1
JEE Main 2023 (Online) 15th April Morning Shift
+4
-1
If $\int\limits_{0}^{1} \frac{1}{\left(5+2 x-2 x^{2}\right)\left(1+e^{(2-4 x)}\right)} d x=\frac{1}{\alpha} \log _{e}\left(\frac{\alpha+1}{\beta}\right), \alpha, \beta>0$, then $\alpha^{4}-\beta^{4}$ is equal to :
A
-21
B
21
C
19
D
0
2
JEE Main 2023 (Online) 13th April Evening Shift
+4
-1

The value of $${{{e^{ - {\pi \over 4}}} + \int\limits_0^{{\pi \over 4}} {{e^{ - x}}{{\tan }^{50}}xdx} } \over {\int\limits_0^{{\pi \over 4}} {{e^{ - x}}({{\tan }^{49}}x + {{\tan }^{51}}x)dx} }}$$ is

A
51
B
50
C
25
D
49
3
JEE Main 2023 (Online) 13th April Morning Shift
+4
-1
Out of Syllabus

Among

(S1): $$\lim_\limits{n \rightarrow \infty} \frac{1}{n^{2}}(2+4+6+\ldots \ldots+2 n)=1$$

(S2) : $$\lim_\limits{n \rightarrow \infty} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots+n^{15}\right)=\frac{1}{16}$$

A
Only (S1) is true
B
Both (S1) and (S2) are true
C
Both (S1) and (S2) are false
D
Only (S2) is true
4
JEE Main 2023 (Online) 13th April Morning Shift
+4
-1

$$\int_\limits{0}^{\infty} \frac{6}{e^{3 x}+6 e^{2 x}+11 e^{x}+6} d x=$$

A
$$\log _{e}\left(\frac{256}{81}\right)$$
B
$$\log _{e}\left(\frac{64}{27}\right)$$
C
$$\log _{e}\left(\frac{32}{27}\right)$$
D
$$\log _{e}\left(\frac{512}{81}\right)$$
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