MCQ
The line $(3x - y + 5) + \lambda (2x - 3y - 4) = 0$ will be parallel to $y$-axis, if $\lambda$ =
- A$\frac{1}{3}$
- ✓$\frac{{ - 1}}{3}$
- C$\frac{3}{2}$
- D$\frac{{ - 3}}{2}$
It is will be parallel to $y$-axis, if
$ - 1 - 3\lambda = 0 $
$\Rightarrow \,\lambda = - \frac{1}{3}$.
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$[A]$ $\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right]$
$[B]$ $\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$
$[C]$ $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$[D]$ $\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$
| $\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
| $\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :