The value of $\mathop {\lim }\limits_{x \to 3}\left[\frac{1}{x-3}+\frac{9 x}{27-x^3}\right]$ is:
$\frac{1}{2}$
1
2
0
$$ \int \sqrt{2 a x-x^2} d x= $$
$$ \frac{x-a}{2} \sqrt{2 a x-x^2}+\frac{a^2}{2} \cos ^{-1}\left(\frac{x-a}{a}\right)+C $$
$$ \frac{a^2}{2} \sqrt{2 a x-x^2}+\frac{x-a}{2} \sin ^{-1}\left(\frac{x-a}{a}\right)+C $$
$$ \frac{x-a}{2} \sqrt{2 a x-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x-a}{a}\right)+C $$
$$ \frac{a^2}{2} \sqrt{2 a x-x^2}+\frac{x-a}{2} \cos ^{-1}\left(\frac{x-a}{a}\right)+C $$
Consider the following list of ordered pairs: $(1,0),(-2,-1),(7,-6),(-3,4)$ and $(0,2)$
Which of the following options correctly identifies only those pairs that are NOT elements of the relation $R=\{(x, y): y=1-|x| ; x, y \in \mathbb{Q}\}$ ?
$(1,0),(-2,-1)$
$(0,2),(-2,-1)$
$(-3,4),(0,2)$
$(-3,4),(7,-6)$
Given $P=\left[\begin{array}{lll}2 & \boldsymbol{\alpha} & 1 \\ 1 & 2 & 2 \\ 1 & 3 & 3\end{array}\right]$ is the adjoint of a $3 \times 3$ matrix A and $|A|=3$, then the value of $\boldsymbol{\alpha}$ is:
7
-25
-8
-26
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