1
JEE Main 2024 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Consider three vectors $$\vec{a}, \vec{b}, \vec{c}$$. Let $$|\vec{a}|=2,|\vec{b}|=3$$ and $$\vec{a}=\vec{b} \times \vec{c}$$. If $$\alpha \in\left[0, \frac{\pi}{3}\right]$$ is the angle between the vectors $$\vec{b}$$ and $$\vec{c}$$, then the minimum value of $$27|\vec{c}-\vec{a}|^2$$ is equal to:

A
124
B
110
C
121
D
105
2
JEE Main 2024 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The differential equation of the family of circles passing through the origin and having centre at the line $$y=x$$ is :

A
$$\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2+2 x y\right) \mathrm{d} y$$
B
$$\left(x^2+y^2-2 x y\right) \mathrm{d} x=\left(x^2+y^2+2 x y\right) \mathrm{d} y$$
C
$$\left(x^2+y^2+2 x y\right) \mathrm{d} x=\left(x^2+y^2-2 x y\right) \mathrm{d} y$$
D
$$\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2-2 x y\right) \mathrm{d} y$$
3
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 ________.

Your input ____
4
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 ________.

Your input ____
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