Limits, Continuity & Differentiability Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Limits, Continuity and Differentiability with step-by-step solutions covering L'Hopital-free limit tricks, points of non-differentiability, greatest-integer discontinuities, functional equations and inverse-trig differentiation.
A curated set of JEE Main 2026 previous-year questions on Limits, Continuity and Differentiability, each solved step by step so you can check both the final answer and the reasoning.
Solutions are AI-generated and pending review.
Solution
First term. Divide numerator and denominator by $4\cos x$:
$$\tan^{-1}\!\left(\dfrac{\tfrac34 - \tan x}{1 + \tfrac34\tan x}\right) = \tan^{-1}\tfrac34 - \tan^{-1}(\tan x) = \tan^{-1}\tfrac34 - x$$so its derivative is $-1$.
Second term. Put $x = \sin\theta$. Then
$$\dfrac{x}{1+\sqrt{1-x^2}} = \dfrac{\sin\theta}{1+\cos\theta} = \tan\dfrac{\theta}{2}$$$$\Rightarrow\ 2\tan^{-1}\!\left(\tan\tfrac{\theta}{2}\right) = \theta = \sin^{-1}x,\qquad \dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$$Combine:
$$\dfrac{dy}{dx} = -1 + \dfrac{1}{\sqrt{1-x^2}}$$At $x = \dfrac{\sqrt3}{2}$: $\sqrt{1-\tfrac34} = \tfrac12$, so
$$\dfrac{dy}{dx} = -1 + 2 = 1$$Answer: C
Solution
Part 1 — the $\max$ term. Compare $6x$ and $2+3x^2$:
$$2+3x^2 - 6x = 3x^2 - 6x + 2 = 0 \ \Rightarrow\ x = 1 \pm \tfrac{1}{\sqrt3}\ (\approx 0.423,\ 1.577)$$The two branches swap dominance at these two points, giving corners at
$$x = 1-\tfrac{1}{\sqrt3},\quad x = 1+\tfrac{1}{\sqrt3}\qquad(\textbf{2 points}).$$Part 2 — the $|x-1|$ factor. $|x-1|$ has a corner at $x=1$, and there $\left|\cos\left|1-\tfrac14\right|\right| = |\cos 0.75| \ne 0$, so the product is non-differentiable at $x=1$ $(\textbf{1 point})$. (The $\max$ term equals $6x$ there and is smooth.)
Part 3 — zeros of $\left|\cos|x^2-\tfrac14|\right|$. Since $\cos|u| = \cos u$ is smooth through $u=0$, the inner $\left|x^2-\tfrac14\right|$ creates no corner at $x=\pm\tfrac12$. But $|\cos(\cdot)|$ has a V-corner wherever $\cos = 0$:
$$\left|x^2 - \tfrac14\right| = \tfrac{\pi}{2} + k\pi \ \Rightarrow\ x^2 = \tfrac14 + \left(\tfrac{\pi}{2}+k\pi\right)$$For $x^2 < \pi^2$ this gives $|x| \approx 1.349,\ 2.228,\ 2.847$, i.e. the six points $x \approx \pm1.349,\ \pm2.228,\ \pm2.847$. At each, $|x-1|\ne 0$, so the product keeps the corner $(\textbf{6 points}).$
Total: $2 + 1 + 6 = 9$.
Answer: 9
Solution
Use $\sin x = x - \dfrac{x^3}{6} + O(x^5)$.
Denominator:
$$x^2 - \sin^2 x = x^2 - \left(x^2 - \dfrac{x^4}{3} + O(x^6)\right) = \dfrac{x^4}{3} + O(x^6)$$Numerator:
$$x^2 \sin^2 x = x^2\left(x^2 + O(x^4)\right) = x^4 + O(x^6)$$Therefore
$$\lim_{x\to 0}\dfrac{x^4 + O(x^6)}{\tfrac{x^4}{3} + O(x^6)} = \dfrac{1}{\tfrac13} = 3$$Answer: B
Solution
Let $t = x - 2$. As $x\to 2$, $\sqrt{x-1}-1 = \sqrt{1+t}-1 \sim \tfrac{t}{2}$ and $\log_e(x-1) = \log(1+t) \sim t$, so the denominator $\sim \dfrac{t^2}{2}$ (order $t^2$).
