MCQ
The solution of the equation $\frac{d y}{d x}=e^{x-y}$ is :
  • A
    $e^x=e^{-y}+c$
  • B
    $e^y=e^{-x}+c$
  • $e^y=e^x+c$
  • D
    $e^{-x}=e^{-y}+c$

Answer

Correct option: C.
$e^y=e^x+c$
$\frac{d y}{d x}=e^{x-y}=e^x \times e^{-y}$
$\Rightarrow \frac{d y}{e^{-y}}=e^x d x$
$\Rightarrow e^y d y=e^x d x$
$\text { So } \int e^y d y=\int e^x d x$
$\Rightarrow e^y=e^x+c$
Hence the correct choice is $(C).$

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Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^2+x-1=0$. Consider the set $T=\{1, \alpha, \beta\}$. For a $3 \times 3$ matrix $M=\left(a_{\ell}\right) 3 \times 3_3$, define $R_l=a_{l 1}+a_{l 2}+a_\beta$ and $C_j=a_{1 j}+a_{2 l}+a_{3 j}$ for $i=1,2,3$ and $j=1,2,3$

Match each entry in $List-I$ to the correct entry in $List-II$.

$List-I$ $List-II$
($P$) The number of matrices $M=\left(a_{i j}\right)_3 \times 3$ with all entries in $T$ such that $R_i=C_j=0$ for all $i, j$ is ($1$) ($1$)
($Q$) The number of symmetric matrices $M=\left(a_{i j}\right) 3 \times 3$ with all entries in $T$ such that $C_j=0$ for all $j$ is ($2$) ($2$)
($R$) Let $M=\left(a_{i j}\right) 3 \times 3$ be a skew symmetric matrix such that $a_{i j} \in T$ for $i>j$. Then the number of elements in the set $\left\{\left(\begin{array}{l}x \\ y \\ z\end{array}\right): x, y \cdot z \in R, M\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}a_{12} \\ 0 \\ -a_{23}\end{array}\right)\right\}$ is is ($3$) Infinite
($S$) Let $M=\left(a_{i j}\right)_3 \times 3$ be a matrix with all entries in $T$ such that $R_i=0$ for all $i$. Then the absolute value of the determinant of $M$ is ($4$) ($6$)
  ($5$) ($0$)

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