List of top Mathematics Questions on limits and derivatives asked in JEE Main

Let $\left\lfloor t \right\rfloor$ be the greatest integer less than or equal to $t$. Then the least value of $p \in \mathbb{N}$ for which

\[ \lim_{x \to 0^+} \left( x \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \dots + \left\lfloor \frac{p}{x} \right\rfloor \right) - x^2 \left( \left\lfloor \frac{1}{x^2} \right\rfloor + \left\lfloor \frac{2}{x^2} \right\rfloor + \dots + \left\lfloor \frac{9^2}{x^2} \right\rfloor \right) \geq 1 \]

is equal to __________.

Let \[ f(x) = \int_{0}^{x} \left( t + \sin\left(1 - e^t\right) \right) \, dt, \, x \in \mathbb{R}. \] Then \[ \lim_{x \to 0} \frac{f(x)}{x^3} \] is equal to:

The integral $\int_{0}^{\pi/4} \frac{136 \sin x}{3 \sin x + 5 \cos x} \, dx$ is equal to

$\lim_{x \to 0} \frac{e - (1 + 2x)^{\frac{1}{2x}}}{x} \quad \text{is equal to:}$

Let \[ \int_{\log_e a}^{4} \frac{dx}{\sqrt{e^x - 1}} = \frac{\pi}{6}. \] Then $e^\alpha$ and $e^{-\alpha}$ are the roots of the equation:

Let $I(x) = \int \frac{6}{\sin^2 x (1 - \cot x)^2} dx$. If $I(0) = 3$, then $I\left(\frac{\pi}{12}\right)$ is equal to:

If the value of the integral \[ \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^x} \, dx = \frac{2}{\pi}, \] then a value of $ \alpha $ is:

The value of $\int_{-\pi}^{\pi} \frac{2y(1 + \sin y)}{1 + \cos^2 y} \, dy$

Let $ P = \{ z \in \mathbb{C} : |z + 2 - 3i| \leq 1 \} $ and $ Q = \{ z \in \mathbb{C} : z(1 + i) + \overline{z}(1 - i) \leq -8 \} $.
Let $ z $ in $ P \cap Q $ have $ |z - 3 + 2i| $ be maximum and minimum at $ z_1 $ and $ z_2 $, respectively.
If $ |z_1|^2 + 2|z_2|^2 = \alpha + \beta \sqrt{2} $, where $ \alpha $ and $ \beta $ are integers, then $ \alpha + \beta $ equals ____.

Let $\{x\}$ denote the fractional part of $x$, and $f(x) = \frac{\cos^{-1}(1 - \{x\}^2) \sin^{-1}(1 - \{x\})}{\{x\} - \{x\}^3}, \quad x \neq 0$.If $L$ and $R$ respectively denote the left-hand limit and the right-hand limit of $f(x)$ at $x = 0$, then $\frac{32}{\pi^2} \left(L^2 + R^2\right)$ is equal to $\_\_\_\_\_\_\_\_$.

If α > β > 0 are the roots of the equation ax2 + bx + 1 = 0, and $\lim\limits_{x \to \frac{1}{\alpha}}\left(\frac{1-cos(x^2+bx+a)}{2(1-ax)^2}\right)^{1/2}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right),$ then k is equal to

If $a =\displaystyle\lim _{ n \rightarrow \infty} \sum_{ k =1}^{ n } \frac{2 n }{ n ^2+ k ^2}$ and $f(x)=\sqrt{\frac{1-\cos x}{1+\cos x}}, x \in(0,1)$, then :

For any real number x, let [ x ] denote the largest integer less than equal to x Let f be a real valued function defined on the interval [-10,10] by $f(x)=\begin{cases} x-[x], & \text { if }(x) \text { is odd } \\ 1+[x]-x & \text { if }(x) \text { is even }\end{cases}$Then the value of$ \frac{\pi^2}{10} \int\limits_{-10}^{10} f(x) \cos \pi x d x$ is :

If $\lim_{n\rightarrow \infty}$ $\frac{(n+1)^{k-1}}{n^{k+1}}[(nk+1)+(nk+2)+....+(nk+n)]=33.\lim_{n\rightarrow \infty}\frac{1}{n^{k+1}}.[1^k+2^k+3^k+....+n^k]$
then the integral value of k is equal to _______.

Let [t] denote the greatest integer ≤ t and {t} denote the fractional part of t. The integral value of α for which the left-hand limit of the function $f(x) = \left\lfloor 1 + x \right\rfloor + \frac{\alpha^{2\left\lfloor x \right\rfloor + \left\{ x \right\}} + \left\lfloor x \right\rfloor - 1}{2\left\lfloor x \right\rfloor + \left\{ x \right\}} $ at x=0 is equal to $\alpha -\frac{4}{3}$, is

Let $f$ be a twice differentiable function defined on $R$ such that $f (0)=1, f'(0)=2$ and $f'(x)\neq 0$ for all $x \in R$. If $\left|\begin{array}{cc}f(x) & f'(x) \\ f'(x) & f''(x)\end{array}\right|=0,$ for all $x \in R,$ then the value of $f (1)$ lies in the interval:

$\displaystyle\lim_{x \to 0} \left(\frac{3x^{2}+2}{7x^{2}+2}\right)^{\frac {1}{x^2}}$ is equal to :

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

Let f:[0,1]$\rightarrow$R be a function. suppose the function f is twice differentiable,f(0)=0=f(1) and satisfies f''(x)-2f'(x)+f(x)$\geq$ex,x$\in$[0,1], which of the following is true?

$\displaystyle\lim_{x \to0} \frac{x \tan2x - 2x \tan x}{\left(1-\cos2x\right)^{2}} $ equals :