1
JEE Main 2024 (Online) 30th January Evening Shift
+4
-1

Let $$\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$$ be defined as $$f(x)=a e^{2 x}+b e^x+c x$$. If $$f(0)=-1, f^{\prime}\left(\log _e 2\right)=21$$ and $$\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}$$, then the value of $$|a+b+c|$$ equals

A
16
B
12
C
8
D
10
2
JEE Main 2024 (Online) 30th January Morning Shift
+4
-1

The value of $$\lim _\limits{n \rightarrow \infty} \sum_\limits{k=1}^n \frac{n^3}{\left(n^2+k^2\right)\left(n^2+3 k^2\right)}$$ is :

A
$$\frac{\pi}{8(2 \sqrt{3}+3)}$$
B
$$\frac{(2 \sqrt{3}+3) \pi}{24}$$
C
$$\frac{13 \pi}{8(4 \sqrt{3}+3)}$$
D
$$\frac{13(2 \sqrt{3}-3) \pi}{8}$$
3
JEE Main 2024 (Online) 30th January Morning Shift
+4
-1

Let $$f:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbf{R}$$ be a differentiable function such that $$f(0)=\frac{1}{2}$$. If the $$\lim _\limits{x \rightarrow 0} \frac{x \int_0^x f(\mathrm{t}) \mathrm{dt}}{\mathrm{e}^{x^2}-1}=\alpha$$, then $$8 \alpha^2$$ is equal to :

A
4
B
2
C
1
D
16
4
JEE Main 2024 (Online) 29th January Morning Shift
+4
-1

$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{1 \over {{{\left( {x - {\pi \over 2}} \right)}^2}}}\int\limits_{{x^3}}^{{{\left( {{\pi \over 2}} \right)}^3}} {\cos \left( {{t^{{1 \over 3}}}} \right)dt} } \right)$$ is equal to

A
$$\frac{3 \pi^2}{4}$$
B
$$\frac{3 \pi^2}{8}$$
C
$$\frac{3 \pi}{4}$$
D
$$\frac{3 \pi}{8}$$
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