1
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If the function

$$ f(x)=\left\{\begin{array}{l} \frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\}, \quad x<0 \\ 4, \quad x=0 \\ \frac{2}{x} \log _e\left(\frac{2+k_1 x}{2+k_2 x}\right), \quad x>0 \end{array}\right. $$

is continuous at $x=0$, then $k_1^2+k_2^2$ is equal to :

A
5
B
10
C
20
D
8
2
JEE Main 2025 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $\lim _\limits{x \rightarrow \infty}\left(\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\frac{1}{\mathrm{e}}-\frac{x}{1+x}\right)\right)^x=\alpha$, then the value of $\frac{\log _{\mathrm{e}} \alpha}{1+\log _{\mathrm{e}} \alpha}$ equals :

A
$e^{-2}$
B
$\mathrm{e}^2$
C
$e$
D
$e^{-1}$
3
JEE Main 2025 (Online) 22nd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $\sum_\limits{r=1}^n T_r=\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)}{64}$, then $\lim _\limits{n \rightarrow \infty} \sum_\limits{r=1}^n\left(\frac{1}{T_r}\right)$ is equal to :

A
$\frac{2}{3}$
B
$\frac{1}{3}$
C
1
D
0
4
JEE Main 2024 (Online) 9th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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

A
$$\frac{-2}{e}$$
B
$$e-e^2$$
C
0
D
$$e$$
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