1
JEE Main 2024 (Online) 9th April Evening Shift
+4
-1

$$\lim _\limits{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$$ is equal to

A
$$\frac{-2}{e}$$
B
$$e-e^2$$
C
0
D
$$e$$
2
JEE Main 2024 (Online) 8th April Evening Shift
+4
-1

For $$\mathrm{a}, \mathrm{b}>0$$, let $$f(x)= \begin{cases}\frac{\tan ((\mathrm{a}+1) x)+\mathrm{b} \tan x}{x}, & x< 0 \\ 3, & x=0 \\ \frac{\sqrt{\mathrm{a} x+\mathrm{b}^2 x^2}-\sqrt{\mathrm{a} x}}{\mathrm{~b} \sqrt{\mathrm{a}} x \sqrt{x}}, & x> 0\end{cases}$$ be a continuous function at $$x=0$$. Then $$\frac{\mathrm{b}}{\mathrm{a}}$$ is equal to :

A
4
B
5
C
8
D
6
3
JEE Main 2024 (Online) 6th April Evening Shift
+4
-1

$$\lim _\limits{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\cdots+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\cdots \cdots+n^3\right)-\left(1^2+2^2+\cdots \cdots+n^2\right)}$$ is equal to :

A
$$\frac{2}{3}$$
B
$$\frac{1}{2}$$
C
$$\frac{3}{4}$$
D
$$\frac{1}{3}$$
4
JEE Main 2024 (Online) 5th April Evening Shift
+4
-1

Let ,$$f:[-1,2] \rightarrow \mathbf{R}$$ be given by $$f(x)=2 x^2+x+\left[x^2\right]-[x]$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. The number of points, where $$f$$ is not continuous, is :

A
5
B
6
C
4
D
3
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