MCQ
The solution of the differential equation ${(x + y)^2}\frac{{dy}}{{dx}} = {a^2}$ is
- A${(x + y)^2} = \frac{{{a^2}}}{2}x + c$
- B${(x + y)^2} = {a^2}x + c$
- C${(x + y)^2} = 2{a^2}x + c$
- ✓None of these
${v^2}\left( {\frac{{dv}}{{dx}} - 1} \right) = {a^2}$
==> $\frac{{dv}}{{dx}} = \frac{{{a^2}}}{{{v^2}}} + 1 = \frac{{{a^2} + {v^2}}}{{{v^2}}}$ ==> $\frac{{{v^2}}}{{{a^2} + {v^2}}}dv = dx$
==> $\left( {1 - \frac{{{a^2}}}{{{a^2} + {v^2}}}} \right)dv = dx$ ==> $v - a{\tan ^{ - 1}}\frac{v}{a} = x + c$
==> $y = a{\tan ^{ - 1}}\left( {\frac{{x + y}}{a}} \right)+ c.$
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Statement $-I$ : ${A^{ - 1}} = \frac{1}{7}\left( {5I - A} \right).$
Statement $II$ : the polynomial $A^3 - 2A^2 - 3A + I$ can be reduced to $5\, (A - 4I)$.
Let f : N → R be the function defined by
$\text{f}(\text{x})=\frac{2\text{x}-1}{2}$ and g : Q → R be another function defined by g(x) = x + 2. Then $(\text{gof})\frac{3}{2}$ is: