- A$ab + bc + ca$
- B$(a+b)(b+c)(c+a)$
- C$2$
- ✓$0$
$\left| {\begin{array}{{}{c}}{ab}&1&{bc + ca}\\{bc}&1&{ca + ab}\\{ca}&1&{ab + bc}\end{array}} \right| = \left| {\begin{array}{{}{c}}{ab}&1&{ab + bc + ca}\\{bc}&1&{bc + ca + ab}\\{ca}&1&{ca + ab + b}\end{array}} \right|$ $(\because C_3\rightarrow C_3+C_1)$
$=(ab+bc+ca) \left| {\begin{array}{{}{c}}{ab}&1&1\\{bc}&1&1\\{ca}&1&1\end{array}} \right|$
$=0$ $(\because C_2=C_3)$
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($I$) $f$ એ $\left(0, \frac{\pi}{2}\right)$ માં વધે છે
($II$) $f^{\prime}$ એ $\left(0, \frac{\pi}{2}\right)$ માં ઘટે છે
$190$ persons had symptom of fever,
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$220$ persons had symptom of breathing problem,
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$350$ 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. . . . .