A five digit number (having all different digits) is formed using the digits $1,2,3,4,5$, 6,78 and 9 . The probability that the formed number either begins or ends with an odd digit is equal to
$\frac{5}{6}$
$\frac{1}{3}$
$\frac{1}{6}$
$\frac{2}{3}$
The function, $f(x)=(3 x-7) x^{2 / 3}, x \in R$ is increasing for all $x$ lying in
$(-\infty, 0) \cup\left(\frac{3}{7}, \infty\right)$
$(-\infty, 0) \cup\left(\frac{14}{15}, \infty\right)$
$\left(-\infty, \frac{14}{15}\right)$
$\left(-\infty, \frac{14}{15}\right) \cup(0, \infty)$
Let $A=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right], x \in R$ and $A^4=\left[a_{i j}\right]$.
It $a_{11}=109$, then $a_{22}$ is equal to
21
10
9
14
If $\alpha$ and $\beta$ are the roots of equation $x^2+p x+2=0$ and $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ are roots of equation $2 x^2+2 q x+1=0$, then $\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$ is equal to
$\frac{9}{4}\left(9+p^2\right)$
$\frac{9}{4}\left(9-q^2\right)$
$\frac{9}{4}\left(9-p^2\right)$
$\frac{9}{4}\left(9+q^2\right)$
VITEEE Papers
All year-wise previous year question papers