For a finite non-zero limit the numerator must also be order $t^2$. Let $P(x) = x^3 - 5x^2 + ax + b$.
Condition $P(2)=0$ (so $\sin P \to 0$): $\ 8 - 20 + 2a + b = 0 \Rightarrow 2a + b = 12$.
Condition $P'(2)=0$ (so $P$ is order $t^2$): $\ P'(x) = 3x^2 - 10x + a$, and $P'(2) = a - 8 = 0 \Rightarrow a = 8$, hence $b = 12 - 16 = -4$.
Then $P(x) = x^3 - 5x^2 + 8x - 4 = (x-2)^2(x-1) = t^2(1+t)$, so $\sin P \sim t^2$.
$$m = \lim_{t\to 0}\dfrac{t^2}{t^2/2} = 2$$$$a + b + m = 8 - 4 + 2 = 6$$Answer: B
Solution
Let $h(x) = \dfrac{\sin x}{x}$, so $f = |h|$. Critical points are where $f'=0$ or $f'$ fails to exist.
Corners (where $h=0$ and changes sign). In $(-2\pi,2\pi)$, $\sin x = 0$ at $x = \pm\pi$ (not $x=0$, since $h(0)=1$). There $|h|$ has a V-corner $(\textbf{2 points}).$
Stationary points of $|h|$ (interior maxima/minima), where $h'(x)=0$:
$$h'(x) = \dfrac{x\cos x - \sin x}{x^2} = 0 \ \Rightarrow\ \tan x = x$$In $(-2\pi, 2\pi)$ the solutions are $x = 0$ (the central maximum, $h(0)=1$) and $x = \pm 4.4934$ $(\textbf{3 points}).$
Total: $2 + 3 = 5$.
Answer: C
Solution
$f(|x|)$. Since $|x|\ge 0$, this is always $|x|^2 - 5|x| + 6 = x^2 - 5|x| + 6$ — continuous everywhere, with a single corner at $x=0$ (from $|x|$).
$|f(x)|$. For $x<0$, $|e^{x-1}| = e^{x-1}$ (smooth, positive). For $x\ge 0$, $|x^2-5x+6| = |(x-2)(x-3)|$, which has V-corners at $x=2,3$.
At $x=0$, $f$ itself jumps: $\displaystyle\lim_{x\to 0^-}e^{x-1} = e^{-1}$ but $f(0)=6$. So $|f(x)|$ (and hence $g$) jumps at $x=0$:
$$g(0^-) = f(0) + e^{-1} = 6 + e^{-1},\qquad g(0^+) = 6 + 6 = 12.$$Continuity. $g$ is discontinuous only at $x=0$ $\Rightarrow \alpha = 1$.
Differentiability. Non-differentiable at $x=0$ (jump/corner) and at $x=2,3$ (corners of $|(x-2)(x-3)|$) $\Rightarrow \beta = 3$.
$$\alpha + \beta = 1 + 3 = 4$$Answer: 4
Solution
The finite limit $\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x^3} = -5$ forces $f(0)=f'(0)=f''(0)=0$ and the $x^3$-coefficient to be $-5$:
$$f(x) = -5x^3 + a x^4 + b x^5$$Extrema at $x=\pm 1$: $f'(x) = -15x^2 + 4a x^3 + 5b x^4$.
$$f'(1) = -15 + 4a + 5b = 0,\qquad f'(-1) = -15 - 4a + 5b = 0$$Adding: $-30 + 10b = 0 \Rightarrow b = 3$. Subtracting: $8a = 0 \Rightarrow a = 0$.
