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} = $a constant.
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$f(x)=\left\{\begin{array}{cc}\min \left\{|x|, 2-x^{2}\right\} & , \quad-2 \leq x \leq 2 \\ {[|x|]} & , \quad 2<|x| \leq 3\end{array}\right.$
where $[x]$ denotes the greatest integer $\leq x .$ The number of points, where $f$ is not differentiable in $(-3,3)$ is
$D^*f(x) =\mathop {Limit}\limits_{h \to 0} \frac{{{f^2}(x + h) - {f^2}(x)}}{h}$ where $f^2(x)$ means $[f(x)]^2.$ If $f(x) = x lnx$ then
${\left. {D^*f(x)} \right|_{x = e}}$ has the value