{"id":20112,"date":"2020-01-17T16:04:51","date_gmt":"2020-01-17T10:34:51","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20112"},"modified":"2021-07-02T15:28:51","modified_gmt":"2021-07-02T09:58:51","slug":"2nd-puc-basic-maths-question-bank-chapter-18-ex-18-7","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-18-ex-18-7\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.7"},"content":{"rendered":"

Students can Download Basic Maths Exercise 18.7 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 18.7<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Differential Calculus Ex 18.7 One Mark Questions and Answers<\/strong><\/p>\n

Question 1.
\nFind \\(\\frac{d^{2} y}{d x^{2}}\\).
\n1. y = 3x3<\/sup> + 4x2<\/sup> + 7
\n2. y = \\(\\sqrt{2 x+3}\\)
\n3. y = e3x + 2<\/sup>
\n4. y = x3<\/sup> . logx
\n5. y = log x + ax<\/sup>
\n6. y = e-x<\/sup> sin 2x
\n7. y = log(log x)
\n8. y = cos 4x cos 2x
\n9. y = sin 3x sin 2x
\n10. y = cos mx sin nx
\nAnswer:
\n1. Given y = 3x3<\/sup> + 4x2<\/sup> + 7
\n\\(\\frac{d y}{d x}\\) = 9x2<\/sup> + 8x
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = 18x + 8<\/p>\n

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

\"KSEEB<\/p>\n

3. y = e3x + 2<\/sup>
\n\\(\\frac{d y}{d x}\\) = 3e3x + 2<\/sup>
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = 9e3x + 2<\/sup> = 9y<\/p>\n

4. y = x3<\/sup> logx
\n\\(\\frac{d y}{d x}\\) = x3<\/sup> . \\(\\frac { 1 }{ x }\\) + log x . 3x2<\/sup> = x2<\/sup> + 3x2<\/sup> log x
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = 2x + 3x2 <\/sup>\\(\\frac { 1 }{ x }\\) + log x . 6x = 5x + 6x log x<\/p>\n

5. y = log x + ax<\/sup>
\n\\(\\frac{d y}{d x}\\) = \\(\\frac { 1 }{ x }\\) + ax<\/sup> . log a
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = \\(\\frac{-1}{x^{2}}\\) + log a . ax<\/sup>log a
\n= \\(\\frac{-1}{x^{2}}\\) + ax<\/sup>(log a)2<\/sup><\/p>\n

6. Given y = e-x<\/sup> sin 2x
\n\\(\\frac{d y}{d x}\\) = e-x<\/sup>(2 cos2x) + sin 2x .(-e-x<\/sup>)
\n= e-x<\/sup>( 2 cos2x – sin2x)
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = e-x<\/sup>(-4 sin 2x – 2 cos 2x) + (2 cos 2x – sin 2x) (-e-x<\/sup>)
\n= e-x<\/sup> [-4 sin 2x – 2 cos 2x – 2 cos 2x + sin 2x]
\n= e-x<\/sup> (-3 sin2x – 4 cos 2x)<\/p>\n

\"KSEEB<\/p>\n

7. y = log(log x)
\n\"2nd<\/p>\n

8. y = cos 4x cpos 2x
\ny = \\(\\frac { 1 }{ 2 }\\) (cos 6x + cos 2x)
\n\\(\\frac{d y}{d x}\\) = \\(\\frac { 1 }{ 2 }\\) (-6 sin6x – 2 sin2x) = -3 sin 6x – sin2x
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = -18 cos 6x – 2 cos 2x<\/p>\n

9. Given y = sin 3x sin 2x
\ny = \\(\\frac { 1 }{ 2 }\\) (cos 5x – cos 7x);
\n\\(\\frac{d y}{d x}\\) = \\(\\frac { 1 }{ 2 }\\) (-5 sin 5x + 7 sin 7x)
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = \\(\\frac { 1 }{ 2 }\\) (- 25 cos 5x + 49 cos 7x)<\/p>\n

10. y = cosmx.sinnx
\ny = \\(\\frac { 1 }{ 2 }\\)[sin (m + n)x – sin(m – n)x]
\n\\(\\frac{d y}{d x}\\) = \\(\\frac { 1 }{ 2 }\\)[(m + n)cos(m – n)x – (m – n)cos (m – n)x]
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = \\(\\frac { 1 }{ 2 }\\)[-sin(m + n)x .(m + n)2<\/sup> + (m – n)2<\/sup> . sin (m – n)x]
\n= \\(\\frac { 1 }{ 2 }\\)[-(m + n)2<\/sup> . sin (m + n)x + (m – n)2<\/sup> sin(m – n)x].<\/p>\n

