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JEE Main Limits Continuity and Differentiability questions

98 Mathematics questions on Limits Continuity and Differentiability as asked in JEE Main, newest first. 47 come with a worked solution.

  1. Let x = 2 be a root of the equation x^2 + px + q = 0 and f(x) = begincases (1 - cos(x^2 - 4px + q^2 + 8q + 16))/((x - 2p)^4), & x ≠ 2p \ 0, & x = 2p endcases Then lim_x → 2p^- [f(x)], where [·] denotes greatest integer…

    2023 · with solution

  2. Let f(x) = begincases x^2 sin (1/x ) & , x ≠ 0 \ 0 & , x = 0 endcases Then at x = 0

    2023 · with solution

  3. If the function f(x) = begincases (1+|cos x|)(λ)/(|cos x|), & 0 < x < (π)/2 \ μ, & x = (π)/2 \ dfraccot 6xe^cot 4x, & (π)/2 < x < π endcases is continuous at x = (π)/2, then 9λ + 6log_e μ + μ^6 - e^6λ is equal to

    2023

  4. The set of all values of a for which lim_x → a([x-5]-[2x+2])=0, where [α] denotes the greatest integer less than or equal to α is equal to

    2023

  5. Let f:(0,1) → R be a function defined by f(x) = dfrac11-e^-x, and g(x) = (f(-x) - f(x)). Consider two statements (I) g is an increasing function in (0, 1) (II) g is one-one in (0, 1) Then,

    2023 · with solution

  6. If f(x) = x^3 - x^2 f'(1) + xf''(2) - f'''(3), x ∈ R, then

    2023 · with solution

  7. Let y(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^16). Then y' - y'' at x = -1 is equal to:

    2023

  8. The equation x^2 - 4x + [x] + 3 = x[x], where [x] denotes the greatest integer function, has:

    2023 · with solution

  9. The value of lim_n → ∞ dfrac1+2-3+4+5-6+......+(3n-2)+(3n-1)-3n√(2n^4+4n+3)-√(n^4+5n+4) is:

    2023

  10. lim_t → 0 (1^1/(sin^2 t) + 2^1/(sin^2 t) + … + n^1/(sin^2 t) )^sin^2 t is equal to

    2023

  11. Let f: R arrow R be a differentiable function that satisfies the relation f(x+y) = f(x) + f(y) - 1, forall x, y ∈ R. If f'(0) = 2, then |f(-2)| is equal to

    2023 · with solution

  12. lim_x → 0 dfracx + 2sin x√(x^2 + 2sin x + 1) - √(sin^2 x - x + 1) is

    2019 · with solution

  13. If the function f defined on ((π)/6, (π)/3 ) by f(x) = begincases dfrac√(2)cos x - 1cot x - 1, & x ≠ (π)/4 \ k, & x = (π)/4 endcases is continuous, then k is equal to

    2019

  14. Let f: R → R be a continuously differentiable function such that f(2) = 6 and f'(2) = 1/48. If int_6^f(x) 4t^3 dt = (x-2)g(x), then lim_x → 2 g(x) is equal to

    2019

  15. Let f: R → R be a differentiable function satisfying f'(3) + f'(2) = 0. Then lim_x → 0 ((1 + f(3+x) - f(3))/(1 + f(2-x) - f(2)) )^1/x is equal to

    2019

  16. If α and β are the roots of the equation 375x^2 - 25x - 2 = 0, then lim_n → ∞ sum_r=1^n α^r + lim_n → ∞ sum_r=1^n β^r is equal to

    2019

  17. If f: R → R is a differentiable function and f(2) = 6, then lim_x → 2 int_6^f(x) (2t dt)/((x-2)) is

    2019 · with solution

  18. The derivative of tan^-1 ((sin x - cos x)/(sin x + cos x) ), with respect to x/2, where (x ∈ (0, (π)/2 ) ) is

    2019 · with solution

  19. If f(x) = begincases (sin(p+1)x + sin x)/x, & x 0 endcases is continuous at x = 0, then the ordered pair (p, q) is equal to

