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
$(2,-2,-2)$
$(-2,-2,2)$
$(-2,2,-2)$
$(2,-2,2)$
Let $A_1, A_2, \ldots, A_6$ are six sets, each with four elements and $B_1, B_2, \ldots ., B_n$ are $n$ sets, each with two elements. Let $S=A_1 \cup A_2 \cup \ldots \cup A_6=B_1 \cup B_2 \cup \ldots \cup B_n$.
Given that each element of $S$ belongs to exactly four of the A's and to exactly three of the B's. Then $n$ is
12
24
6
9
A figure is bounded by the curves $y=x^2+1, y=0, x=0$ and $x=1$. The point at which a tangent should be drawn to the curve $y=x^2+1$ for it to cut off trapezium of the greatest area from the figure is
$(1,2)$
$(-1,2)$
$\left(\frac{1}{2}, \frac{5}{4}\right)$
$(0,1)$
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