{"id":19111,"date":"2020-01-09T16:01:43","date_gmt":"2020-01-09T10:31:43","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19111"},"modified":"2021-07-02T15:29:50","modified_gmt":"2021-07-02T09:59:50","slug":"2nd-puc-maths-question-bank-chapter-5-ex-5-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-5-ex-5-1\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1"},"content":{"rendered":"

Students can Download Maths Chapter 5 Continuity and Differentiability Ex 5.1 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1<\/h2>\n

2nd PUC Maths Continuity and Differentiability NCERT Text Book Questions and Answers Ex 5.1<\/h3>\n

Question 1.
\nProve that the function f (x) = 5x – 3 is Continuous at x = 0, at x = – 3 and at x = 5.
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\nExamine the continuity of the function f (x) = 2x2<\/sup> – 1 at x = 3.
\nAnswer:
\n\"2nd<\/p>\n

Question 3.
\nExamine the following functions for continuity.<\/p>\n

(a) f (x) = x – 5
\nAnswer:
\nLet \u2018a\u2019 be any real, then
\n\"2nd
\nit is continuous at x = a, i.e. f is continuous function as it is continuous at every point of its domain.<\/p>\n

\"KSEEB<\/p>\n

(b)
\n\\((x)=\\frac{1}{x-5}, x \\neq 5\\)
\nAnswer:
\n\"2nd<\/p>\n

(c)
\n\\(f(x)=\\frac{x^{2}-25}{x+5}, x \\neq-5\\)
\nAnswer:
\nHere f(x) is a rational function and its domain D1<\/sub> = R – {-5}
\n\"2nd<\/p>\n

(d) \\(f(x)=|x-5|\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 4.
\nProve that the function f (x) = xn<\/sup> is continuous at x = a, where is a positive integer.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nIt is the function f defined by
\n\\(f(x)=\\left\\{\\begin{array}{l}{x, \\text { if } x \\leq 1} \\\\{5, \\text { if }x>1}\\end{array}\\right.\\)
\ncontinous at x = 0? At x = 1? At x = 2?
\nAnswer:
\n\"2nd<\/p>\n

Question 6.
\nFind all points of discontinuity of f, where f is defined by
\n\\(f(x)=\\left\\{\\begin{array}{l}{2 x+3, \\text { if } x \\leq 2} \\\\{2 x-3, \\text { if } x>2}\\end{array}\\right.\\)
\nAnswer:
\n\"2nd
\nsince LHL \u2260 RHL
\nlimit does not exist and is discontinuous at x = 2 The function is continuous every where
\nexcept x = 2
\n\u2234 x = 2 is the point of discontinuity.<\/p>\n

Question 7.
\n\\(f(x)=\\left\\{\\begin{array}{r}{|x|+3, \\text { if } x \\leq-3} \\\\{-2 x, \\text { if }-3<x<3} \\\\{6 x+2, \\text { if } x \\geq 3}\\end{array}\\right.\\)
\nAnswer:
\n\"2nd
\nsince LHL \u2260 RHL
\nlimit does not exist and hence discontinuous at x = 3
\nHowever we can prove that the function is continuous at
\na < -3, a > 3 and also at -3 < a <3<\/p>\n

Question 8.
\n\\(\\mathbf{f}(\\mathbf{x})=\\left\\{\\begin{array}{l}{\\frac{|\\mathbf{x}|}{\\mathbf{x}}, \\text { if } \\mathbf{x} \\neq \\mathbf{0}} \\\\{\\mathbf{0}, \\text { if } \\mathbf{x}=\\mathbf{0}}\\end{array}\\right.\\)
\nAnswer:
\n\"2nd
\n\"2nd
\nSince LHL \u2260 RHL\u00a0 \u2260 f(0), limit does not exist
\n\u2234 It is discontinuous at x = 0.<\/p>\n

