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Limits, Continuity & Differentiability

25 JEE Main questions · Maths

  1. MathsJEE Main 21 Jan 2026, Shift 1·Q10·MCQ·Moderate
    Open

    Limits of indeterminate forms

    Let f:R→(0,∞)f : \mathbb{R} \to (0, \infty) be a twice differentiable function such that f(3)=18f(3) = 18, f′(3)=0f'(3) = 0 and f′′(3)=4f''(3) = 4. Then lim⁡x→1(log⁡e(f(2+x)f(3))18(x−1)2)\lim_{x\to1}\left(\log_e\left(\dfrac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^2}}\right) is equal to:
    1. (1)1
    2. (2)9
    3. (3)2
    4. (4)18
  2. MathsJEE Main 21 Jan 2026, Shift 2·Q3·MCQ·Easy–moderate
    Open

    Derivatives of a polynomial

    Let f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3)f(x) = x^3 + x^2 f'(1) + 2x f''(2) + f'''(3), x∈Rx \in \mathbb{R}. Then the value of f′(5)f'(5) is:
    1. (1)625\dfrac{62}{5}
    2. (2)6575\dfrac{657}{5}
    3. (3)25\dfrac{2}{5}
    4. (4)1175\dfrac{117}{5}
  3. MathsJEE Main 21 Jan 2026, Shift 2·Q21·Numerical·Tough
    Open

    Sandwich theorem with greatest integer

    Let [⋅][\cdot] denote the greatest integer function and f(x)=lim⁡n→∞1n3∑k=1n[k23x]f(x) = \displaystyle\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^{n}\left[\frac{k^2}{3^x}\right]. Then 12∑j=1∞f(j)12\displaystyle\sum_{j=1}^{\infty} f(j) is equal to .
  4. MathsJEE Main 22 Jan 2026, Shift 2·Q5·MCQ·Moderate
    Open

    Limits using series expansion

    If lim⁡x→0e(a−1)x+2cos⁡bx+(c−2)e−xxcos⁡x−log⁡e(1+x)=2\displaystyle\lim_{x\to0}\frac{e^{(a-1)x} + 2\cos bx + (c-2)e^{-x}}{x\cos x - \log_e(1+x)} = 2, then a2+b2+c2a^2 + b^2 + c^2 is equal to:
    1. (1)5
    2. (2)3
    3. (3)7
    4. (4)9
  5. MathsJEE Main 22 Jan 2026, Shift 2·Q15·MCQ·Moderate
    Open

    Discontinuity of greatest integer function

    Let [ ∙ ][\,\bullet\,] denote the greatest integer function, and let f(x)=min⁡{2x,x2}f(x) = \min\left\{\sqrt2 x, x^2\right\}. Let S={x∈(−2,2):the function g(x)=∣x∣[x2] is discontinuous at x}S = \{x \in (-2, 2) : \text{the function } g(x) = \lvert x\rvert\left[x^2\right] \text{ is discontinuous at } x\}. Then ∑x∈Sf(x)\displaystyle\sum_{x\in S} f(x) equals:
    1. (1)2−22 - \sqrt2
    2. (2)26−322\sqrt6 - 3\sqrt2
    3. (3)1−21 - \sqrt2
    4. (4)6−22\sqrt6 - 2\sqrt2
  6. MathsJEE Main 23 Jan 2026, Shift 1·Q9·MCQ·Moderate
    Open

    Continuity of piecewise function

    Let f(x)={ax2+2ax+34x2+4x−3,x≠−32,12b,x=−32,12f(x) = \begin{cases} \dfrac{ax^2 + 2ax + 3}{4x^2 + 4x - 3}, & x \neq -\dfrac32, \dfrac12\\[2ex] b, & x = -\dfrac32, \dfrac12\end{cases} be continuous at x=−32x = -\dfrac32. If f∘f(x)=75f\circ f(x) = \dfrac75, then xx is equal to:
    1. (1)2
    2. (2)1
    3. (3)0
    4. (4)1.4
  7. MathsJEE Main 23 Jan 2026, Shift 2·Q1·MCQ·Easy–moderate
    Open

    Continuity at a point

    If f(x)={a∣x∣+x2−2(sin⁡∣x∣)(cos⁡∣x∣)x,x≠0b,x=0f(x) = \begin{cases} \dfrac{a\lvert x\rvert + x^2 - 2(\sin\lvert x\rvert)(\cos\lvert x\rvert)}{x}, & x \neq 0\\[2mm] b, & x = 0\end{cases} is continuous at x=0x = 0, then a+ba + b is equal to:
    1. (1)1
    2. (2)2
    3. (3)0
    4. (4)4
  8. MathsJEE Main 24 Jan 2026, Shift 1·Q1·MCQ·Moderate
    Open

