If ' $a$ ' is an integer lying in $[-5,30]$, then the probability that the graph of $y=x^2+2(a+4) x-5 a+64$ lies above the $x-$ axis is
$\frac{1}{6}$
$\frac{7}{36}$
$\frac{2}{9}$
$\frac{3}{5}$
Consider a square $A B C D$ of diagonal length 2a. The square is folded along the diagonal $A C$ so that the plane of $\triangle A B C$ is perpendicular to the plane of $\triangle A D C$. In this case the shortest distance between $A B$ and $C D$ is
$\frac{2 a}{\sqrt{3}}$
$\frac{a}{2 \sqrt{3}}$
$\frac{\mathrm{a}}{\sqrt{3}}$
$\frac{\sqrt{3} a}{2}$
If $\int \frac{\left(1-x^2\right)}{\sqrt{x} \sqrt{\left(1+x^2\right)^3}}=\alpha \frac{x^\beta}{\left(1+x^2\right)^\gamma}+C ; \alpha, \beta, \gamma \in \mathbb{R}$ and $C$ is constant of integration, then $\alpha: \beta: \gamma$ will be
$4: 1: 1$
$2: 2: \frac{1}{2}$
$\frac{1}{6}: 2: \frac{1}{2}$
$1: 2: \frac{1}{2}$
Let $\vec{a}=(x, y, z)$ be the vector with $|\vec{a}|=2 \sqrt{3}$, which makes equal angles with the vector $\vec{b}=(y,-2 z, 3 x)$ and $\vec{c}=(2 z, 3 x,-y)$ and is perpendicular to the vector $\vec{d}=(1,-1,2)$. If the angle between $\vec{a}$ and the unit vector $\hat{j}$ is obtuse, then $\vec{a}$ is
$(2,-2,-2)$
$(-2,-2,2)$
$(-2,2,-2)$
$(2,-2,2)$
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