MCQ
If ${y^2} = a{x^2} + bx + c$, then ${y^3}{{{d^2}y} \over {d{x^2}}}$ is
- ✓A constant
- BA function of $ x$ only
- CA function of $ y$ only
- DA function of $ x$ and $y$
==> $2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 2y\frac{{{d^2}y}}{{d{x^2}}} = 2a $
$\Rightarrow y\frac{{{d^2}y}}{{d{x^2}}} = a - {\left( {\frac{{dy}}{{dx}}} \right)^2}$
==>$y\frac{{{d^2}y}}{{d{x^2}}} = a - {\left( {\frac{{2ax + b}}{{2y}}} \right)^2}$
==> $y\frac{{{d^2}y}}{{d{x^2}}} = \frac{{4a{y^2} - {{(2ax + b)}^2}}}{{4{y^2}}}$
==> $4{y^3}\frac{{{d^2}y}}{{d{x^2}}} = 4a(a{x^2} + bx + c) - (4{a^2}{x^2} + 4abx + {b^2})$
==>$4{y^3}\frac{{{d^2}y}}{{d{x^2}}} = 4ac - {b^2} \Rightarrow {y^3}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{4ac - {b^2}}}{4} = $a constant.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Statement $-1 :$ $f\left( c \right) = \frac{1}{3}$ for some $c\; \in R$
Statement $-2 :$$0 < f\left( x \right) < \frac{1}{{2\sqrt 2 }}\;,\forall x\; \in R$