    Limits using series expansion

    If the function f(x)=ex(etan⁡x−x−1)+log⁡e(sec⁡x+tan⁡x)−xtan⁡x−xf(x) = \dfrac{e^x\left(e^{\tan x - x} - 1\right) + \log_e(\sec x + \tan x) - x}{\tan x - x} is Continuous at x=0x = 0, then the value of f(0)f(0) is equal to
    1. (1)2
    2. (2)23\dfrac23
    3. (3)12\dfrac12
    4. (4)32\dfrac32
  9. MathsJEE Main 24 Jan 2026, Shift 1·Q16·MCQ·Easy–moderate
    Open

    Differentiability of piecewise function

    Let α,β∈R\alpha, \beta \in \mathbb{R} be such that the function f(x)={2α(x2−2)+2βx,x<1(α+3)x+(α−β),x≥1f(x) = \begin{cases} 2\alpha(x^2 - 2) + 2\beta x, & x < 1\\ (\alpha + 3)x + (\alpha - \beta), & x \ge 1\end{cases} be differentiable at all x∈Rx \in \mathbb{R}. Then 34(α+β)34(\alpha + \beta) is equal to
    1. (1)84
    2. (2)48
    3. (3)36
    4. (4)24
  10. MathsJEE Main 24 Jan 2026, Shift 2·Q20·MCQ·Moderate
    Open

    Continuity of piecewise functions

    Let [t][t] denote the greatest integer less than or equal to tt. If the function f(x)={b2sin⁡(π2[π2(cos⁡x+sin⁡x)cos⁡x]),x<0sin⁡x−12sin⁡2xx3,x>0a,x=0f(x) = \begin{cases} b^2\sin\left(\dfrac{\pi}{2}\left[\dfrac{\pi}{2}(\cos x + \sin x)\cos x\right]\right), & x < 0\\[10pt] \dfrac{\sin x - \frac12\sin 2x}{x^3}, & x > 0\\[10pt] a, & x = 0\end{cases} is continuous at x=0x = 0, then a2+b2a^2 + b^2 is equal to
    1. (1)58\dfrac58
    2. (2)916\dfrac{9}{16}
    3. (3)34\dfrac34
    4. (4)12\dfrac12
  11. MathsJEE Main 28 Jan 2026, Shift 1·Q14·MCQ·Moderate
    Open

    Limits using standard expansions

    The value of lim⁡x→0log⁡e(sec⁡(ex)⋅sec⁡(e2x)⋅…⋅sec⁡(e10x))e2−e2cos⁡x\lim_{x\to0}\frac{\log_e\left(\sec(ex)\cdot\sec\left(e^2x\right)\cdot\ldots\cdot\sec\left(e^{10}x\right)\right)}{e^2 - e^{2\cos x}} is equal to
    1. (1)(e10−1)2e2(e2−1)\dfrac{(e^{10} - 1)}{2e^2(e^2 - 1)}
    2. (2)(e20−1)2e2(e2−1)\dfrac{(e^{20} - 1)}{2e^2(e^2 - 1)}
    3. (3)(e20−1)2(e2−1)\dfrac{(e^{20} - 1)}{2(e^2 - 1)}
    4. (4)(e10−1)2(e2−1)\dfrac{(e^{10} - 1)}{2(e^2 - 1)}
  12. Continuity of a limit-defined function

    Let f(x)=lim⁡n→∞(cos⁡πx−x2nsin⁡(x−1)1+x2n(x−1)), x∈R.f(x) = \lim_{n\to\infty}\left(\dfrac{\cos\pi x - x^{2n}\sin(x-1)}{1 + x^{2n}(x-1)}\right),\ x \in \mathbb{R}. Consider the following two statements :

    (I) f(x)f(x) is discontinuous at x=1x = 1.

    (II) f(x)f(x) is continuous at x=−1x = -1.

    Then,

    1. (1)Neither (I) nor (II) is True
    2. (2)Both (I) and (II) are True
    3. (3)Only (II) is True
    4. (4)Only (I) is True
    We corrected a printing error in the original paper

    Source printed “discontinous” and “continous”; source printed the limit as θ→0\theta\to0 with x(2/θ)x^{(2/\theta)} in place of x2nx^{2n} (undefined for x<0x<0, and the one-sided limits in θ\theta differ for 0<∣x∣<10<\lvert x\rvert<1); corrected to the intended n→∞n\to\infty with x2nx^{2n}, the reading the key uses.

