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    Previous Year Questions

JEE MAIN Previous Year Questions (PYQs)

JEE MAIN Limit PYQ

JEE MAIN PYQ
$\displaystyle \lim_{x\to 0}\frac{\sin(\pi \cos^{2}x)}{x^{2}}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2014 (Offline) PYQ

Solution

JEE MAIN PYQ
$\lim_{x \to 1} \left( \dfrac{\int_{0}^{(x-1)^{2}} t \cos(t^{2}) \, dt}{(x-1)\sin(x-1)} \right)$





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 6 September 2020 (Morning) PYQ

Solution

JEE MAIN PYQ
Let $\beta=\lim_{x\to 0}\dfrac{\alpha x-(e^{3x}-1)}{\alpha x(e^{3x}-1)}$ for some $\alpha\in\mathbb{R}$. Then the value of $\alpha+\beta$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (26 July Evening Shift) PYQ

Solution

JEE MAIN PYQ
$\lim_{x\to 0}\dfrac{x+2\sin x}{\sqrt{x^{2}+2\sin x+1}-\sqrt{\sin^{2}x-x+1}}$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (12 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
Let f : R $\to$ R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of $\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (27 July Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to\infty}\frac{(2x^{2}-3x+5),(3x-1)^{x/2}}{(3x^{2}+5x+4),\sqrt{(3x+2)^{x}}}$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (23 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to 0}\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2013 (Offline) PYQ

Solution

JEE MAIN PYQ
Let $f$ be a differentiable function on $\mathbb{R}$ such that $f(2)=1,\ f'(2)=4$. Let $\displaystyle \lim_{x\to 0}\big(f(2+x)\big)^{\frac{3}{x}}=e^{\alpha}$. Then the number of times the curve $y=4x^3-4x^2-4(\alpha-7)x-\alpha$ meets the $x$-axis is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (4 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
Let $[x]$ denote the greatest integer less than or equal to $x$. Then $\displaystyle \lim_{x\to 0}\frac{\tan(\pi\sin^{2}x)+\left(|x|-\sin(x[x])\right)^{2}}{x^{2}}$ :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (11 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to 0^{+}}\frac{\tan\big(5x^{1/5}\big),\ln(1+3x^{2})}{\big(\tan^{-1}(3\sqrt{2})\big),\big(e^{x\sqrt{3}}-1\big)}$ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (7 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
Let $a_{1},a_{2},a_{3},\ldots,a_{n}$ be $n$ positive consecutive terms of an arithmetic progression. If $d>0$ is its common difference, then \[ \lim_{n\to\infty}\sqrt{\frac{d}{n}} \left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}} +\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}} +\cdots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}\right) \] is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (6 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \lim_{x \to 0} \csc x \left( \sqrt{2\cos^2 x + 3\cos x} - \sqrt{\cos^2 x + \sin x + 4} \right)$ is :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (24 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
If the function $f(x)=\dfrac{\sin 3x+\alpha\sin x-\beta\cos 3x}{x^{3}},; x\in\mathbb{R},$ is continuous at $x=0$, then $f(0)$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (5 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $f(x) = \begin{vmatrix} \cos x & x & 1 \\ 2\sin x & x^{2} & 2x \\ \tan x & x & 1 \end{vmatrix}$, then $\lim_{x \to 0} \dfrac{f'(x)}{x}$





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (15 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
\[ \lim_{n\to\infty} \left\{ \left(2^{\tfrac12}-2^{\tfrac13}\right)\left(2^{\tfrac12}-2^{\tfrac15}\right)\cdots\left(2^{\tfrac12}-2^{\tfrac{1}{2n+1}}\right) \right\} \] is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (6 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $a=\displaystyle\lim_{x\to 0}\dfrac{\sqrt{\,1+\sqrt{\,1+x^{2}\,}\,}-\sqrt{2}}{x^{2}}$ and $b=\displaystyle\lim_{x\to 0}\dfrac{\sin^{2}x}{\sqrt{2}-\sqrt{\,1+\cos x\,}}$, then the value of $ab^{3}$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (27 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
$ \displaystyle \lim_{x\to 0} \left( \frac{1-\cos^2(3x)}{\cos^3(4x)} \right)\left( \frac{\sin^3(4x)}{(\log_e(2x+1))^5} \right) \text{ is equal to } \underline{\ \ \ \ \ \ \ \ \ \ }. $





