{"id":29983,"date":"2022-03-29T11:30:32","date_gmt":"2022-03-29T06:00:32","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=29983"},"modified":"2022-03-29T12:04:49","modified_gmt":"2022-03-29T06:34:49","slug":"ncert-solutions-for-class-12-maths-chapter-5-ex-5-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-5-ex-5-1\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> 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\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.1<\/h2>\n

\"NCERT<\/p>\n

Ex 5.1 Class 12 NCERT Solutions Question 1.<\/strong>
\nProve that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
\nSolution:
\n(i) At x = 0. limx\u21920<\/sub> f (x) = limx\u21920<\/sub> (5x – 3) = – 3 and
\nf(0) = – 3
\n\u2234 f is continuous at x = 0<\/p>\n

(ii) At x = – 3, limx\u21923<\/sub> f(x)= limx\u2192-3<\/sub>\u00a0(5x – 3) = – 18
\nand f( – 3) = – 18
\n\u2234 f is continuous at x = – 3<\/p>\n

(iii) At x = 5, limx\u21925<\/sub> f(x) = limx\u21925<\/sub>\u00a0(5x – 3) = 22 and
\nf(5) = 22
\n\u2234 f is continuous at x = 5<\/p>\n

Class 12 Maths Chapter 5 Exercise 5.1 Question 2.<\/strong>
\nExamine the continuity of the function f(x) = 2x\u00b2 – 1 at x = 3.
\nSolution:
\nlimx\u21923<\/sub> f(x) = limx\u21923<\/sub>\u00a0(2x\u00b2 – 1) = 17 and f(3)= 17
\n\u2234 f is continuous at x = 3<\/p>\n

\"NCERT<\/p>\n

Ex.5.1 Class 12 NCERT Solutions Question 3.<\/strong>
\nExamine the following functions for continuity.
\n(a) f(x) = x – 5
\n(b) f(x) = \\(\\\\ \\frac { 1 }{ x-5 } \\), x \u2260 5
\n(c) f(x) = \\(\\frac { { x }^{ 2 }-25 }{ x+5 } \\),x\u22605
\n(d) f(x) = |x – 5|
\nSolution:
\n(a) f(x) = x – 5
\nf(x) is a polynomial.
\nHence f(x) is continuous in R.<\/p>\n

(b) f(x) = \\(\\\\ \\frac { 1 }{ x-5 } \\)
\nf(x) is not defined at x = 5.
\nHence f(x) is not continuous at x = 5.
\nAt x \u2260 5, f(x) is a rational function.
\nHence f(x) is continuous at x \u2260 5.<\/p>\n

(c) f(x) = \\(\\frac { { x }^{ 2 }-25 }{ x+5 } \\)
\nf(x) is not defined at x = – 5.
\nHence f(x) is not continuous at x = – 5.
\nAt x \u2260 – 5, f(x) is a rational function.
\nHence f(x) is continuous at x \u2260 – 5.<\/p>\n

(d) f(x) = |x – 5|
\nf(x) is redefined as
\nf(x) = \\(\\begin{cases}5-x ; & x<5 \\\\ x-5 ; & x \\geq 5\\end{cases}\\)
\nFor x < 5 and x > 5, f(x) is a polynomial function. Hence f(x) is continuous at x < 5 and x > 5.
\nAt x = 5, f(x) is not defined uniquely, so take the left and right hand limits at x = 5
\n\"Ex
\n\u2234 f is continuous at x = 5
\nThus f is continuous since it is continuous at all real values.
\nAnother method
\nLet g(x) = |x| and h(x) = x – 5
\n(goh)(x) = g(h(x)) = g(x – 5)
\n= |x – 5| = f(x) .
\nAlso g(x) and h(x) are continuous functions since the modulus function and the polynomial functions are continuous in R.
\n\u2234 f is a continuous function as it is the com-position of two continuous functions.<\/p>\n

\"NCERT<\/p>\n

Exercise 5.1 Class 12 Maths Ncert Solutions Question 4.<\/strong>
\nProve that the function f (x) = xn<\/sup> is continuous at x = n, where n is a positive integer.
\nSolution:
\nf (x) = xn<\/sup> is a polynomial which is continuous for all x \u2208 R.
\nHence f is continuous at x = n, n \u2208 N.<\/p>\n