  13. MathsJEE Main 2 Apr 2026, Shift 1·Q16·MCQ·Moderate
    Open

    Limits of indeterminate forms

    If lim⁡x→2sin⁡(x3−5x2+ax+b)(x−1−1)log⁡e(x−1)=m\displaystyle\lim_{x\to2}\frac{\sin\left(x^3 - 5x^2 + ax + b\right)}{\left(\sqrt{x-1} - 1\right)\log_e(x-1)} = m, then a+b+ma + b + m is equal to:
    1. (1)5
    2. (2)6
    3. (3)8
    4. (4)10
  14. MathsJEE Main 2 Apr 2026, Shift 2·Q25·Numerical·Easy–moderate
    Open

    Discontinuity of greatest integer function

    The number of points in the interval [2,4][2, 4], at which the function f(x)=[x2−x−12]f(x) = \left[x^2 - x - \dfrac12\right], where [ ⋅ ][\,\cdot\,] denotes the greatest integer function, is discontinuous, is .
  15. MathsJEE Main 4 Apr 2026, Shift 1·Q16·MCQ·Moderate
    Open

    Derivative of inverse trigonometric functions

    If y=tan⁡−1(3cos⁡x−4sin⁡x4cos⁡x+3sin⁡x)+2tan⁡−1(x1+1−x2)y = \tan^{-1}\left(\dfrac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) + 2\tan^{-1}\left(\dfrac{x}{1 + \sqrt{1 - x^2}}\right), then dydx\dfrac{dy}{dx} at x=32x = \dfrac{\sqrt3}{2} is equal to:
    1. (1)3
    2. (2)−1-1
    3. (3)1
    4. (4)2
  16. MathsJEE Main 4 Apr 2026, Shift 1·Q17·MCQ·Moderate
    Open

    Polynomial satisfying a derivative identity

    Let ff be a real polynomial of degree nn such that f(x)=f′(x) f′′(x)f(x) = f'(x)\,f''(x), for all x∈Rx \in \mathbb{R}. If f(0)=0f(0) = 0, then 36(f′(2)+f′′(2)+∫02f(x) dx)36\left(f'(2) + f''(2) + \displaystyle\int_0^2 f(x)\,dx\right) is equal to:
    1. (1)42
    2. (2)46
    3. (3)56
    4. (4)66
  17. MathsJEE Main 4 Apr 2026, Shift 1·Q25·Numerical·Moderate
    Open

    Points of non-differentiability

    The number of points, at which the function f(x)=max⁡{6x,2+3x2}+∣x−1∣cos⁡∣x2−14∣f(x) = \max\{6x, 2 + 3x^2\} + \lvert x - 1\rvert\cos\left\lvert x^2 - \dfrac14\right\rvert, x∈(−π,π)x \in (-\pi, \pi), is not differentiable, is .
    We corrected a printing error in the original paper

    Source printed “fucntion”.

  18. MathsJEE Main 4 Apr 2026, Shift 2·Q22·Numerical·Moderate
    Open

    Continuity and differentiability of modulus functions

    Let f(x)={ex−1,x<0x2−5x+6,x≥0f(x) = \begin{cases} e^{x-1}, & x < 0\\ x^2 - 5x + 6, & x \ge 0\end{cases} and g(x)=f(∣x∣)+∣f(x)∣g(x) = f(\lvert x\rvert) + \lvert f(x)\rvert. If the number of points where gg is not continuous and is not differentiable are α\alpha and β\beta respectively, then α+β\alpha + \beta is equal to .
  19. MathsJEE Main 5 Apr 2026, Shift 1·Q17·MCQ·Moderate
    Open

    Trigonometric limits

    The product of all possible values of α\alpha, for which lim⁡x→0(1−cos⁡(αx)cos⁡((α+1)x)cos⁡((α+2)x)sin⁡2((α+1)x))=2,\lim_{x\to0}\left(\frac{1 - \cos(\alpha x)\cos\big((\alpha+1)x\big)\cos\big((\alpha+2)x\big)}{\sin^2\big((\alpha+1)x\big)}\right) = 2, is:
    1. (1)−2-2
    2. (2)1
    3. (3)−1-1
    4. (4)54\dfrac54
  20. MathsJEE Main 5 Apr 2026, Shift 1·Q19·MCQ·Moderate
    Open