    2019

  20. If lim_x → 1 (x^2 - ax + b)/(x - 1) = 5, then a + b is equal to

    2019 · with solution

  21. lim_x → 1^- dfrac√(π) - sqrt2sin^-1x√(1-x) is equal to

    2019 · with solution

  22. Let f(x) = 15 - |x - 10|; x ∈ R. Then, the set of all values of x, at which the function, g(x) = f(f(x)) is not differentiable, is

    2019 · with solution

  23. If lim_x → 1 (x^4 - 1)/(x - 1) = lim_x → k (x^3 - k^3)/(x^2 - k^2), then k is

    2019 · with solution

  24. Let [x] denote the greatest integer less than or equal to x. Then, lim_x → 0 (tan(π sin^2 x) + (|x| - sin(x[x]))^2)/(x^2)

    2019

  25. If e^y + xy = e, the ordered pair (dy/dx, (d^2y)/(dx^2) ) at x = 0 is equal to

    2019

  26. If the function f(x) = begincases a|π - x| + 1, x ≤ 5 \ b|x - π| + 3, x > 5 endcases is continuous at x = 5, then the value of a - b is

    2019 · with solution

  27. Let S be the set of all points in (-π, π) at which the function, f(x) = min\sin x, cos x\ is not differentiable. Then, S is a subset of which of the following?

    2019

  28. For each t ∈ R, let [t] be the greatest integer less than or equal to t. Then, lim_x → 1^+ ((1-|x|+sin|1-x|)sin ((π)/2[1-x] ))/(|1-x|[1-x])

    2019

  29. lim_x → 0 dfracsin^2 x√(2) - √(1 + cos x) equals

    2019 · with solution

  30. If f(x) = [x] - [x/4 ], x ∈ R where [x] denotes the greatest integer function, then (a) lim_x → 4+ f(x) exists but lim_x → 4- f(x) does not exist (b) f is continuous at x = 4 (c) Both lim_x → 4- f(x) and lim_x → 4+ f(x)…

    2019

  31. If f(1) = 1, f'(1) = 3, then the derivative of f(f(f(x))) + (f(x))^2 at x = 1 is

    2019

  32. Let K be the set of all values of x, where the function f(x) = sin|x| - |x| + 2(x - π)cos|x| is not differentiable. Then, the set K is equal to

    2019

  33. lim_x → (π)/4 (cot^3 x - tan x)/(cos (x + (π)/4 )) is

    2019 · with solution

  34. Let f: [-1, 3] arrow R be defined as f(x) = begincases |x| + [x], & -1 ≤ x < 1 \ x + |x|, & 1 ≤ x < 2 \ x + [x], & 2 ≤ x ≤ 3 endcases where, [t] denotes the greatest integer less than or equal to t. Then, f is…

    2019

  35. For each x ∈ R, let [x] be the greatest integer less than or equal to x. Then, lim_x → 0^- (x([x] + |x|) sin[x])/(|x|) is equal to

    2019

  36. Let f(x) = begincases -1, & -2 ≤ x < 0 \ x^2 - 1, & 0 ≤ x ≤ 2 endcases and g(x) = |f(x)| + f(|x|). Then, in the interval (-2, 2), g is

    2019

  37. For x > 1, if (2x)^2y = 4e^2x-2y, then (1 + log_e 2x)^2 dy/dx is equal to

    2019 · with solution

  38. lim_x → 0 (xcot(4x))/(sin^2 x cot^2(2x)) is equal to

    2019 · with solution

  39. Let f: R → R be a function defined as f(x) = begincases 5, & if x ≤ 1 \ a + bx, & if 1 < x < 3 \ b + 5x, & if 3 ≤ x < 5 \ 30, & if x ≥ 5 endcases Then, f is

    2019

  40. Let f:(-1, 1) → R be a function defined by f(x) = max\-|x|, -√(1-x^2)\. If K be the set of all points at which f is not differentiable, then K has exactly

    2019 · with solution

  41. If xlog_e(log_e x) - x^2 + y^2 = 4 (y > 0), then dy/dx at x = e is equal to

    2019

  42. lim_y → 0 dfrac√(1 + √(1 + y^4)) - √(2)y^4 \hspace{2cm}

    2019

  43. Let f(x) = begincases max\|x|, x^2\, & |x| ≤ 2 \ 8 - 2|x|, & 2 < |x| ≤ 4 endcases Let S be the set of points in the interval (-4, 4) at which f is not differentiable. Then, S

