\u21922+<\/sub> f(x), f is not continuous \nat x = 2 \nHence f is discontinuous at x = 2<\/p>\nQuestion 7. \n\\(f(x)=\\begin{cases} |x|+3,if\\quad x\\le -3 \\\\ -2x,if\\quad -3<x<3 \\\\ 6x+2,if\\quad x\\ge 3 \\end{cases}\\) \nSolution: \nFor x \u2264 – 3, f(x) = – x + 3 \nFor – 3 < x < 3, f(x) = – 2x For x > 3, f(x) = 6x + 2 \nThus for x < – 3, x \u2208 (- 3, 3) and x > 3, \nf(x) is a polynomial function. Hence f(x) is continuous. \n \n\u2234 f is not continuous at x = 3<\/p>\n
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Question 8. \n\\(f(x)=\\begin{cases} \\frac { |x| }{ x } ;x\\neq 0 \\\\ 0;x=0 \\end{cases}\\) \nSolution: \nThe function f can be redefined as \n \nHence the point of discontinuity of f is at x = 0.<\/p>\n
Question 9. \nf(x) = \\(\\begin{cases} \\frac { x }{ |x| } ;if\\quad x<0 \\\\ -1,if\\quad x\\ge 0 \\end{cases}\\) \nSolution: \nMethod I \nFor x < 0, f(x) is the quotient of two continuous functions. Hence f(x) is continuous for x < 0. For x > 0, f(x) is a constant function. Hence\/x) is continuous for x > 0. \nAt x = 0 \n \nHence f has no point of discontinuity.<\/p>\n
Method II \nThe function f can be redefined as \nf(x) = \\(\\begin{cases}-1 & \\text { if } x<0 \\\\ -1 & \\text { if } x \\geq 0\\end{cases}\\) \ni.e., f(x) = – 1 for x \u2208 R \n\u2234 f is a continuous function since f(x) is a constant function. \nHence f has no point of discontinuity.<\/p>\n
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Question 10. \nf(x) = \\(\\begin{cases} x+1,if\\quad x\\ge 1 \\\\ { x }^{ 2 }+1,if\\quad x<1 \\end{cases}\\) \nSolution: \nFor x < 1 and x > 1, f(x) is a polynomial function. Hence f(x) is continuous at x < 1 and x > 1. \n \n\u2234 f is a continuous at x = 1 \nHence f has no point of discontinuity.<\/p>\n
Question 11. \nf(x) = \\(\\begin{cases} { x }^{ 3 }-3,if\\quad x\\le 2 \\\\ { x }^{ 2 }+1,if\\quad x>2 \\end{cases} \\) \nSolution: \nFor x < 2 and x > 2, f(x) is a polynomial function. Hence f(x) is continuous at x < 2 and x > 2. \n \n\u2234 f is a continuous at x = 2 \nHence f has no point of discontinuity.<\/p>\n
Question 12. \nf(x) = \\(\\begin{cases} { x }^{ 10 }-1,if\\quad x\\le 1 \\\\ { x }^{ 2 },if\\quad x>1 \\end{cases} \\) \nSolution: \nFor x < 1 and x > 1, f(x) is a polynomial function. Hence f(x) is continuous at x < 1 and x > 1. \n \n\u2234 f is a continuous at x = 1 \nHence f has no point of discontinuity of f is at x = 1<\/p>\n
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Question 13. \nIs the function defined by f(x) = \\(\\begin{cases} x+5,if\\quad x\\le 1 \\\\ x-5,if\\quad x>1 \\end{cases} \\) a continuous function? \nSolution: \nFor x < 1 and x > 1, f(x) is a polynomial function. Hence f(x) is continuous at x < 1 and x > 1. \nAt x = 1, L.H.L.= limx\u21921-<\/sub> f(x) = limx\u21921-<\/sub> (x + 5) = 6, \nR.H.L. = limx\u21921+<\/sub> f(x) = limx\u21921+<\/sub> (x – 5) = – 4 \nf(1) = 1 + 5 = 6, \nf(1) = L.H.L. \u2260 R.H.L. \nf is not continuous at x = 1 \nAt x = c < 1, limx\u2192c<\/sub> (x + 5) = c + 5 = f(c) \nAt x = c > 1, limx\u2192c<\/sub> (x – 5) = c – 5 = f(c) \n\u2234 f is continuous at all points x \u2208 R except x = 1.