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (8 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to 0}\frac{x\cot(4x)}{\sin^{2}x\;\cot^{2}(2x)}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (11 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
For some $a, b,$ let $f(x)=\left|\begin{array}{ccc}\mathrm{a}+\frac{\sin x}{x} & 1 & \mathrm{~b} \\ \mathrm{a} & 1+\frac{\sin x}{x} & \mathrm{~b} \\ \mathrm{a} & 1 & \mathrm{~b}+\frac{\sin x}{x}\end{array}\right|, x \neq 0, \lim \limits_{x \rightarrow 0} f(x)=\lambda+\mu \mathrm{a}+\nu \mathrm{b}.$ Then $(\lambda+\mu+v)^2$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (24 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \lim_{x\to0}\frac{3+a\sin x+b\cos x+\log_e(1-x)}{3\tan^2 x}=\frac{1}{3}$, then $2a-b$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (27 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to 0} \frac{x\tan 2x - 2x\tan x}{(1-\cos 2x)^{2}}$ equals :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (15 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $\alpha$, $\beta$ are the distinct roots of x2 + bx + c = 0, then $\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (27 August Morning Shift) PYQ

Solution

JEE MAIN PYQ
Let f(x) be a polynomial function such that $f(x) + f'(x) + f''(x) = {x^5} + 64$. Then, the value of $\mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x - 1}}$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (25 June Morning Shift) PYQ

Solution

JEE MAIN PYQ
The value of $\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}$ is :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (26 February Morning Shift) PYQ

Solution

JEE MAIN PYQ
The numbers $\alpha>\beta>0$ are the roots of the equation $a x^{2}+b x+1=0$, and $\displaystyle \lim_{x\to \frac{1}{\alpha}} \left( \frac{1-\cos!\big(x^{2}+bx+a\big)}{2(1-a x)^{2}} \right)^{\tfrac{1}{2}} = \frac{1}{k}!\left(\frac{1}{\beta}-\frac{1}{\alpha}\right).$ Then $k$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (8 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $\lim_{x \to 0} \dfrac{\alpha e^{x^2} + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \dfrac{2}{3}$, where $\alpha, \beta, \gamma \in \mathbb{R}$, then which of the following is NOT correct?





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (29 July Morning Shift) PYQ

Solution

JEE MAIN PYQ
For each $t \in \mathbb{R}$, let $[t]$ be the greatest integer less than or equal to $t$. Then $\displaystyle \lim_{x \to 0^{+}} x\left(\left[\frac{1}{x}\right] + \left[\frac{2}{x}\right] + \cdots + \left[\frac{15}{x}\right]\right)$





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (Offline) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to \pi/4}\frac{\cot^{3}x-\tan x}{\cos\!\left(x+\frac{\pi}{4}\right)}$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (12 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{{\tan }^2}x\left( {{{(2{{\sin }^2}x + 3\sin x + 4)}^{{1 \over 2}}} - {{({{\sin }^2}x + 6\sin x + 2)}^{{1 \over 2}}}} \right)} \right)$ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (25 June Evening Shift) PYQ

Solution

JEE MAIN PYQ
Let f(x) be a differentiable function at x = a with f'(a) = 2 and f(a) = 4. Then $\mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}$ equals :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (26 February Morning Shift) PYQ

