List of top Questions asked in JEE Main

In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance $R_1$, and $R_2$ respectively, are:

$\int\frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx $ is equal to : (where $c$ is a constant of integration)

${K_2HgI_4}$ is $40\%$ ionised in aqueous solution. The value of its van't Hoff factor (i) is :

Let $a_1, a_2, ....., a_{10}$ be a G.P. If $\frac{a_3}{a_1} = 25 $, then $\frac{a_9}{a_5}$ equals :

Let $a_1, a_2 , ....... , a_{30}$ be an $A. P$., $S =\displaystyle \sum^{30}_{i=1} a_i $ and $T = \displaystyle\sum^{15}_{i=1} a_{(2i -1)}$. If $a_5 = 27$ and $S - 2T = 75, $ then $a_{10}$ is equal to :

Let $A$ and $B$ be two invertible matrices of order $3 \times 3$. If $\text{det} (ABA^T) = 8$ and $\text{det} (AB^{-1}) = 8$, then $\text{det} (BA^{-1} BT) $ is equal to :

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

Let $f \left(x\right)= \int\limits_0^{x} g (t) dt$ , where g is a non-zero even function. If $f \left(x+5\right)=g\left(x\right)$, then $ \int\limits_0^{x}f (t) dt $ equals-

Let $N$ be the set of natural numbers and two functions $f$ and $g$ be defined as $f,g : N \to N$ such that : $f(n) = \begin{cases} \frac{n+1}{2} & \quad \text{if } n \text{ is odd}\\ \frac{n}{2} & \quad \text{if } n \text{ is even} \end{cases}$ and $g(n) = n-(-1)^n$. The $fog$ is :

Let $S_n$ denote the sum of the first $n$ terms of an $A.P$. If $S_4 = 16$ and $S_6 = -48$, then $S_{10}$ is equal to :

Let $(x + 10)^{50} + (x - 10)^{50} = a_0 + a_1 x + a_2x^2 + ..... + a_{50} \; x^{50} , $ for all $x \in R , $ the $\frac{a_2}{a_0} $ is equal to :-

Let $z_1$ and $z_2$ be any two non-zero complex numbers such that $3|z_1| = 4 |z_2|$. If $z = \frac{3z_{1}}{2z_{2}} + \frac{2z_{2}}{3z_{1}}$ then :

Let $z$ be a complex number such that $|z| + z = 3 + i$ (where i = $\sqrt{-1}$ ). Then $|z|$ is equal to :

Let $Z$ be the set of integers. If $A \, = \, \{ x \in Z : 2^{ (x+2) (x^2 - 5x + 6)} \}=1$ and $B \, = \, \{ \, x \in Z: -3 <2x -1 <9 \}$, then the number of subsets of the set $A \times B$, is :

Let $z \in C$ with Im(z) = 10 and it satisfies $\frac{2z-n}{2z+n} = 2i -1$ for some natural number $n$.Then :

$\displaystyle\lim_{n\to\infty} \left(\frac{\left(n+1\right)^{\frac{1}{3}} }{n^{\frac{4}{3}}} + \frac{\left(n+2\right)^{\frac{1}{3}}}{n^{\frac{4}{3}}} + ..... + \frac{\left(2n\right)^{\frac{1}{3}}}{n^{\frac{4}{3}}}\right) $ equal to :

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

The actual value of resistance $R$, shown in the figure is $30 \Omega $. This is measured in an experiment as shown using the standard formula $R = \frac{V}{I}$ , where $V$ and $I$ are the readings of the voltmeter and ammeter, respectively. If the measured value of $R$ is $5\%$ less, then the internal resistance of the voltmeter is :

The compound used in the treatment of lead poisoning is :

The concentration of dissolved oxygen $(DO)$ in cold water can go upto :

The correct statements among I to III are : (I) Valence bond theory cannot explain the color exhibited by transition metal complexes. (II) Valence bond theory can predict quantitatively the magnetic properties of transtition metal complexes. (III) Valence bond theory cannot distinguish ligands as weak and strong field ones.

The correct structure of histidine in a strongly acidic solution (pH=2) is

The decreasing order of electrical conductivity of the following aqueous solutions is : 0.1 M Formic acid (A), 0.1 M Acetic acid (B) 0.1 M Benzoic acid (C)

The increasing order of nucleophilicity of the following nucleophiles is : (a) ${CH_3CO^{\ominus}_2}$ (b) ${H_2O} $ (c) ${CH_3SO^{\ominus}_3}$ (d) ${O^{\ominus}H}$

The integral $\int sec^{2/3} x cosec^{4/3}x\, dx$ is equal to (Hence $C$ is a constant of integration)