- A${1 \over {{x^2} + {y^2}}}$
- ✓$0$
- C${{{x^2} - {y^2}} \over {{{({x^2} + {y^2})}^2}}}$
- D${{{y^2} - {x^2}} \over {{{({x^2} + {y^2})}^2}}}$
$\therefore \frac{{\partial u}}{{\partial x}} = \frac{1}{{{x^2} + {y^2}}}.2x$
$\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{{({x^2} + {y^2}).2 - 2x.2x}}{{{{({x^2} + {y^2})}^2}}}$$ = \frac{{2({y^2} - {x^2})}}{{{{({x^2} + {y^2})}^2}}}$’
$\frac{{\partial u}}{{\partial y}} = \frac{1}{{{x^2} + {y^2}}}.2y$
$\therefore \frac{{{\partial ^2}u}}{{\partial {y^2}}} = \frac{{({x^2} + {y^2})\,.\,2 - 2y.2y}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2({x^2} - {y^2})}}{{{{({x^2} + {y^2})}^2}}}$
$\therefore $ $\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0$.
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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 $\lambda$ .........