Solution

JEE MAIN PYQ
Let $f:(-\infty,\infty)\setminus{0}\to\mathbb{R}$ be differentiable such that $f'(1)=\lim_{a\to\infty} a^{2}f!\left(\tfrac{1}{a}\right)$. Then $\displaystyle \lim_{a\to\infty}\left(\frac{a(a+1)}{2}\tan^{-1}!\frac{1}{a}+a^{2}-2\log_{e}a\right)$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (6 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b$, then the ordered pair (a, b) is :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (27 August Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b$, then the ordered pair (a, b) is :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (31 August Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle\lim_{x\to0} \dfrac{(27+x)^{1/3}-3}{9-(27+x)^{2/3}}$ equals :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (16 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to\frac{\pi}{2}} \left( \frac{1}{(x-\frac{\pi}{2})^{2}}\, \frac{\left(\frac{\pi}{3}\right)^{3}}{x^{3}} \int_{0}^{x}\cos\!\left(t^{1/3}\right)\,dt \right)$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (29 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to1^-}\frac{\sqrt{x}-\sqrt{2\sin^{-1}x}}{\sqrt{1-x}}$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (12 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (26 June Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (26 June Evening Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{t\to 0}\left(1^{\frac{1}{\sin^2 t}}+2^{\frac{1}{\sin^2 t}}+\cdots+n^{\frac{1}{\sin^2 t}}\right)^{\sin^2 t}$ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (24 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
The value of $\displaystyle \lim_{n\to\infty}\sum_{k=1}^{n}\frac{n^{3}}{(n^{2}+k^{2})(n^{2}+3k^{2})}$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (30 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
Given below are two statements: Statement I: $\displaystyle \lim_{x \to 0} \left( \tan^{-1}x + \log_e \dfrac{\sqrt{1+x}}{1-x} - 2x \right) = \dfrac{2}{5}$ Statement II: $\displaystyle \lim_{x \to 1} \left( x^{\frac{1}{x-1}} \right) = \dfrac{1}{e^2}$ In the light of the above statements, choose the correct answer from the options given below:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (8 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to0}\frac{\sin^{2}x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (8 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
Let a be an integer such that $\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$ exists, where [t] is greatest integer $\le$ t. Then a is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (27 June Morning Shift) PYQ

Solution

JEE MAIN PYQ
Let $f:[-\tfrac{\pi}{2}, \tfrac{\pi}{2}] \to \mathbb{R}$ be a differentiable function such that $f(0)=\tfrac{1}{2}$. If $\displaystyle \lim_{x \to 0} \frac{x \int_0^x f(t),dt}{e^{x^2} - 1} = \alpha$, then $8\alpha^2$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (30 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
The set of all values of $a$ for which $\displaystyle \lim_{x\to a}\big([x-5]-[2x+2]\big)=0$, where $[\alpha]$ denotes the greatest integer less than or equal to $\alpha$, is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (24 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
$\lim_{x\to \frac{\pi}{2}} \dfrac{\cot x - \cos x}{(\pi - 2x)^3}$ equals :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2017 (Offline) PYQ

Solution

JEE MAIN PYQ
$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}$ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2 September 2020 (Evening) PYQ

Solution

JEE MAIN PYQ
Let [t] denote the greatest integer$ \le $ t. If for some $\lambda $ $ \in $ R - {1, 0}, $\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$ = L, then L isequal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 3 September 2020 (Morning) PYQ

Solution

JEE MAIN PYQ
The value of $\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}$, where [ x ] denotes the greatest integer $ \le $ x is :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (17 March Morning Shift) PYQ

Solution

JEE MAIN PYQ
The value of $\displaystyle \lim_{n\to\infty} \frac{1+2-3+4+5-6+\cdots+(3n-2)+(3n-1)-3n} {\sqrt{2n^{4}+4n+3}-\sqrt{n^{4}+5n+4}}$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (25 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{1 \over {{{\left( {x - {\pi \over 2}} \right)}^2}}}\int\limits_{{x^3}}^{{{\left( {{\pi \over 2}} \right)}^3}} {\cos \left( {{t^{{1 \over 3}}}} \right)dt} } \right)$ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (29 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function satisfying $f'(3)+f'(2)=0$. Then $\displaystyle \lim_{x\to0}\left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\right)^{!1/x}$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (8 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
$ \displaystyle \lim_{x\to 3} \frac{\sqrt{3x}-3}{\sqrt{2x}-\sqrt{6}} $ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2017 (8 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {\left( {\sin \sqrt t } \right)dt} } \over {{x^3}}}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (24 February Morning Shift) PYQ

Solution

JEE MAIN PYQ
The value of the limit $\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (17 March Evening Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to 0}\frac{e^{\,2|\sin x|}-2|\sin x|-1}{x^{2}}$





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (31 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
The value of $\mathop {\lim }\limits_{n \to \infty } {{[r] + [2r] + ... + [nr]} \over {{n^2}}}$, where r is a non-zero real number and [r] denotes the greatest integer less than or equal to r, is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (17 March Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $\mathop {\lim }\limits_{x \to 0} {{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x} \over {3{x^3}}}$ is equal to L, then the value of (6L + 1) is