So $f(x) = -5x^3 + 3x^5$, which is odd, hence $f(-2) = -f(2)$:
$$f(2) - f(-2) = 2f(2) = 2\left(-5\cdot 8 + 3\cdot 32\right) = 2(56) = 112$$Answer: D
Solution
Let $u(x) = x^2 - x - \tfrac12$. On $[2,4]$, $u'(x) = 2x - 1 > 0$, so $u$ is strictly increasing and continuous with
$$u(2) = 4 - 2 - \tfrac12 = \tfrac32,\qquad u(4) = 16 - 4 - \tfrac12 = \tfrac{23}{2}.$$$[u(x)]$ jumps exactly where $u(x)$ passes through an integer. As $u$ rises continuously from $1.5$ to $11.5$, it attains each integer in
$$\{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$$once, giving $10$ discontinuity points. (Endpoints $u=1.5,\ 11.5$ are non-integers, so no boundary issue.)
Answer: 10
Solution
Find $r,p$. As $x\to 2$, $\tan(x-2)\to 0$, so the quadratic must vanish at $x=2$:
$$4r + 2(p-2) - 2p = 4r - 4 = 0 \ \Rightarrow\ r = 1.$$Then $x^2 + (p-2)x - 2p = (x-2)(x+p)$, and using $\tan(x-2)\sim (x-2)$:
$$\lim_{x\to 2}\dfrac{(x-2)\,(x-2)(x+p)}{(x-2)^2} = (2+p) = 5 \ \Rightarrow\ p = 3.$$Roots of $x^2 - 3x + q = 0$ in $(0,2)$. Let $\phi(x) = x^2 - 3x + q$. Both roots lie in $(0,2)$ iff:
$$\phi(0) > 0:\ q > 0,\qquad \phi(2) > 0:\ 4 - 6 + q > 0 \Rightarrow q > 2,$$$$\text{discriminant} \ge 0:\ 9 - 4q \ge 0 \Rightarrow q \le \tfrac94,\qquad 0 < \tfrac32 < 2\ (\text{vertex, OK}).$$So $q \in (2, \tfrac94)$, giving $\alpha = 2,\ \beta = \tfrac94$:
$$4(\alpha + \beta) = 4\left(2 + \tfrac94\right) = 4\cdot\tfrac{17}{4} = 17$$Answer: C
Solution
Discontinuities of $g(f(x))$ arise where $f$ jumps, or where a continuous $f$ crosses $0$ (the switch/discontinuity input of $g$, since $\displaystyle\lim_{u\to 0^-}(u-8)^{1/3} = -2 \ne 2 = \lim_{u\to 0^+}(u+4)^{1/2}$).
$x = 0$ ($f$ jumps). $f(0^-) = 8 \Rightarrow g = \sqrt{12}$; $\ f(0^+) = -4 \Rightarrow g = (-12)^{1/3} = -\sqrt[3]{12}$. Left and right limits differ $\Rightarrow$ discontinuous.
$f(x) = 0$ with $f$ continuous.
- $x<0$: $x^3 + 8 = 0 \Rightarrow x = -2$. As $x$ passes $-2$, $f$ crosses $0$, so $g\circ f$ jumps between $-2$ and $2 \Rightarrow$ discontinuous.
- $x\ge 0$: $x^2 - 4 = 0 \Rightarrow x = 2$. Again $f$ crosses $0$, so $g\circ f$ jumps between $-2$ and $2 \Rightarrow$ discontinuous.
Discontinuity points: $x = -2,\ 0,\ 2$ — a total of $3$.
Answer: 3
Solution
Use the standard expansion (each $\cos(ax) \approx 1 - \tfrac{a^2x^2}{2}$, cross terms are $O(x^4)$):
$$1 - \cos(a_1 x)\cos(a_2 x)\cos(a_3 x) \approx \dfrac{a_1^2 + a_2^2 + a_3^2}{2}\,x^2,\qquad \sin^2((\alpha+1)x) \sim (\alpha+1)^2 x^2.$$With $a_1=\alpha,\ a_2=\alpha+1,\ a_3=\alpha+2$:
$$\dfrac{\tfrac12\left(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2\right)}{(\alpha+1)^2} = 2$$$$\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2 = 3\alpha^2 + 6\alpha + 5$$$$3\alpha^2 + 6\alpha + 5 = 4(\alpha+1)^2 = 4\alpha^2 + 8\alpha + 4 \ \Rightarrow\ \alpha^2 + 2\alpha - 1 = 0$$Product of roots $= \dfrac{-1}{1} = -1$ (roots are $-1 \pm \sqrt2$).