\"KSEEB<\/p>\n

Part – B<\/span><\/p>\n

2nd PUC Basic Maths Differential Calculus Ex 18.7 Two and Three Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nFind \\(\\frac{d^{2} y}{d x^{2}}\\) if.
\n1. x = a cos \u03b8, y = a sin \u03b8
\n2. x = a(\u03b8 + sin\u03b8), y = a(1 – cos \u03b8)
\n3. x = a cos3<\/sup>t, y = -a sin3<\/sup>t
\nAnswer:
\n1. Given x = a cos \u03b8, y = a sin \u03b8
\nDifferentiate both W.r.t. \u03b8
\n\"2nd<\/p>\n

2. Given x = a(\u03b8 + sin\u03b8), y = a(1 – cos \u03b8)
\nDifferentiate both w.r.t \u03b8
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

3. Given x = a cos3<\/sup>t, y = -a sin3<\/sup>t
\nDifferentiate bpth w.r.t t
\n\"2nd<\/p>\n

Question 2.
\nIf y = sin mx, show that \\(\\frac{d^{2} y}{d x^{2}}+m^{2} y=0\\)
\nAnswer:
\nGiven y = sin mx
\n\\(\\frac{d y}{d x}\\) = -m cos mx
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = m2<\/sup> sin mx = -m2<\/sup>y
\n\u2234 \\(\\frac{d^{2} y}{d x^{2}}+m^{2} y=0\\)<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nIf y = 500e7x<\/sup> + 600e-7x<\/sup>, show that \\(\\frac{d^{2} y}{d x^{2}}\\) = 49y.
\nAnswer:
\ny = 500e7x<\/sup> + 600e-7x<\/sup>
\n\\(\\frac{d y}{d x}\\) = 500.7e7x<\/sup> + 600(-7. e-7x<\/sup>)
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = 500(49e7x<\/sup>) + 600(+49 .e-7x<\/sup>)
\n= 49 (5007x<\/sup> + 600e-7x<\/sup>) = 49y
\n\u2234 \\(\\frac{d^{2} y}{d x^{2}}\\) = 49y<\/p>\n

Question 4.
\nIf y = eax<\/sup> + e-ax<\/sup>, Show that y2<\/sub> – a2<\/sup>y = 0
\nAnswer:
\ny = eax<\/sup> + e-ax<\/sup>
\ny1<\/sub> = aeax<\/sup> – ae-ax<\/sup>
\nY2<\/sub> = a2<\/sup>eax<\/sup> + a2<\/sup>e-ax<\/sup>
\n= a2<\/sup>(eax<\/sup> + e-ax<\/sup>) = a2<\/sup>y
\n\u2234 y2<\/sub> – a2<\/sup>y = 0.<\/p>\n

Question 5.
\nIf y = 2 + log x, show that xy2<\/sub> + y1<\/sub> = 0.
\nAnswer:
\nyi<\/sub> = \\(\\frac { 1 }{ x }\\) \u21d2 xy1<\/sub> = 1
\n\u2234 = xy2<\/sub> + y1<\/sub> = 0<\/p>\n

\"KSEEB<\/p>\n

Part – C<\/span><\/p>\n

2nd PUC Basic Maths Differential Calculus Ex 18.7 Five Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nIf x2<\/sup> – xy + y2<\/sup> = a2<\/sup>, show that \\(\\frac{d^{2} y}{d x^{3}}=\\frac{6 a^{2}}{(x-2 y)^{3}}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\nIf x2<\/sup> + 2xy + 3y2<\/sup> = 1, show that y2<\/sup> = \\(\\frac{-2}{(x+3 y)^{3}}\\)
\nAnswer:
\nx2<\/sup> + 2xy + 3y2<\/sup> = 1
\nDifferentiate w.r.t x we get
\n2x + 2(x.y1<\/sub> + y.1) + 6y.y1<\/sub> = 0
\ny1<\/sub>(2x + 6y) = -2x -2y
\n\"2nd
\nHence proved.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nIf y = a cos mx + b sin mx, show that \\(\\frac{d^{2} y}{d x^{2}}+m^{2} y=0\\)
\nAnswer:
\nGiven y = a cos mx + b sin mx
\n\\(\\frac{d y}{d x}\\) = -am sin mx + bm cos mx
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = -am2<\/sup> cos mx – bm2<\/sup> sin mx
\n= m2<\/sup>(a cos mx + b sin mx) = -m2<\/sup>y
\n\u21d2 \\(\\frac{d^{2} y}{d x^{2}}+m^{2} y=0\\)<\/p>\n