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (18 March Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \lim_{x\to0}\frac{e^{ax}-\cos(bx)-\dfrac{e^{x}-e^{-x}}{2}}{1-\cos(2x)}=17$, then $5a^2+b^2$ is equal to:






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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (13 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
Let $ p = \displaystyle \lim_{x \to 0^{+}} \left(1 + \tan^{2}\sqrt{x}\right)^{\tfrac{1}{x}} $ then $\log p$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2016 (Offline) PYQ

Solution

JEE MAIN PYQ
For $\alpha, \beta, \gamma \in \mathbf{R}$, if $\lim _\limits{x \rightarrow 0} \frac{x^2 \sin \alpha x+(\gamma-1) \mathrm{e}^{x^2}}{\sin 2 x-\beta x}=3$, then $\beta+\gamma-\alpha$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (2 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
If the function $f(x) = \left\{ {\matrix{ {a|\pi - x| + 1,x \le 5} \cr {b|x - \pi | + 3,x > 5} \cr } } \right.$
is continuous at x = 5, then the value of a – b is :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (9 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (31 August Morning Shift) PYQ

Solution

JEE MAIN PYQ
Let $x=2$ be a root of the equation $x^{2}+px+q=0$ and define \[ f(x)= \begin{cases} \dfrac{1-\cos\!\big(x^{2}-4px+q^{2}+8q+16\big)}{(x-2p)^{4}}, & x\ne 2p,\\[6pt] 0, & x=2p. \end{cases} \] Then $\displaystyle \lim_{x\to 2p^{+}} \big[\,f(x)\,\big]$, where $[\cdot]$ denotes the greatest integer function, is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (29 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (9 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \lim_{x \to 0} \frac{\cos(2x) + a\cos(4x) - b}{x^4}$ is finite, then $(a + b)$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (2 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$ and $\beta = \mathop {\lim }\limits_{x \to 0 } {(\cos x)^{\cot x}}$ are the roots of the equation, ax2 + bx $-$ 4 = 0, then the ordered pair (a, b) is :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (31 August Evening Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to 0}\ \frac{e^{-(1+2x)^{\tfrac{1}{2x}}}}{x}$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (9 April Evening Shift) PYQ

Solution

JEE MAIN PYQ
Let $f,g:(0,\infty)\to\mathbb{R}$ be defined by $f(x)=\int_{-x}^{x}\big(|t|-t^{2}\big)e^{-t^{2}}\,dt,\qquad g(x)=\int_{0}^{x^{2}} t^{1/2}e^{-t}\,dt.$ Then the value of $g\!\left( f\!\big(\sqrt{\log_e 9}\,\big)+g\!\big(\sqrt{\log_e 9}\,\big)\right)$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (31 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $f:\mathbb{R}\to\mathbb{R}$ is a differentiable function and $f(2)=6$, then $\displaystyle \lim_{x\to 2}\dfrac{\int_{1}^{f(x)}2t,dt}{\dfrac{6}{x-2}}$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (9 April Evening Shift) PYQ

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JEE MAIN PYQ
Let f : R $\to$ R be a continuous function. Then $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (1 September Evening Shift) PYQ

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JEE MAIN PYQ
$\displaystyle \lim_{y\to 0}\frac{\sqrt{\,1+\sqrt{1+y^{4}}\,}-\sqrt{2}}{y^{4}}$:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (9 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
If $f:R \to R$ is given by $f(x) = x + 1$, then the value of $\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)} \right]$ is :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (20 July Evening Shift) PYQ

Solution

JEE MAIN PYQ
Let $f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$ be a differentiable function such that f(1) = e and
$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$. If f(x) = 1, then x is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 4 September 2020 (Evening) PYQ

Solution

JEE MAIN PYQ
Let $f,g,h$ be the real valued functions defined on $\mathbb{R}$ as \[ f(x)= \begin{cases} \dfrac{x}{|x|}, & x\neq 0,\\[6pt] 1, & x=0, \end{cases} \qquad g(x)= \begin{cases} \dfrac{\sin(x+1)}{x+1}, & x\neq -1,\\[6pt] 1, & x=-1, \end{cases} \] and $h(x)=2\lfloor x\rfloor - f(x)$, where $\lfloor x\rfloor$ is the greatest integer $\le x$. Then the value of $\displaystyle \lim_{x\to 1} g\!\big(h(x-1)\big)$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (30 January Evening Shift) PYQ