Question 9.
\n\\(f(x)=\\left\\{\\begin{array}{l}{\\frac{x}{|x|}, \\text { if } x<0} \\\\{-1, \\text { if } x \\geq 0}\\end{array}\\right.\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 10.
\n\\(f(x)=\\left\\{\\begin{array}{l}{x+1, \\text { if } x \\geq 1} \\\\{x^{2}+1, \\text { if } x<1}\\end{array}\\right.\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 11.
\n\\(f(x)=\\left\\{\\begin{array}{l}{x^{3}-3, \\text { if } x \\leq 2} \\\\{x^{2}+1, \\text { if } x>1}\\end{array}\\right.\\)
\nAnswer:
\n\"2nd
\nLHL = RHL function is continuous at x = 2
\n\u2234 function is continuous every where
\n\u2234 There is no point of discontinuity<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\n\\(f(x)=\\left\\{\\begin{array}{ll}{x^{10}-1,} & {\\text { if } x \\leq 1} \\\\{x^{2},} & {\\text { if } x>1}\\end{array}\\right.\\)
\nAnswer:
\nThe domain of the function is R
\n\"2nd
\nLHL \u2260 RHL, function is not continuous
\n\u2234 only point of discontinuity is at x = 1<\/p>\n

Question 13.
\nIs the function defined by
\n\\(f(x)=\\left\\{\\begin{array}{l}{x+5, \\text { if } x \\leq 1} \\\\{x-5, \\text { if } x>1}\\end{array}\\right.\\)
\na continuous function ?
\nAnswer:
\nThe domain of the function is R
\nLet a \u2208 R be any arbitrary point on R
\n\"2nd
\nsince LHL \u2260 RHL, function is not continuous.
\nHence the point of discontinuity is x = 1.<\/p>\n

Question 14.
\nDiscuss the continuity of the function f, where f is defined by.
\n\\(f(x)=\\left\\{\\begin{array}{l}{3, \\text { if } 0 \\leq x \\leq 1} \\\\{4, \\text { if } 1<x<3} \\\\{5, \\text { if } 3 \\leq x \\leq 10}\\end{array}\\right.\\)
\nAnswer:
\nHere domain [0,1] U [2] U [3, 10]
\nCase (i): Let a domain and a = 0 The function is defined only on the right of a and the function is 3 for all values right of a hence continuous at a = 0.<\/p>\n

Case (ii): 0 < a < 1
\nThe function take the value 3 for all a between and 1, hence continuous 0 < a < 1 Case<\/p>\n

\"KSEEB<\/p>\n

Case (iii): a = 1
\nThe function is defined only on the left of and it takes the value 3 every where hence continuous when a = 1.<\/p>\n

Case (iv): a = 2 \/ f (a) = f (2) = 4. But it is not defined either in the left of 2 or in the right of 2 hence it can\u2019t be continuous at x = 2.<\/p>\n

Case (v): a = 3
\nThe function is defined only to the right of a and takes the value of 5 through out. Hence it is continuous.<\/p>\n

Case (vi): 3 < a < 10
\nThe function take the value 5 between 3 and 10 and hence continuous when 3 < a < 10.<\/p>\n

Case (vii): a = 10
\nThe function is defined only on the left of 10 and takes the value 5 through out. Hence the function is continuous at a = 10.
\n\u2234 2 is the only point of discontinuity.<\/p>\n

Question 15.
\n\\(f(x)=\\left\\{\\begin{array}{l}{2 x, \\text { if } x<0} \\\\{0, \\text { if } 0 \\leq x \\leq 1} \\\\{4 x, \\text { if } x>1}\\end{array}\\right.\\)
\nAnswer:
\nCase (i) : The domain of the function is R
\n\"2nd
\nf (1) = 0, LHL \u2260 RHL hence discontinuous at x = 1
\nHow ever the function is continuous
\nwhen a < 0, 0 < a < 1 and a > 1
\nhence x = 1 is the only point of discontinuity.<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\n\\(f(x)=\\left\\{\\begin{aligned}-2, & \\text { if } x \\leq-1 \\\\2 x, & \\text { if }-1<x \\leq 1 \\\\2, & \\text { if } x>1\\end{aligned}\\right.\\)
\nAnswer:
\n\"2nd
\nLHL = RHL = f (1), fn<\/sup> is continuous we can also prove that the function is continuous when a < -1, -1 < a < 1 and a > 1. The function is continuous at all point in R.<\/p>\n

Question 17.
\nFind the relationship between a and b so that the function f defined by
\n\\(f(x)=\\left\\{\\begin{array}{l}{a x+1, \\text { if } x \\leq 3} \\\\{b x+3, \\text { if } x>3}\\end{array}\\right.\\)
\nis continuos at x = 3.
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 18.
\nFor what value of \u03bb is the function defined by
\n\\(f(x)=\\left\\{\\begin{array}{ll}{\\lambda\\left(x^{2}-2 x\\right),} & {\\text { if } x \\leq 0} \\\\{4 x+1,} & {\\text { if } x>0}\\end{array}\\right.\\)
\ncontinous at x = 0 ? what about continuity at x = 1?
\nAnswer:
\n\"2nd
\n\u2234 For any value of \u03bb, the function is discontinuous, how ever at x = 1 and x = -1.
\nwe can prove that the function is continuous.<\/p>\n