Question 4.
\nIf y = a cos (log x) + b sin (log x), show that x2<\/sup>y2<\/sub> + xy1<\/sub> + y = 0
\nAnswer:
\ny = a cos(log x) + b sin (log x)
\n\\(y_{1}=\\frac{-a \\sin (\\log x)}{x}+\\frac{b \\cos (\\log x)}{x}\\)
\nxy1<\/sub> = -a sin (log x) + b cos (log x) Again diff w.r.t x
\nxy2<\/sub> + y1<\/sub>1 = \\(\\frac{-a \\cos (\\log x)}{x}-\\frac{b \\sin (\\log x)}{x}\\)
\nx2<\/sup>y2<\/sub> + xy1<\/sub> = -(a cos logx + b sin (log x)) = =y
\n\u21d2 x2<\/sup>y2<\/sub> + xy1<\/sub> + y = 0<\/p>\n

Question 5.
\nIf y = log (x – \\(\\sqrt{x^{2}+1}\\)), show that (x2<\/sup> + 1)y2<\/sub> + xy1<\/sub> = 0
\nAnswer:
\nIf y = log (x – \\(\\sqrt{x^{2}+1}\\))
\n\"2nd
\n2y1<\/sub> . y2<\/sub>(1 + x2<\/sup>) + y1<\/sub>2<\/sup>(2x) = 0 \u00f7 2y1<\/sub>
\n\u21d2 (1 + x2<\/sup>)y2<\/sub> + xy1<\/sub> = 0.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nIf y = x + \\(\\sqrt{x^{2} – 1}\\) , show that (x2<\/sup> – 1)y2<\/sub> + xy1<\/sub> – y = 0
\nAnswer:
\ny = x + \\(\\sqrt{x^{2} – 1}\\) differentiate w.r.t x
\n\"2nd
\n\u2234 (x2<\/sup> – 1) y1<\/sub>2<\/sup> = y2<\/sup> Differentiate agin w.r.t x
\n(x2<\/sup> – 1)2y1<\/sub>y2<\/sub> + y1<\/sub>2<\/sup> (2x) = 2y .y1<\/sub> (\u00f7 2y1<\/sub> we get)
\n(x2<\/sup> – 1)y2<\/sub> + xy1<\/sub> – y = 0 Hence proved<\/p>\n

Question 7.
\nIf y = \\((x+\\sqrt{x^{2}+1})^{m}\\) , show that (x2<\/sup> + 1)y2<\/sub> + xy1<\/sub> – m2<\/sup>y = 0
\nAnswer:
\ny = \\((x+\\sqrt{x^{2}+1})^{m}\\)
\n\"2nd
\n(x2<\/sup>)y1<\/sub>2<\/sup> = m2<\/sup>y2<\/sup> ; (x2<\/sup> + 1) . 2y1<\/sub>y2<\/sub> + y2<\/sup>1<\/sub>(2x) = m2<\/sup>.2yy1<\/sub> \u00f7 by 2y1<\/sub>
\n(x2<\/sup> + 1)y2<\/sub> + xy1<\/sub> – m2<\/sup> = 0<\/p>\n

Question 8.
\nIf y = sin(log x), show that x2<\/sup>y2<\/sub> + xy1<\/sub> + y = 0\/.
\nAnswers:
\ny = sin (log x)
\ny1<\/sub> = \\(\\frac{\\cos (\\log x)}{x}\\)
\nxy1<\/sub> = cos (log x).
\nAgain differentiate w.r.t. x
\nxy2<\/sub>\u00a0+ y1<\/sub>1 = \\(\\frac{-\\sin (\\log x)}{x}\\)
\nx2<\/sup>y2<\/sub> + xy1<\/sub> = -y
\n\u21d2 x2<\/sup>y2<\/sub> + xy1<\/sub> + y = 0.<\/p>\n

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

Students can Download Basic Maths Exercise 18.7 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 18.7 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\/20112"}],"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=20112"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20112\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}