Answer: C
Solution
Determine $f$. Put $x=y=0$: $f(0) = \tfrac{2}{3}f(0) \Rightarrow f(0) = 0$. Put $y=0$: $f\!\left(\tfrac{x}{3}\right) = \tfrac{f(x)}{3}$. A linear $f(x) = cx$ satisfies the equation, and $f'(0)=c=3$, so
$$f(x) = 3x.$$Minimise $g$. $g(x) = 3 + 3x e^x$, so
$$g'(x) = 3e^x(1 + x) = 0 \ \Rightarrow\ x = -1,$$and $g''(-1) = 3e^{-1}(2 + x)\big|_{x=-1} = 3e^{-1} > 0$ (minimum).
$$g(-1) = 3 + 3(-1)e^{-1} = 3 - \dfrac{3}{e} = 3\left(\dfrac{e-1}{e}\right)$$Answer: B
Solution
Evaluate $f(x)$. As $y\to 0$: $\ 1 - \cos(xy) \sim \dfrac{(xy)^2}{2} = \dfrac{x^2 y^2}{2}$ and $\tan(xy) \sim xy$, so
$$f(x) = \lim_{y\to 0}\dfrac{\tfrac{x^2 y^2}{2}\cdot xy}{y^3} = \dfrac{x^3}{2}.$$Solve $\dfrac{x^3}{2} = \sin x$, i.e. $x^3 = 2\sin x$. Both sides are odd, so solutions are symmetric about $0$.
- $x = 0$ is a solution.
- For $x>0$: since $\sin x \le 1$, any root needs $x^3 \le 2 \Rightarrow x \le 2^{1/3} \approx 1.26$. On $(0, 1.26]$, $x^3 - 2\sin x$ goes from $0^-$ to positive, giving exactly one root $x \approx 1.236$.
- By oddness, one matching root at $x \approx -1.236$.
Total: $3$ solutions.
Answer: C
Solution
$f'' = g''$ means $(f - g)'' = 0$, so $f - g$ is linear: $f(x) - g(x) = cx + d$.
Slope $c$. $f'(1) = 4$ and $2g'(1) = 4 \Rightarrow g'(1) = 2$, so
$$c = (f-g)'(1) = f'(1) - g'(1) = 4 - 2 = 2.$$Intercept $d$. $g(2) = 9$ and $3f(2) = 9 \Rightarrow f(2) = 3$, so
$$(f-g)(2) = 3 - 9 = -6 = 2(2) + d \Rightarrow d = -10.$$Hence $f(x) - g(x) = 2x - 10$, and
$$f(25) - g(25) = 2(25) - 10 = 40.$$Answer: B
Solution
Statement I. The only possible corner is at $x=0$ (from $|x|$). Check one-sided derivatives:
- Right ($x>0$): $\dfrac{d}{dx}\left(e^{\sin x} - x\right) = e^{\sin x}\cos x - 1 \to 1\cdot 1 - 1 = 0$.
- Left ($x<0$, so $|x|=-x$): $\dfrac{d}{dx}\left(e^{-\sin x} + x\right) = -e^{-\sin x}\cos x + 1 \to -1 + 1 = 0$.
Both equal $0$, so $f$ is differentiable at $x=0$ (and smooth elsewhere). Statement I is true.
Statement II. For $x<0$, $f(x) = e^{-\sin x} + x$, so
$$f'(x) = -e^{-\sin x}\cos x + 1.$$On $\left(-\pi, -\tfrac{\pi}{2}\right)$, $\cos x < 0$, hence $-e^{-\sin x}\cos x > 0$ and $f'(x) > 1 > 0$ throughout. So $f$ is increasing there. Statement II is true.
Both statements are true.
Answer: A