1
JEE Main 2024 (Online) 5th April Evening Shift
Numerical
+4
-1
Change Language

Let $$y=y(x)$$ be the solution of the differential equation

$$\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 x}{\left(1+x^2\right)^2} y=x \mathrm{e}^{\frac{1}{\left(1+x^2\right)}} ; y(0)=0.$$

Then the area enclosed by the curve $$f(x)=y(x) \mathrm{e}^{-\frac{1}{\left(1+x^2\right)}}$$ and the line $$y-x=4$$ is ________.

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2
JEE Main 2024 (Online) 5th April Evening Shift
Numerical
+4
-1
Change Language

Let the mean and the standard deviation of the probability distribution

$$\mathrm{X}$$ $$\alpha$$ 1 0 $$-$$3
$$\mathrm{P(X)}$$ $$\frac{1}{3}$$ $$\mathrm{K}$$ $$\frac{1}{6}$$ $$\frac{1}{4}$$

be $$\mu$$ and $$\sigma$$, respectively. If $$\sigma-\mu=2$$, then $$\sigma+\mu$$ is equal to ________.

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3
JEE Main 2024 (Online) 5th April Evening Shift
Numerical
+4
-1
Change Language

Let the maximum and minimum values of $$\left(\sqrt{8 x-x^2-12}-4\right)^2+(x-7)^2, x \in \mathbf{R}$$ be $$\mathrm{M}$$ and $$\mathrm{m}$$, respectively. Then $$\mathrm{M}^2-\mathrm{m}^2$$ is equal to _________.

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4
JEE Main 2024 (Online) 5th April Evening Shift
Numerical
+4
-1
Change Language

Let the point $$(-1, \alpha, \beta)$$ lie on the line of the shortest distance between the lines $$\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$$ and $$\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$$. Then $$(\alpha-\beta)^2$$ is equal to _________.

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