- A$e^{-x}+e^{y}=C$
- B$e^{x}+e^{y}=C$
- ✓$ e^{x}+e^{-y}=C$
- D$e^{-x}+e^{-y}=C$
$\Rightarrow \frac{d y}{e^{y}}=e^{x} d x$
$\Rightarrow \mathrm{e}^{-\mathrm{y}} \mathrm{dy}=\mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
Intergrating both sides, we get:
$\int e^{-y} d y=\int e^{x} d x$
$\Rightarrow-e^{-y}=e^{x}+k$
$\Rightarrow \mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{y}}=-\mathrm{k}$
$\Rightarrow e^{x}+e^{-y}=c \quad(c=-k)$
Hence, the correct answer is $\mathrm{C}.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$T_{\mathrm{I}}=V_{\mathrm{r}+1}-V_{\mathrm{I}}-2 \text { and } \mathrm{Q}_{\mathrm{I}}=T_{\mathrm{r}+1}-\mathrm{T}_{\mathrm{r}} \text { for } \mathrm{r}=1,2, \ldots$
$1.$ The sum $V_1+V_2+\ldots+V_n$ is
$(A)$ $\frac{1}{12} n(n+1)\left(3 n^2-n+1\right)$
$(B)$ $\frac{1}{12} n(n+1)\left(3 n^2+n+2\right)$
$(C)$ $\frac{1}{2} n\left(2 n^2-n+1\right)$
$(D)$ $\frac{1}{3}\left(2 n^3-2 n+3\right)$
$2.$ $\mathrm{T}_{\mathrm{T}}$ is always
$(A)$ an odd number $(B)$ an even number
$(C)$ a prime number $(D)$ a composite number
$3.$ Which one of the following is a correct statement?
$(A)$ $Q_1, Q_2, Q_3, \ldots$ are in $A.P.$ with common difference $5$
$(B)$ $\mathrm{Q}_1, \mathrm{Q}_2, \mathrm{Q}_3, \ldots$ are in $A.P.$ with common difference $6$
$(C)$ $\mathrm{Q}_1, \mathrm{Q}_2, \mathrm{Q}_3, \ldots$ are in $A.P.$ with common difference $11$
$(D)$ $Q_1=Q_2=Q_3=\ldots$
Give the answer question $1,2$ and $3.$
$x^2+y^2+5 x-3 y+4=0 .$
Then which of the following statements is (are) TRUE?
$(A)$ $\alpha=-1$ $(B)$ $\alpha \beta=4$ $(C)$ $\alpha \beta=-4$ $(D)$ $\beta=4$