MCQ
વિકલ સમીકરણ $3\frac{{{d^2}y}}{{d{x^2}}} = {\left\{ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right\}^{3/2}}$ નો પરિમાણ મેળવો
- A$1$
- ✓$2$
- C$3$
- D$6$
On squaring, we get $9{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} = {\left\{ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right\}^3}$
Obviously the highest derivative $\frac{{{d^2}y}}{{d{x^2}}}$ contains a degree $2.$
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વિધાન $-2$ : સમીકરણ કે જે $\alpha $ સ્વરૂપ માં છે
$\left| {\begin{array}{*{20}{c}}
{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\
{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\
{\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha }
\end{array}} \right| = 0$
નું એક માત્ર બીજ અંતરાલ $\left( {0\,,\,\frac{\pi }{2}} \right)$ માં છે .