\"KSEEB<\/p>\n

Question 19.
\nShow that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest less than or equal to x.
\nAns:
\nLet g (x) = h (x) – q (x)
\nwhere h (x) = x and q (x) = [x]
\nh (x) = x, being a polynomial function continuous every where q (x) = [x]
\n\"2nd
\nsince LHL \u2260 RHL, function is not continuous
\n\u2234 q (x) is not continuous
\n\u2234 h (x) – q (x) is not continuous
\n\u2234 g (x) is not continuous
\n\"2nd<\/p>\n

Question 20.
\nIs the function defined by f (x) = x2<\/sup> – sin x + 5 continuous at x = \u03c0?
\nAnswer:
\n\"2nd<\/p>\n

Question 21.
\nDiscuss the continuity of the following functions:
\n(a) f (x) = sin x + cos x
\nhere the domain is R
\nAnswer:
\nLet a \u2208 R,
\n\\(\\underset { x\\longrightarrow a }{ Lt } f(x)=sin\\quad x\\quad =\\underset { x\\longrightarrow a }{ Lt } (sin\\quad x\\quad + cos\\quad x)\\)
\n= sin a + cos a
\n= f (a) = RHL = LHL
\n\u2234 function is continuous everywhere.<\/p>\n

(b) f (x) = sin x – cos x
\nAnswer:
\nHere the domain is R. Let a \u2208 R
\n\\(\\underset { x\\longrightarrow a }{ Lt } f(x)=sin\\quad x\\quad =\\underset { x\\longrightarrow a }{ Lt } (sin\\quad x\\quad – cos\\quad x)\\)
\n= sin a – cos a
\n= f (a) = RHL = LHL
\nfunction is continuous at x = a, a \u2208 R<\/p>\n

\"KSEEB<\/p>\n

(c) f (x) = sin x. cos x
\nAnswer:
\nLet a \u2208 R where the domain is R
\nLt f (x<\/sub>) = Lt gin x cos x = sin a cos a = f (a)
\n\\(\\underset { x\\longrightarrow a }{ Lt } f(x)=sin\\quad x\\quad =\\underset { x\\longrightarrow a }{ Lt } (sin\\quad x\\quad cos\\quad x)\\)
\n= sin a cos a = f(a)
\nhence the function is continuous at x = a, a \u2208 R<\/p>\n

Question 22.
\nDiscuss the continuity of the cosine, cosecant, secant and cotangent functions.
\nAns:
\nCosine: Domain of the function is R.
\nLet x = a + h
\n\\(\\underset { x\\longrightarrow a }{ Lt } f(x)=\\underset { h\\longrightarrow 0 }{ Lt } (a+h)\\)
\n\"2nd
\n\"2nd
\n\u2234 function is continuous every where similarly we can prove that
\nSecant and cotangent are also continuous ^where ever it is defined.<\/p>\n

\"KSEEB<\/p>\n

Question 23.
\nFind all points of discontinuity of f, where
\n\\(\\mathbf{f}(\\mathbf{x})=\\left\\{\\begin{array}{l}{\\frac{\\sin \\mathbf{x}}{\\mathbf{x}}, \\text { if } \\mathbf{x}<\\mathbf{0}} \\\\{\\mathbf{x}+\\mathbf{1}, \\text { if } \\mathbf{x} \\geq \\mathbf{0}}\\end{array}\\right.\\)
\nAnswer:
\nThe domain is R, Let a \u2208 R
\nIt can be proved that the function is continuous when a < 0
\n\"2nd<\/p>\n

Question 24.
\nDetermine if f defined by
\n\\(\\mathbf{f}(\\mathbf{x})=\\left\\{\\begin{array}{ll}{\\mathbf{x}^{2} \\sin \\frac{1}{\\mathbf{x}},} & {\\text { if } \\mathbf{x} \\neq 0} \\\\{\\mathbf{0},} & {\\text { if } \\mathbf{x}=\\mathbf{0}}\\end{array}\\right.\\)
\nis a continuous functions ?
\nAnswer:
\nLet a \u2208 R where R is the domain of f(x)
\n\"2nd
\n\"2nd
\nHence the function is continous when a \u2260 0
\nLet a = 0 f (x) = 0
\n\u2234 function is continuous at x = 0
\n\u2234 f is continuous every where.<\/p>\n