<\/p>\nQuestion 14. \nf(x) = \\(\\begin{cases} 3,if\\quad 0\\le x\\le 1 \\\\ 4,if\\quad 1<x<3 \\\\ 5,if\\quad 3\\le x\\le 10 \\end{cases}\\) \nSolution: \nFor x \u2208 (0, 1), x \u2208 (1, 3) and x \u2208 (3, 10), f(x) is a constant function. Hence f(x) is continuous at x \u2208 (0, 1), x \u2208 (1, 3) and x \u2208 (3, 10) \nAt x = 1 \nL.H.L. = lim f(x) = 3, \nR.H.L. = limx\u21921+<\/sub> f(x) = 4 => f is discontinuous at \nx = 1 \nAt x = 3, L.H.L. = limx\u21923-<\/sub> f(x)=4, \nR.H.L. = limx\u21923+<\/sub> f(x) = 5 => f is discontinuous at \nx = 3 \n\u2234 f is not continuous at x = 1 and x = 3.<\/p>\nQuestion 15. \nf(x) = \\(\\begin{cases} 2x,if\\quad x<0 \\\\ 0,if\\quad 0\\le x\\le 1 \\\\ 4x,if\\quad x>1 \\end{cases}\\) \nSolution: \nFor x < 0 and x > 1, f(x) is a polynomial function. \nHence f(x) is continuous at x < 0, x > 1 and x \u2208 (0, 1) \n \n\u2234 f is a continuous at x = 1 \nHence f is not continuous at x = 1<\/p>\n
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Question 16. \n\\(f(x)=\\begin{cases} -2,if\\quad x\\le -1 \\\\ 2x,if\\quad -1<x\\le 1 \\\\ 2,if\\quad x>1 \\end{cases}\\) \nSolution: \nFor x < – 1 and x > 1, f(x) is a constant function. \nFor x \u2208 (-1, 1), f(x) is constant polynomial function. Hence f(x) is continuous at x < – 1, x > 1 and x \u2208 (- 1, 1) \nAt x = – 1, L.H.L. = limx\u21921-<\/sub> f(x) = – 2, f(-1) = – 2, \nR.H.L. = limx\u21921+<\/sub> f(x) = – 2 \n\u2234 f is continuous at x = – 1 \nAt x= 1, L.H.L. = limx\u21921-<\/sub> f(x) = 2,f(1) = 2 \n\u2234 f is continuous at x = 1, \nR.H.L. = limx\u21921+<\/sub> f(x) = 2 \nHence, f is continuous function.<\/p>\nQuestion 17. \nFind the relationship between a and b so that the function f defined by \n\\(f(x)=\\begin{cases} ax+1,if\\quad x\\le 3 \\\\ bx+3,if\\quad x>3 \\end{cases}\\) \nis continuous at x = 3 \nSolution: \n <\/p>\n
Question 18. \nFor what value of \u03bb is the function defined by \n\\(f(x)=\\begin{cases} \\lambda ({ x }^{ 2 }-2x),if\\quad x\\le 0 \\\\ 4x+1,if\\quad x>0 \\end{cases} \\) \ncontinuous at x = 0? What about continuity at x = 1? \nSolution: \nAt x = 0, L.H.L. = limx\u21920-<\/sub> \u03bb (x\u00b2 – 2x) = 0 , \nR.H.L. = limx\u21920+<\/sub> (4x+ 1) = 1, f(0)=0 \nf (0) = L.H.L. \u2260 R.H.L. \n\u2234 f is not continuous at x = 0, \nwhatever value of \u03bb \u2208 R may be \nAt x = 1, limx\u21921<\/sub> f(x) = limx\u21921<\/sub> (4x + l) = f(1) \n\u2234 f is not continuous at x = 0 for any value of \u03bb but f is continuous at x = 1 for all values of \u03bb.<\/p>\nQuestion 19. \nShow that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x. \nSolution: \nLet c be an integer \n \n\u2234 g is discontinuous at c. \nSince c is an integer, g is discontinuous at all integral points.<\/p>\n
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Question 20. \nIs the function defined by f (x) = x\u00b2 – sin x + 5 continuous at x = \u03c0? \nSolution: \n \n\u2234 f is a continuous at x = \u03c0 \nAnother method \nx\u00b2 + 5 and since are continuous functions. \n\u2234 Hence x\u00b2 + 5 – sinx = x\u00b2 – sinx + 5 is continuous. Since sum of continuous func-tions is continuous. \n\u2234 f(x) = 2 – since + 5 is continuous at x = \u03c0<\/p>\n
