{"id":31851,"date":"2022-03-29T15:30:38","date_gmt":"2022-03-29T10:00:38","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=31851"},"modified":"2022-03-29T15:41:09","modified_gmt":"2022-03-29T10:11:09","slug":"ncert-solutions-for-class-12-maths-chapter-7-ex-7-4","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-4\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 7 Integrals Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-4\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.4<\/h2>\n

\"NCERT<\/p>\n

Maths Class 12 Chapter 7 Exercise 7.4 Question 1.<\/strong>
\n\\(\\frac { { 3x }^{ 2 } }{ { x }^{ 6 }+1 } \\)
\nSolution:
\nLet x3<\/sup> = t \u21d2 3x\u00b2dx = dt
\n\\(\\int { \\frac { { 3x }^{ 2 } }{ { x }^{ 6 }+1 } dx } =\\int { \\frac { dt }{ { t }^{ 2 }+1 } } ={ tan }^{ -1 }t+C\\)
\n= tan-1<\/sup> (x3<\/sup>) + C<\/p>\n

Question 2.
\n\\(\\frac { 1 }{ \\sqrt { 1+{ 4x }^{ 2 } } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 3.
\n\\(\\frac { 1 }{ \\sqrt { { (2-x) }^{ 2 }+1 } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 4.
\n\\(\\frac { 1 }{ \\sqrt { 9-{ 25x }^{ 2 } } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 5.
\n\\(\\frac { 3x }{ 1+{ 2x }^{ 4 } } \\)
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 6.
\n\\(\\frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } } \\)
\nSolution:
\nput x\u00b3 = t,so that 3x\u00b2dx = dt
\n\\(\\int { \\frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } } dx } \\quad =\\frac { 1 }{ 3 } \\int { \\frac { dt }{ 1-{ t }^{ 2 } } \\quad =\\frac { 1 }{ 6 } log } \\left| \\frac { 1+t }{ 1-t } \\right| +C\\)
\n\\(=\\frac { 1 }{ 6 } log\\left| \\frac { 1+{ x }^{ 3 } }{ 1-{ x }^{ 3 } } \\right| \\) + C<\/p>\n

Question 7.
\n\\(\\frac { x-1 }{ \\sqrt { { x }^{ 2 }-1 } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 8.
\n\\(\\frac { { x }^{ 2 } }{ \\sqrt { { x }^{ 6 }+{ a }^{ 6 } } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 9.
\n\\(\\frac { { sec }^{ 2 }x }{ \\sqrt { { tan }^{ 2 }x+4 } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 10.
\n\\(\\frac { 1 }{ \\sqrt { { x }^{ 2 }+2x+2 } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 11.
\n\\(\\frac { 1 }{ { 9x }^{ 2 }+6x+5 } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 12.
\n\\(\\frac { 1 }{ \\sqrt { 7-6x-{ x }^{ 2 } } } \\)
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 13.
\n\\(\\frac { 1 }{ \\sqrt { (x-1)(x-2) } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 14.
\n\\(\\frac { 1 }{ \\sqrt { 8+3x-{ x }^{ 2 } } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 15.
\n\\(\\frac { 1 }{ \\sqrt { (x-a)(x-b) } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 16.
\n\\(\\frac { 4x+1 }{ \\sqrt { { 2x }^{ 2 }+x-3 } } \\)
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 17.
\n\\(\\frac { x+2 }{ \\sqrt { { x }^{ 2 }-1 } } \\)
\nSolution:
\n\"NCERT<\/p>\n

Question 18.
\n\\(\\frac { 5x-2 }{ 1+2x+{ 3x }^{ 2 } } \\)
\nSolution:
\nLet 5x – 2 = A(6x + 2) + B … (1)
\nEquating the coefficients of x, we get
\n6A = 5 \u21d2 A = \\(\\frac { 5 }{ 6 }\\)
\nEquating the constant terms, we get
\n2A + B = – 2 \u21d2 B = – 2 – 2A
\n= – 2 – \\(\\frac { 5 }{ 3 }\\) = \\(\\frac { – 11 }{ 3 }\\)
\n\u2234 (1) \u2192 5x – 2 = \\(\\frac { 5 }{ 6 }\\) (6x + 2) – \\(\\frac { 11 }{ 3 }\\)
\n\"NCERT<\/p>\n

