MCQ
$\sin \frac{{dy}}{{dx}} = a$ ; $y(0) = 1$ નો ઉકેલ મેળવો.
- A${\sin ^{ - 1}}\frac{{(y - 1)}}{x} = a$
- ✓$\sin \frac{{(y - 1)}}{x} = a$
- C$\sin \frac{{(1 - y)}}{{(1 + x)}} = a$
- D$\sin \frac{y}{{(x + 1)}} = a$
Integrating both sides,$\int_{}^{} {dy} = \int_{}^{} {{{\sin }^{ - 1}}a\,dx} $
$y = x{\sin ^{ - 1}}a + c$ and $y(0) = 0 + c = 1$, $\therefore c = 1$
$\therefore y = x{\sin ^{ - 1}}a + 1$ ==> $a = \sin \frac{{y - 1}}{x}$.
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$\frac{{{x^4} - 4{x^3} + 3{x^2}}}{{({x^2} - 4)({x^2} - 7x + 10)}} \ge 0$
$|\overline {AB} \,\, \times \,\,\overline {CD} \,\, + \;\,\overline {BC} \,\, \times \,\,\overline {AD} \,\, + \;\,\overline {CA} \,\, \times \,\,\overline {BD} |\,\, = \,\,\lambda $ $×$( ${\Delta \,\,ABC}$ નું ક્ષેત્રફળ )
તો $\lambda \,\, = \,\,......$