Let the ellipse $E: \frac{x^2}{144} + \frac{y^2}{169} = 1$ and the hyperbola $H: \frac{x^2}{16} - \frac{y^2}{\lambda^2} = -1$ have the same foci. If $e$ and $L$
respectively denote the eccentricity and the length of the latus rectum of $H$, then the value of $24(e+L)$ is :
The probability distribution of a random variable X is given below :
| X | 4k | $\frac{30}{7}k$ | $\frac{32}{7}k$ | $\frac{34}{7}k$ | $\frac{36}{7}k$ | $\frac{38}{7}k$ | $\frac{40}{7}k$ | 6k |
|---|---|---|---|---|---|---|---|---|
| P(X) | $\frac{2}{15}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{1}{5}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{1}{5}$ | $\frac{1}{15}$ |
If E(X) = $\frac{263}{15}$, then P(X < 20) is equal to :
Let $P_1 : y = 4x^2$ and $P_2 : y = x^2 + 27$ be two parabolas. If the area of the bounded region enclosed between $P_1$ and $P_2$ is six times the area of the bounded region enclosed between the line $y = \alpha x$, $\alpha > 0$ and $P_1$, then $\alpha$ is equal to :
The sum of all the elements in the range of $f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x)$, $x \neq \frac{n\pi}{2}, n \in \mathbb{Z}$, where
$\text{Sgn}(t) = \begin{cases} 1, & \text{if } t > 0 \\ -1, & \text{if } t < 0 \end{cases}$
is :
JEE Main Papers
All year-wise previous year question papers