1
JEE Main 2023 (Online) 10th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$f$$ be a continuous function satisfying $$\int_\limits{0}^{t^{2}}\left(f(x)+x^{2}\right) d x=\frac{4}{3} t^{3}, \forall t > 0$$. Then $$f\left(\frac{\pi^{2}}{4}\right)$$ is equal to :

A
$$-\pi\left(1+\frac{\pi^{3}}{16}\right)$$
B
$$\pi\left(1-\frac{\pi^{3}}{16}\right)$$
C
$$-\pi^{2}\left(1+\frac{\pi^{2}}{16}\right)$$
D
$$\pi^{2}\left(1-\frac{\pi^{2}}{16}\right)$$
2
JEE Main 2023 (Online) 10th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

For $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$, if $$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$ , where $$e=\sum_\limits{n=0}^{\infty} \frac{1}{n !}$$ and $$\mathrm{C}$$ is constant of integration, then $$\alpha+2 \beta+3 \gamma-4 \delta$$ is equal to :

A
$$-8$$
B
$$-4$$
C
1
D
4

3
JEE Main 2023 (Online) 10th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$\mu$$ be the mean and $$\sigma$$ be the standard deviation of the distribution

$${x_i}$$ 0 1 2 3 4 5
$${f_i}$$ $$k + 2$$ $$2k$$ $${k^2} - 1$$ $${k^2} - 1$$ $${k^2} + 1$$ $$k - 3$$

where $$\sum f_{i}=62$$. If $$[x]$$ denotes the greatest integer $$\leq x$$, then $$\left[\mu^{2}+\sigma^{2}\right]$$ is equal to :

A
9
B
8
C
6
D
7
4
JEE Main 2023 (Online) 10th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k}$$ and $$\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$$. Let $$\vec{d}$$ be a vector which is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$, and $$\vec{c} \cdot \vec{d}=12$$. Then $$(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$$ is equal to :

A
24
B
42
C
44
D
48
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