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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
$f(x) = \left| {\begin{array}{*{20}{c}}
{{x^3}}&{{x^2}}&{3{x^2}}\\
1&{ - 6}&4\\
p&{{p^2}}&{{p^3}}
\end{array}} \right|$ , here $ p $ is a constant, then ${{{d^3}f(x)} \over {d{x^3}}}$ is
  • A
    Proportional to ${x^2}$
  • B
    Proportional to $ x$
  • C
    Proportional to ${x^3}$
  • ✓
    A constant
✓

Answer

Correct option: D.
A constant
d
(d) $f(x) = \left| {\,\begin{array}{*{20}{c}}{{x^3}}&{{x^2}}&{3{x^2}}\\1&{ - 6}&4\\p&{{p^2}}&{{p^3}}\end{array}\,} \right|$

==>$f(x) = {x^3}( - 6{p^3} - 4{p^2}) - {x^2}({p^3} - 4p) + 3{x^2}({p^2} + 6p)$

 ==>$f(x) = - 6{p^3}{x^3} - 4{p^2}{x^3} - {x^2}{p^3} + 4p{x^2} + 3{p^2}{x^2} + 18p{x^2}$

$\therefore$ $\frac{d}{dx}f(x)=-18{{p}^{3}}{{x}^{2}}-12{{p}^{2}}{{x}^{2}}-2x{{p}^{3}}+8px+6{{p}^{2}}x+36px$

and $\frac{{{d}^{2}}}{d{{x}^{2}}}\,f(x)=-36{{p}^{3}}x-24{{p}^{2}}x-2{{p}^{3}}+8p+6{{p}^{2}}+36p$

and $\frac{{{d}^{3}}f(x)}{d{{x}^{3}}}=-36{{p}^{3}}-24{{p}^{2}}$ = a constant.

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