Question 21. \nDiscuss the continuity of the following functions: \n(a) f (x) = sin x + cos x \n(b) f (x) = sin x – cos x \n(c) f (x) = sin x \u00b7 cos x \nSolution: \n(a) f(x) = sinx + cosx \n <\/p>\n
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Question 22. \nDiscuss the continuity of the cosine, cosecant, secant and cotangent functions. \nSolution: \n(a) Let f(x) = cosx \nAt x = c, c \u2208 R, limx\u2192c<\/sub> cos x = cos c = f(c) \n\u2234 f is continuous for all values of x \u2208 R<\/p>\n(b) Let f(x) = cosecx \nDomain of f is R- {n\u03c0 : n \u2208 Z} \nLet c \u2208 Domain of f \nlimx\u2192c<\/sub> f(x) = limx\u2192c<\/sub> cosec x \n= limx\u2192c<\/sub>\\(\\frac { 1 }{ sinx }\\) = \\(\\frac { 1 }{ sin c }\\) \nSince sinx is continuous \n= cosec c = f(c) \n\u2234 f is continuous at x = c \nSince c is any real number in the domain, the cosecant function is continuous in R except for x = n\u03c0, n \u2208 Z<\/p>\n(c) Let f(x) = secx \nDomain of f is R – \\(\\left\\{(2 n+1) \\frac{\\pi}{2}, n \\in \\mathbf{Z}\\right\\}\\) \nLet c \u2208 Domain of f \nlimx\u2192c<\/sub> f(x) = limx\u2192c<\/sub>sec x = limx\u2192c<\/sub>\\(\\frac { 1 }{ cosx }\\) \n= \\(\\frac { 1 }{ cos c }\\) since cosx is continuous \n= sec c = f(c) \n\u2234 f is continuous at x = c \nSince c is any real number in the domain, the secant function is continuous in R \nexcept for x = (2n +1)\\(\\frac { \u03c0 }{ 2 }\\), n \u2208 Z<\/p>\n(d) Let f(x) = cot x \nDomain of f, R – {n\u03c0, n Z} \nLet c \u2208 Domain of f \n \n\u2234 f is continuous at x = c \nSince c is any real number in the domain, the cotangent function is continuous in R except for x = n\u03c0, n \u2208 Z<\/p>\n
Question 23. \nFind all points of discontinuity of f, where \n\\(f(x)=\\begin{cases} \\frac { sinx }{ x } ,if\\quad x<0 \\\\ x+1,if\\quad x\\ge 0 \\end{cases}\\) \nSolution: \nFor x < 0, sinx and x are continuous functions. \n\u2234 f(x) = \\(\\frac { sin x }{ x }\\) is continuous when x < 0. For x > 0, f(x) = x + 1 is a polynomial function \n\u2234 f(x) = x + 1 is continuous. \nAt x = 0 \n \nf is continuous at x = 0 \nHence f is continuous in R and has no point of discontinuity.<\/p>\n
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Question 24. \nDetermine if f defined by f(x) = \\(\\begin{cases} { x }^{ 2 }sin\\frac { 1 }{ x } ,if\\quad x\\neq 0 \\\\ 0,if\\quad x=0 \\end{cases}\\) is a continuous function? \nSolution: \nAt x = 0 \n <\/p>\n
Question 25. \nExamine the continuity of f, where f is defined by f(x) = \\(\\begin{cases} sinx-cosx,if\\quad x\\neq 0 \\\\ -1,if\\quad x=0 \\end{cases}\\) \nSolution: \nsinx and cosx are continuous functions. Hence sinx – cosx is a continuous function at x \u2260 0. \nAt x = 0 \n \nf is continuous at x = 0 \nThus f is continuous function in R.<\/p>\n
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Question 26. \nf(x) = \\(\\begin{cases} \\frac { k\\quad cosx }{ \\pi -2x } ,\\quad if\\quad x\\neq \\frac { \\pi }{ 2 } \\quad at\\quad x=\\frac { \\pi }{ 2 } \\qquad \\\\ 3,if\\quad x=\\frac { \\pi }{ 2 } \\quad at\\quad x=\\frac { \\pi }{ 2 } \\end{cases} \\) \nSolution: \n <\/p>\n
