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5. continuity and differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths5. continuity and differentiation1 Mark
MCQ
If ${y^2} = a{x^2} + bx + c$, then ${y^3}{{{d^2}y} \over {d{x^2}}}$ is
  • ✓
    A constant
  • B
    A function of $ x$  only
  • C
    A  function of $ y$ only
  • D
    A function of $ x$ and $y$
✓

Answer

Correct option: A.
A constant
a
(a) ${y^2} = a{x^2} + bx + c \Rightarrow 2y\frac{{dy}}{{dx}} = 2ax + b$

==> $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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