MCQ
જો $u = {\tan ^{ - 1}}(x + y),$ તો $x{{\partial u} \over {\partial x}} + y{{\partial u} \over {\partial y}} = $
- A$\sin 2u$
- ✓${1 \over 2}\sin 2u$
- C$2\tan u$
- D${\sec ^2}u$
$\therefore $ $\tan u$ is homogeneous in $x,\,y$ of order $ 1$.
$\therefore $ $x\frac{\partial }{{\partial x}}(\tan u) + y\frac{\partial }{{\partial y}}(\tan u) = \tan u$
$\therefore $ $x{\sec ^2}u\frac{{\partial u}}{{\partial x}} + y{\sec ^2}u\frac{{\partial u}}{{\partial y}} = \tan u$
$\therefore $ $x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = \tan u.{\cos ^2}u = \sin u\cos u$ = $\frac{1}{2}\sin 2u$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
કારણ $(R) : \,{\left( {\vec x \,\, + \;\,\vec y } \right)^2}\, = \,\,|\vec x {|^2}\,\, + \,\,|\vec y {|^2}\,\, + \;\,2\,\,\left( {\vec x .\,\,\vec y } \right)$
(કે જ્યાં $p$ એ અચળ છે)
$f(x)=\left[\begin{array}{ll}{\left[e^{x}\right],} \,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,x<0 \\ a e^{x}+[x-1], \,\,\,\,\,\,\,\,\,0 \leq x<1 \\ b+[\sin (\pi x)], \,\,\,\,\,\,\,\,\,\,\,\,1 \leq x<2 \\ {\left[e^{-x}\right]-c,} \,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,x \geq 2\end{array}\right.$
પ્રમાણે વ્યાખ્યાયિત છે, જ્યાં $a, b, c \in R$ અને $[t]$ એ $t$ અથવા તેથી નાનો મહત્તમ પૂર્ણક દર્શાવે છે. તો નીચેના પૈકી કયું વિધાન સાયું છે $?$