If $\boldsymbol{k}$ is the arithmetic mean of two given quantities and $\boldsymbol{p}, \boldsymbol{q}$ are the geometric means between the same two quantities, then $\boldsymbol{p}^{\mathbf{3}}+\boldsymbol{q}^{\mathbf{3}}$ is:
$2 k p q$
$2 k \sqrt{p q}$
$2 k(p+q)$
$\frac{2 k}{p q}$
The feasible region represented by the constraints:
$$ \begin{aligned} & x+2 y \leq 120 \\ & x+y \geq 60 \\ & x-2 y \geq 0 \\ & x \geq 0 \text { and } y \geq 0 \end{aligned} $$
A
C
B
D
$$ \text { The range of the relation } R=\left\{(x, y): y=x+\frac{6}{x} \text {; where } x, y \in \mathbb{N} \text { and } x<6\right\} \text { is: } $$
$\{1,2,3\}$
$\{5,6\}$
$\left\{5, \frac{11}{2}\right\}$
$\{5,7\}$
Let A be a square matrix of order $3 \times 3$. If $|A|=-4$, then the value of $\left|\frac{A^{-1}}{-2}\right|$ is:
$-1$
$2$
$\frac{1}{32}$
$-\frac{1}{16}$
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