- A$e^x+e^y=\frac{x^3}{3}+c$
- B$e^{-x}+e^y+\frac{x^3}{3}=c$
- C$e^{-x}+e^{-y}-\frac{x^3}{3}=c$
- ✓$e^x+e^{-y}+\frac{x^3}{3}=c$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$ \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R} $
$ \mathrm{L}_2: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{k}, \mu \in \mathbb{R}$
is $\frac{\mathrm{m}}{\sqrt{\mathrm{n}}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then the value of $\mathrm{m}+\mathrm{n}$ equals.
$190$ persons had symptom of fever,
$220$ persons had symptom of cough,
$220$ persons had symptom of breathing problem,
$330$ persons had symptom of fever or cough or both,
$350$ persons had symptom of cough or breathing problem or both,
$340$ persons had symptom of fever or breathing problem or both,
$30$ persons had all three symptoms (fever, cough and breathing problem).
If a person is chosen randomly from these 900 persons, then the probability that the person has at most one symptom is. . . . .