    Functional equation and differentiability

    Let f:R→Rf : \mathbb{R} \to \mathbb{R} be a differentiable function such that f(x+y3)=f(x)+f(y)3for all x,y∈Rf\left(\frac{x + y}{3}\right) = \frac{f(x) + f(y)}{3}\quad\text{for all } x, y \in \mathbb{R} and f′(0)=3f'(0) = 3. Then the minimum value of the function g(x)=3+exf(x)g(x) = 3 + e^x f(x), is
    1. (1)3(e+1e)3\left(\dfrac{e + 1}{e}\right)
    2. (2)3(e−1e)3\left(\dfrac{e - 1}{e}\right)
    3. (3)3−ee\dfrac{3 - e}{e}
    4. (4)3e3e
  21. MathsJEE Main 5 Apr 2026, Shift 2·Q17·MCQ·Moderate
    Open

    Standard trigonometric limits

    Let f(x)=lim⁡y→0(1−cos⁡(xy))tan⁡(xy)y3f(x) = \displaystyle\lim_{y\to0}\frac{(1 - \cos(xy))\tan(xy)}{y^3}. Then the number of solutions of the equation f(x)=sin⁡xf(x) = \sin x, x∈Rx \in \mathbb{R} is:
    1. (1)0
    2. (2)2
    3. (3)3
    4. (4)1
  22. MathsJEE Main 6 Apr 2026, Shift 1·Q17·MCQ·Easy–moderate
    Open

    Limits by series expansion

    The value of lim⁡x→0(x2sin⁡2xx2−sin⁡2x)\displaystyle\lim_{x\to0}\left(\frac{x^2\sin^2x}{x^2 - \sin^2x}\right) is:
    1. (1)2
    2. (2)3
    3. (3)4
    4. (4)6
  23. MathsJEE Main 6 Apr 2026, Shift 2·Q16·MCQ·Tough
    Open

    Evaluation of limits

    Let lim⁡x→2(tan⁡(x−2))(rx2+(p−2)x−2p)(x−2)2=5\lim_{x\to2}\dfrac{\left(\tan(x-2)\right)\left(rx^2 + (p-2)x - 2p\right)}{(x-2)^2} = 5 for some r,p∈Rr, p \in \mathbb{R}. If the set of all possible values of qq, such that the roots of the equation rx2−px+q=0rx^2 - px + q = 0 lie in (0,2)(0, 2), be the interval (α,β](\alpha, \beta], then 4(α+β)4(\alpha + \beta) equals.
    1. (1)11
    2. (2)13
    3. (3)17
    4. (4)21
  24. MathsJEE Main 6 Apr 2026, Shift 2·Q25·Numerical·Moderate
    Open

    Continuity of composite function

    Let f(x)={x3+8x<0x2−4x≥0andg(x)={(x−8)1/3;x<0(x+4)1/2;x≥0f(x) = \begin{cases} x^3 + 8 & x < 0\\ x^2 - 4 & x \ge 0\end{cases} \quad\text{and}\quad g(x) = \begin{cases} (x - 8)^{1/3}; & x < 0\\ (x + 4)^{1/2}; & x \ge 0\end{cases} Then the number of points, where the function gofg\mathrm{o}f is discontinuous, is .
  25. MathsJEE Main 8 Apr 2026, Shift 2·Q13·MCQ·Moderate
    Open

    Differentiability of modulus functions

    For the function f(x)=esin⁡∣x∣−∣x∣f(x) = e^{\sin\lvert x\rvert} - \lvert x\rvert, x∈Rx \in \mathbb{R}, consider the following statements:

    Statement I: ff is differentiable for all x∈Rx \in \mathbb{R}.

    Statement II: ff is increasing in (−π,−π2)\left(-\pi, -\dfrac{\pi}{2}\right).

    In the light of the above statements, choose the correct answer from the options given below:

    1. (1)Both Statement I and Statement II are true
    2. (2)Both Statement I and Statement II are false
    3. (3)Statement I is true but Statement II is false
    4. (4)Statement I is false but Statement II is true
    We corrected a printing error in the original paper

    Answer corrected; the source key is (4) — for f(x)=esin⁡∣x∣−∣x∣f(x) = e^{\sin\lvert x\rvert} - \lvert x\rvert both one-sided derivatives at x=0x = 0 are 00 and ff is smooth elsewhere, so Statement I is true, and f′(x)=1−cos⁡x e−sin⁡x>0f'(x) = 1 - \cos x\, e^{-\sin x} > 0 on (−π,−π/2)(-\pi, -\pi/2), so Statement II is true; the source's solution works with e∣sin⁡x∣e^{\lvert\sin x\rvert} instead.