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JEE MAIN PYQ
If $\displaystyle \lim_{x \to \infty} \left(1 + \frac{a}{x} - \frac{4}{x^{2}}\right)^{2x} = e^{3}$, then $a$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2016 (9 April Morning Shift) PYQ

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JEE MAIN PYQ
If $\displaystyle \lim_{x\to1}\frac{x^{4}-1}{x-1}=\lim_{x\to k}\frac{x^{3}-k^{3}}{x^{2}-k^{2}}$, then $k$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (10 April Morning Shift) PYQ

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JEE MAIN PYQ
If $\displaystyle \lim_{x\to1}\frac{x^{2}-ax+b}{x-1}=5$, then $a+b$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (10 April Evening Shift) PYQ

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JEE MAIN PYQ
If $\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$, then $8(\alpha+\beta)$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (25 July Morning Shift) PYQ

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JEE MAIN PYQ
For each $x\in\mathbb{R}$, let $[x]$ be the greatest integer less than or equal to $x$. Then $\displaystyle \lim_{x\to 0^-}\frac{x\left([x]+|x|\right)\sin|x|}{|x|}$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (9 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
$\displaystyle \lim_{x\to 0} \frac{(1-\cos 2x)^{2}}{2x\tan x - x\tan 2x}$ is :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2016 (10 April Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\lim_{x \to \tfrac{\pi}{4}} \dfrac{8\sqrt{2} - (\cos x + \sin x)^7}{\sqrt{2} - \sqrt{2}\sin 2x}$ is equal to






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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (25 July Evening Shift) PYQ

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JEE MAIN PYQ
If $\displaystyle \lim_{x\to\infty}\left(\frac{e}{1-e}\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^{x}=\alpha$, then the value of $\displaystyle \frac{\log_e \alpha}{1+\log_e \alpha}$ equals:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (22 January Evening Shift) PYQ

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JEE MAIN PYQ
$\displaystyle \lim_{x\to 0} \frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}$ is equal to :





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2015 (Offline) PYQ

Solution

JEE MAIN PYQ
Let $f(x)= \begin{cases} x-1, & x \text{ is even},\\ 2x, & x \text{ is odd}, \end{cases}\quad x\in\mathbb N.$ If for some $a\in\mathbb N$, $f(f(f(a)))=21$, then $\displaystyle \lim_{x\to a}\Big\{\dfrac{|x|^{3}}{a}-\Big\lfloor\dfrac{x}{a}\Big\rfloor\Big\}$ is equal to:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (1 February Evening Shift) PYQ

Solution

JEE MAIN PYQ
For each $t\in\mathbb{R}$, let $[t]$ be the greatest integer less than or equal to $t$. Then $\displaystyle \lim_{x\to 1^{+}}\frac{\big(1-|x|+|\sin|1-x||\big)\,\sin\!\left(\tfrac{\pi}{2}[\,1-x\,]\right)}{|1-x|\,[\,1-x\,]}$ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (10 January Morning Shift) PYQ

Solution

JEE MAIN PYQ
$\lim_{x \to 0} \dfrac{x \left( e^{\tfrac{\sqrt{1+x^{2}+x^{4}}-1}{x}} - 1 \right)}{\sqrt{1+x^{2}+x^{4}} - 1}$





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 5 September 2020 (Evening) PYQ

Solution

JEE MAIN PYQ
$ \displaystyle \lim_{x\to\infty} \left\{ \frac{\big(\sqrt{3x+1}+\sqrt{3x-1}\big)^{6}+\big(\sqrt{3x+1}-\sqrt{3x-1}\big)^{6}}{\big(x+\sqrt{x^{2}-1}\big)^{6}+\big(x-\sqrt{x^{2}-1}\big)^{6}} - x^{3} \right\} $ is:





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (31 January Evening Shift) PYQ

Solution

JEE MAIN PYQ
If $\displaystyle \lim_{x\to 1^{+}}\frac{(x-1)\big(6+\lambda\cos(x-1)\big)+\mu\sin(1-x)}{(x-1)^{3}}=-1$, where $\lambda,\mu\in\mathbb{R}$, then $\lambda+\mu$ is equal to





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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (4 April Morning Shift) PYQ

Solution


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