    2019

  44. Let f: R → R be a function such that f(x) = x^3 + x^2 f'(1) + xf''(2) + f'''(3), x ∈ R. Then, f(2) equals

    2019

  45. Let f be a differentiable function from R to R such that |f(x) - f(y)| ≤ 2|x - y|^3/2, for all x, y ∈ R. If f(0) = 1, then int_0^1 f^2(x) dx is equal to

    2019 · with solution

  46. Let f be a differentiable function such that f'(x) = 7 - 3/4 (f(x))/x, (x > 0) and f(1) ≠ 4. Then, lim_x → 0^+ x f (1/x )

    2019

  47. For each t ∈ R, let [t] be the greatest integer less than or equal to t. Then, lim_x → 0^+ x ( [1/x ] + [2/x ] + … + [15/x ] )

    2018

  48. Let S = \t ∈ R : f(x) = |x - π| · (e^|x| - 1)sin|x|\ is not differentiable at t. Then, the set S is equal to

    2018

  49. lim_x → π/2 (cot x - cos x)/((π - 2x)^3) equals

    2017 · with solution

  50. For x ∈ (0, 1/4 ), if the derivative of tan^-1 (frac6x√(x)1-9x^3 ) is √(x) · g(x), then g(x) equals

    2017 · with solution

  51. Let p = lim_x → 0^+ (1 + tan^2 √(x))^1/2x, then log p is equal to

    2016 · with solution

  52. For x ∈ R, f(x) = |log 2 - sin x| and g(x) = f(f(x)), then

    2016

  53. If g is the inverse of a function f and f'(x) = 1/(1+x^5), then g'(x) is equal to

    2015

  54. lim_x → 0 (sin(π cos^2 x))/(x^2) is equal to

    2014 · with solution

  55. If f and g are differentiable functions in (0, 1) satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then for some c ∈ [0, 1]

    2014 · with solution

  56. lim_x → 0 ((1 - cos 2x)(3 + cos x))/(x tan 4x) is equal to

    2013 · with solution

  57. If y = sec(tan^-1 x), then dy/dx at x = 1 is equal to

    2013 · with solution

  58. Let α(a) and β(a) be the roots of the equation (3√(1+a)-1)x^2-(√(1+a)-1)x+(6√(1+a)-1)=0, where a>-1. Then lim_a→ 0^+α(a) and lim_a→ 0^+β(a) are

    2012

  59. If lim_x → ∞ ( (x^2 + x + 1)/(x + 1) - ax - b ) = 4, then

    2012 · with solution

  60. If lim_x → 0 [1 + x log(1 + b^2)]^1/x = 2bsin^2θ, b > 0 and θ ∈ (-π, π], then the value of θ is

    2011 · with solution

  61. Let g(x) = ((x-1)^n)/(logcos^m(x-1)); 0 0 and let p be the left hand derivative of |x-1| at x = 1. If lim_x → 1^- g(x) = p, then

    2008

  62. (d^2x)/(dy^2) equals

    2007 · with solution

  63. Let f(x) be differentiable on the interval (0, ∞) such that f(1) = 1, and lim_t → x (t^2 f(x) - x^2 f(t))/(t - x) = 1 for each x > 0. Then, f(x) is

    2007

  64. For x > 0, lim_x → 0 [(sin x)^1/x + (1/x )^sin x ]

    2006

  65. If f''(x) = -f(x), where f(x) is a continuous double differentiable function and g(x) = f'(x). If F(x) = \f (x/2 ) \^2 + \g (x/2 ) \^2 and F(5) = 5, then F(10) is

    2006 · with solution

  66. If f is a differentiable function satisfying f (1/n ) = 0, forall n ≥ 1, n ∈ I, then

    2005

  67. Let f be twice differentiable function satisfying f(1) = 1, f(2) = 4, f(3) = 9, then

    2005

  68. Let f(x) = ||x| - 1|, then points where, f(x) is not differentiable is/are

    2005

  69. If y is a function of x and log(x + y) = 2xy, then the value of y'(0) is

    2004 · with solution

  70. lim_h → 0 (f(2h + 2 + h^2) - f(2))/(f(h - h^2 + 1) - f(1)), given that f'(2) = 6 and f'(1) = 4.

