MCQ
વિકલ સમીકરણ $(1 + {x^2})\frac{{dy}}{{dx}} = x(1 + {y^2})$ નો ઉકેલ મેળવો.
- ✓$2{\tan ^{ - 1}}y = \log (1 + {x^2}) + c$
- B${\tan ^{ - 1}}y = \log (1 + {x^2}) + c$
- C$2{\tan ^{ - 1}}y + \log (1 + {x^2}) + c = 0$
- Dએકપણ નહી.
On integrating, we get ${\tan ^{ - 1}}y = \frac{1}{2}\log (1 + {x^2}) + c$
==> $2{\tan ^{ - 1}}y = \log (1 + {x^2}) + c$.
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ધારો કે $a \in S$ અને $A =\left[\begin{array}{ccc}1 & 0 & a \\ -1 & 1 & 0 \\ - a & 0 & 1\end{array}\right]$ છે.
જો $\sum_{ a \in S } \operatorname{det}(\operatorname{adj} A )=100 \lambda$ હોય, તો $\lambda$ .........