Question 27. \nf(x) = \\(\\begin{cases} { kx }^{ 2 },if\\quad x\\le 2\\quad at\\quad x=2 \\\\ 3,if\\quad x>2\\quad at\\quad x=2 \\end{cases} \\) \nSolution: \n <\/p>\n
Question 28. \nf(x) = \\(\\begin{cases} kx+1,if\\quad x\\le \\pi \\quad at\\quad x=\\pi \\\\ cosx,if\\quad x>\\pi \\quad at\\quad x=\\pi \\end{cases} \\) \nSolution: \n <\/p>\n
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Question 29. \nf(x) = \\(\\begin{cases} kx+1,if\\quad x\\le 5\\quad at\\quad x=5 \\\\ 3x-5,if\\quad x>5\\quad at\\quad x=5 \\end{cases} \\) \nSolution: \n <\/p>\n
Question 30. \nFind the values of a and b such that the function defined by \nf(x) = \\(\\begin{cases} 5,if\\quad x\\le 2 \\\\ ax+b,if\\quad 2<x<10 \\\\ 21,if\\quad x\\ge 10 \\end{cases} \\) \nto is a continuous function. \nSolution: \nIt is clear that, when x < 2, 2 < x < 10 and x > 10, the function f is continuous. \n \ni.e., 10a + b = 21 …. (2) \nSolving (1) and (2), we get a = 2 and b = 1<\/p>\n
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Question 31. \nShow that the function defined by f(x)=cos (x\u00b2) is a continuous function. \nSolution: \nNow, f (x) = cosx\u00b2, let g (x)=cosx and h (x) x\u00b2 \n\u2234 goh(x) = g (h (x)) = cos x\u00b2 \nNow g and h both are continuous \u2200 x \u2208 R. \nf (x) = goh (x) = cos x\u00b2 is also continuous at all x \u2208 R.<\/p>\n
Question 32. \nShow that the function defined by f (x) = |cos x| is a continuous function. \nSolution: \nLet g(x) =|x|and h (x) = cos x, f(x) = goh(x) = g (h (x)) = g (cosx) = |cos x | \nNow g (x) = |x| and h (x) = cos x both are continuous for all values of x \u2208 R. \n\u2234 (goh) (x) is also continuous. \nHence, f (x) = goh (x) = |cos x| is continuous for all values of x \u2208 R.<\/p>\n
Question 33. \nExamine that sin |x| is a continuous function. \nSolution: \nLet g (x) = sin x, h (x) = |x|, goh (x) = g (h(x)) \n= g(|x|) = sin|x| = f(x) \nNow g (x) = sin x and h (x) = |x| both are continuous for all x \u2208 R. \n\u2234 f (x) = goh (x) = sin |x| is continuous at all x \u2208 R.<\/p>\n
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Question 34. \nFind all the points of discontinuity of f defined by f(x) = |x|-|x+1|. \nSolution: \nf(x) = |x| – |x + 1|. The domain of f is R. \nLet g(x) = |x| and h(x) = x + 1 \nThen g(x) and h(x) are continuous functions. \n(goh)(x) = g(h(x)) = g(x + 1) = |x + 1| \nSince g and h are continuous functions, goh is also continuous. \n\u2234 |x +1| is a continuous function in R. \nHence |x| – |x + 1| is also continuous in R \nsince difference of continuous functions is also continuous. \nThat is, f is continuous in R. \n\u2234 f has no point of discontinuity.<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-1\/ NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.1 Ex 5.1 Class 12 NCERT Solutions Question 1. Prove that the function f (x) = 5x …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1<\/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 5 Continuity and Differentiability Ex 5.1 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n