Question 25.
\nExamine the continuity of f, where f is defined by
\n\\(f(x)=\\left\\{\\begin{array}{ll}{\\sin x-\\cos x,} & {\\text { if } x \\neq 0} \\\\{-1,} & {\\text { if } x=0}\\end{array}\\right.\\)
\nAnswer:
\n\"\"<\/p>\n

Find the value of k so that the function f is continuous at the indicated point in Exercises 26 to 29.<\/p>\n

\"KSEEB<\/p>\n

Question 26.
\n\\(f(x)=\\left\\{\\begin{array}{ll}{\\frac{k \\cos x}{\\pi-2 x},} & {\\text { if } x \\neq \\frac{\\pi}{2}} \\\\{3,} & {\\text { if } x=\\frac{\\pi}{2}}\\end{array} \\text { at } x=\\frac{\\pi}{2}\\right.\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 27.
\n\\(f(x)=\\left\\{\\begin{array}{ll}{k x^{2},} & {\\text { if } x \\leq 2} \\\\{3,} & {\\text { if } x>2}\\end{array} \\text { at } x=2\\right.\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 28.
\n\\(f(x)=\\left\\{\\begin{array}{ll}{k x+1,} & {\\text { if } x \\leq \\pi} \\\\{\\cos x,} & {\\text { if } x>\\pi}\\end{array} \\text { at } x=\\pi\\right.\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 29.
\n\\(f(x)=\\left\\{\\begin{array}{ll}{k x+1,} & {\\text { if } x \\leq 5} \\\\{3 x-5,} & {\\text { if } x>5}\\end{array} \\text { at } x=5\\right.\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 30.
\nFind the values of a and b such that the function defined by
\n\\(f(x)=\\left\\{\\begin{aligned}5, & \\text { if } x \\leq 2 \\\\a x+b, & \\text { if } 2<x<10 \\\\21, & \\text { if } x \\geq 10\\end{aligned}\\right.\\)
\nis a continous function.
\nAnswer:
\n\"2nd<\/p>\n

Question 31.
\nShow that the function defined by f (x) = cos (x2<\/sup>) is a continuous function.
\nAnswer:
\nLet h (x) = cos x, g (x) x2<\/sup>, x\u2208 R
\nhog (x) = h [x2<\/sup>] = cos x2<\/sup> = f (x) .
\nBeing polynomial function x2<\/sup> is continuous on R.
\nBeing trigonometric function cos x is continuous on R in the domain.
\nSince h (x) is continuous and g (x) is continuous hog (x) is also continuous on R.
\n\u2234 f (x) is also continuous on R.<\/p>\n

Question 32.
\nShow that the function defined by is a continuous function.
\nAnswer:
\nThe domain of f (x) is R
\nLet h (x) = cos x, g (x) = |x|
\ngoh (x) = g (cos x) = |cos x| = f (x)
\ncos x and |x| are continuous on R as they are trigonometric function and modulus function. g(x) and h (x) are continuous and hence
\ngoh (x) is also continuous, hence |cosx| is continuous on R.<\/p>\n

\"KSEEB<\/p>\n

Question 33.
\nExamine that sin |x| is a continuous function. 3
\nAnswer:
\nLet g (x) = sin x and h (x) = |x|.
\nBoth g (x) and h (x) are continuous on R
\ngoh (x) = g |x| = sin |x| = f (x) is continuous on R.<\/p>\n

Question 34.
\nFind all the points of discontinuity of f defined by f (x) = x – x +1.
\nAnswer:
\nThe domain is R Let g (x) = |x|, and h (x) = |x +1|
\n\u2234 g(x)-h(x)= |x| – |x + 1| =f(x)
\ng (x) and h (x) are continuous on R
\n\u2234 g (x) – h (x) are also continuous on R
\nf (x) is continuous on R.
\n\u2234 So no points of discontinuity present on R.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 5 Continuity and Differentiability Ex 5.1 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.1 2nd PUC Maths …<\/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\/19111"}],"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=19111"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19111\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19111"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19111"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19111"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}