – 5)]
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nx3<\/sup> . e2x<\/sup> . sec2<\/sup>x.
\nAnswer:
\nLet y = x3<\/sup> . e2x<\/sup> . sec2<\/sup>x.
\nTaking logm<\/sup> boyh sides
\nlog y = log x3<\/sup> + log 2x<\/sup> + log sec2<\/sup>x
\nDifferentiate w.r.t x \u2192 log y = 3 log x + 2x log e + 2 log sec x
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 18.5 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 18.5 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\/20100"}],"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=20100"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20100\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20100"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20100"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20100"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}