Class 12 Maths Exercise 5.1 Question 5.<\/strong>
\nIs the function f defined by \\(f(x)=\\begin{cases} x,ifx\\le 1 \\\\ 5,ifx>1 \\end{cases}\\) continuous at x = 0? At x = 1? At x = 2?
\nSolution:
\n(i) At x = 0
\nlimx\u21920-<\/sub> f(x) = lim\u21920-<\/sub> x = 0 and
\nlimx\u21920+<\/sub> f(x) = lim\u21920+<\/sub> x = 0 => f(0) = 0
\n\u2234 f is continuous at x = 0<\/p>\n

(ii) At x = 1
\nlimx\u21921-<\/sub> f(x) = lim\u21921-<\/sub> (x) = 1 and
\nlimx\u21921+<\/sub> f(x) = lim\u21921+<\/sub>(x) = 5
\n\u2234 limx\u21921-<\/sub> f(x) \u2260 lim\u21921+<\/sub> f(x)
\n\u2234 f is discontinuous at x = 1<\/p>\n

(iii) At x = 2
\nlimx\u21922<\/sub> f(x) = 5, f(2) = 5
\n\u2234 f is continuous at x = 2<\/p>\n

Ex 5.1 Maths Class 12 NCERT Solutions Question 6.<\/strong>
\n\\(f(x)=\\begin{cases} 2x+3,if\\quad x\\le 2 \\\\ 2x-3,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.
\nAt x = 2
\nlimx\u21922+<\/sub> f(x) = lim\u21922+<\/sub> 2x – 3 = 2(2) – 3 = 1
\nlimx\u21922-<\/sub> f(x) = lim\u21922-<\/sub>(x) = 2(2) + 3 = 7
\nlimx\u21922-<\/sub> f(x) \u2260 lim\u21922+<\/sub> f(x), f is not continuous
\nat x = 2
\nHence f is discontinuous at x = 2<\/p>\n

Question 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\"Class
\n\u2234 f is not continuous at x = 3<\/p>\n

\"NCERT<\/p>\n

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\"Ex.5.1
\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\"Exercise
\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

\"NCERT<\/p>\n

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\"Class
\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\"Ex
\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\"NCERT
\n\u2234 f is a continuous at x = 1
\nHence f has no point of discontinuity of f is at x = 1<\/p>\n

\"NCERT<\/p>\n

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>\n

Question 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>\n

Question 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\"NCERT
\n\u2234 f is a continuous at x = 1
\nHence f is not continuous at x = 1<\/p>\n

\"NCERT<\/p>\n

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>\n

Question 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\"NCERT<\/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>\n

Question 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\"NCERT
\n\u2234 g is discontinuous at c.
\nSince c is an integer, g is discontinuous at all integral points.<\/p>\n

\"NCERT<\/p>\n

Question 20.
\nIs the function defined by f (x) = x\u00b2 – sin x + 5 continuous at x = \u03c0?
\nSolution:
\n\"NCERT
\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\"NCERT<\/p>\n

\"NCERT<\/p>\n

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\"NCERT
\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\"NCERT
\nf is continuous at x = 0
\nHence f is continuous in R and has no point of discontinuity.<\/p>\n

\"NCERT<\/p>\n

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\"NCERT<\/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\"NCERT
\nf is continuous at x = 0
\nThus f is continuous function in R.<\/p>\n

\"NCERT<\/p>\n

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\"NCERT<\/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\"NCERT<\/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\"NCERT<\/p>\n

\"NCERT<\/p>\n

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\"NCERT<\/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\"NCERT
\ni.e., 10a + b = 21 …. (2)
\nSolving (1) and (2), we get a = 2 and b = 1<\/p>\n

\"NCERT<\/p>\n

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

\"NCERT<\/p>\n

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<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-5-ex-5-1\/\" \/>\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 5 Continuity and Differentiability Ex 5.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"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. 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