MCQ
The general solution of differential equation is $(y + c)^2= cx$ where ccis an arbitrary constant. The order and degree of the differential equation are respectively:
- ✓$1, 2$
- B$2, 2$
- C$1, 1$
- D$2, 1$
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$l_1: \overrightarrow{ r }=(\hat{ i }-11 \hat{ j }-7 \hat{ k })+\lambda(\hat{ i }+2 \hat{ j }+3 \hat{ k }), \lambda \in R$
and $l_2: \overrightarrow{ r }=(-\hat{ i }+\hat{ k })+\mu(2 \hat{ i }+2 \hat{ j }+\hat{ k }), \mu \in R$.
If $P$ is the point of intersection of $l$ and $l_1$, and $Q (\alpha$ $, \beta, \gamma)$ is the foot of perpendicular from $P$ on $l_2$, then $9(\alpha+\beta+\gamma)$ is equal to $..........$.
$P_6=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$ and $X=\sum_{k=1}^6 P_k\left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right] P_k^{\top}$
where $P _{ K }^{ T }$ denotes the transpose of the matrix $P _{ K }$. Then which of the following options is/are correct?
$(1)$ $X -30 I$ is an invertible matrix
$(2)$ The sum of diagonal entries of $X$ is 18
$(3)$ If $X \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\alpha\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$, then $\alpha=30$
$(4)$ $X$ is a symmetric matrix