Question 19.
\n\\(\\frac { 6x+7 }{ \\sqrt { (x-5)(x-4) } } \\)
\nSolution:
\n\u222b\\(\\frac { 6x+7 }{ \\sqrt { (x-5)(x-4) } } \\)dx = \\(\\int \\frac{6 x+7}{\\sqrt{x^{2}-9 x+20}} d x\\)
\nLet 6x + 7 = A\\(\\frac { d }{ dx }\\)(x\u00b2 – 9x + 20) + B
\n6x + 7 = A(2x – 9) + B
\nEquating coefficients of x, we get
\n2A = 6 \u21d2 A = 3
\nEquating the constant terms,
\nwe get 7 = – 9A + B
\n= B = 7 + 9A \u21d2 B = 7 + 27 = 34
\ni.e, 6x + 7 = 3(2x – 9) + 34
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 20.
\n\\(\\frac { x+2 }{ \\sqrt { 4x-{ x }^{ 2 } } } \\)
\nSolution:
\nLet x + 2 = A\\(\\frac { d }{ dx }\\)(4x – x\u00b2) + B
\nx + 2 = A(4 – 2x) + B.
\nEquating coeffleints of x and constants an both sides, we get
\n– 2A = 1 \u21d2 A = \\(\\frac {- 1 }{ 2 }\\) and 4A + B = 2
\n\u21d2 B = 2 – 4A
\n\u21d2 B = 2 – 4\\(\\frac {- 1 }{ 2 }\\) = 4
\ni.e., x + 2 = \\(\\frac {- 1 }{ 2 }\\)(4 – 2x) + 4
\n\"NCERT<\/p>\n

Question 21.
\n\\(\\frac { x+2 }{ \\sqrt { { x }^{ 2 }+2x+3 } } \\)
\nSolution:
\n\u222b\\(\\frac { x+2 }{ \\sqrt { { x }^{ 2 }+2x+3 } } \\)dx
\nLet x + 2 = A\\(\\frac { d }{ dx }\\)(x\u00b2 + 2x + 3) + B
\nx + 2 = A(2x + 2) + B
\nEquating coefficients of x and constant terms on both sides, we get
\n2A = 1 \u21d2 A = \\(\\frac { 1 }{ 2 }\\) and 2A + B = 2
\n\u21d2 B = 2 – 4A
\n\u21d2 B = 2 – 2\\(\\frac { 1 }{ 2 }\\) = 2 – 1 = 1
\ni.e., x + 2 = \\(\\frac { 1 }{ 2 }\\)(2x + 2) + 1
\n\"NCERT<\/p>\n

Question 22.
\n\\(\\frac { x+3 }{ { x }^{ 2 }-2x-5 } \\)
\nSolution:
\nLet x + 3 = A(2x – 1) + B
\nEquating coeffleints of x and constants an both sides, we get 2A = 1 \u21d2 A = \\(\\frac { 1 }{ 2 }\\)
\nEquating the constant terms, we get – 2A + B = 3 \u21d2 B = 4 \u2234 A = \\(\\frac { 1 }{ 2 }\\), B = 4
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 23.
\n\\(\\frac { 5x+3 }{ \\sqrt { { x }^{ 2 }+4x+10 } } \\)
\nSolution:
\nLet 5x + 3 = A\\(\\frac { d }{ dx }\\)(x\u00b2 + 4x + 10) + B
\n5x + 3 = A(2x + 4) + B
\nEquating the coefficients ofx, we get
\n2A = 5 or A = \\(\\frac { 5 }{ 2 }\\)
\nEquating the constant terms, we get
\n4A + B = 3 \u21d2 B = 3 – 4A
\n= 3 – \\(\\frac { 4\u00d75 }{ 2 }\\) = – 7
\n\u2234 5x + 3 = \\(\\frac { 5 }{ 2 }\\)(2x + 4) – 7
\n\"NCERT<\/p>\n

Question 24.
\n\\(\\int { \\frac { dx }{ { x }^{ 2 }+2x+2 } equals } \\)
\n(a) xtan-1<\/sup>(x+1)+c
\n(b) (x+1)tan-1<\/sup>x+c
\n(c) tan-1<\/sup>(x+1)+c
\n(d) tan-1<\/sup>x+c
\nSolution:
\n(b) (x+1)tan-1<\/sup>x+c
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 25.
\n\\(\\int { \\frac { dx }{ \\sqrt { 9x-{ 4x }^{ 2 } } } equals } \\)
\n(a) \\(\\frac { 1 }{ 9 } { sin }^{ -1 }\\left( \\frac { 9x-8 }{ 8 } \\right) +c\\)
\n(b) \\(\\frac { 1 }{ 2 } { sin }^{ -1 }\\left( \\frac { 8x-9 }{ 9 } \\right) +c\\)
\n(c) \\(\\frac { 1 }{ 3 } { sin }^{ -1 }\\left( \\frac { 9x-8 }{ 8 } \\right) +c\\)
\n(d) \\({ sin }^{ -1 }\\left( \\frac { 9x-8 }{ 9 } \\right) +c\\)
\nSolution:
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-4\/ NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.4 Maths Class 12 Chapter 7 Exercise 7.4 Question 1. Solution: Let x3 = t \u21d2 3x\u00b2dx = dt = tan-1 …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-4\/ NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.4 Maths Class 12 Chapter 7 Exercise 7.4 Question 1. 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