    2003 · with solution

  71. Let f: R → R be such that f(1) = 3 and f'(1) = 6. Then, lim_x → 0 [(f(1+x))/(f(1)) ]^1/x equals

    2002

  72. The integer n for which lim_x → 0 ((cos x - 1)(cos x - e^x))/(x^n) is a finite non-zero number, is

    2002 · with solution

  73. The domain of the derivative of the functions f(x) = begincases tan^-1x, & if |x| ≤ 1 \ 1/2(|x|-1), & if |x| > 1 endcases is

    2002

  74. Which of the following functions is differentiable at x = 0?

    2001

  75. The left hand derivative of f(x) = [x] sin(π x) at x = k, k is an integer, is

    2001 · with solution

  76. Let f: R arrow R be a function defined by f(x) = max\x, x^3\. The set of all points, where f(x) is not differentiable, is

    2001 · with solution

  77. If x^2 + y^2 = 1, then

    2000 · with solution

  78. The function f(x) = [x]^2 - [x^2] (where, [x] is the greatest integer less than or equal to x), is discontinuous at

    1999

  79. lim_x → 0 (x tan 2x - 2x tan x)/((1 - cos 2x)^2) is

    1999 · with solution

  80. The function f(x) = (x^2 - 1)|x^2 - 3x + 2| + cos(|x|) is not differentiable at

    1999 · with solution

  81. lim_x → 1 dfrac√(1 - cos 2(x-1))x-1

    1998

  82. Let f(x) = beginvmatrix x^3 & sin x & cos x \ 6 & -1 & 0 \ p & p^2 & p^3 endvmatrix, where p is constant. Then (d^3)/(dx^3) f(x) at x = 0 is

    1997

  83. Let [.] denotes the greatest integer function and f(x) = [tan^2 x], then

    1993

  84. The value of lim_x → 0 dfrac√(1/2(1-cos^2 x))x is

    1991

  85. If f(x) = 1/2 - x - 1, then on the interval [0, π]

    1989

  86. If y^2 = P(x) is a polynomial of degree 3, then 2d/dx (y^3(d^2y)/(dx^2) ) equals

    1988

  87. The set of all points, where the function f(x) = x/(1+|x|) is differentiable, is

    1987 · with solution

  88. If f(x) = x (√(x) + √(x+1) ), then

    1985 · with solution

  89. If f(x) = begincases (sin[x])/([x]), & [x] ≠ 0 \ 0, & [x] = 0 endcases where, [x] denotes the greatest integer less than or equal to x, then lim_x → 0 f(x) equals hspace2cm

    1985 · with solution

  90. lim_n → ∞ ( 1/(1-n^2) + 2/(1-n^2) + … + n/(1-n^2) ) is equal to

    1984 · with solution

  91. The function f(x) = (log(1 + ax) - log(1 - bx))/x is not defined at x = 0. The value which should be assigned to f at x = 0, so that it is continuous at x = 0, is

    1983 · with solution

  92. If f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2, then the value of lim_x → a (g(x)f(a) - g(a)f(x))/(x - a) is

    1983

  93. If G(x) = -√(25 - x^2), then lim_x → 1 (G(x) - G(1))/(x - 1) has the value

    1983 · with solution

  94. There exists a function f(x) satisfying f(0)=1, f'(0)=-1, f(x)>0, forall x and

    1982

  95. For a real number y, let [y] denotes the greatest integer less than or equal to y. Then, the function f(x) = (tan π [(x-π)])/(1+[x]^2) is \\

    1981

  96. lim_x → (π)/4 dfracint_2^sec^2 x f(t) dtx^2 - (π^2)/16 equals

  97. lim_n → ∞ 1/n sum_r=1^2n fracr√(n^2 + r^2) equals

    with solution

  98. Let f(x) = (1 - x(1 + |1 - x|))/(|1 - x|) cos (1/(1 - x) ) for x ≠ 1 Then (a) lim_x → 1^- f(x) = 0 (b) lim_x → 1^- f(x) does not exist (c) lim_x → 1^+ f(x) = 0 (d) lim_x → 1^+ f(x) does not exist

    with solution

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