Let f(x) and g(x) be two functions having finite non-zero 3^ rd o… - Vidyadip

MCQ

Let $f(x)$ and $g(x)$ be two functions having finite non-zero $3^{rd}$ order derivatives $f'''(x)$ and $g'''(x)$ for all, $x \in R$. If $f(x)g(x) = 1$ for all $x \in R$, then ${{f'''} \over {f'}} - {{g'''} \over {g'}}$ is equal to

A
$3{\rm{ }}\left( {{{f''} \over g} - {{g''} \over f}} \right)$
✓
$3{\rm{ }}\left( {{{f''} \over f} - {{g''} \over g}} \right)$
C
$3{\text{ }}\left( {\frac{{g}}{g} - \frac{{f}}{g}} \right)$
D
$3{\rm{ }}\left( {{{f''} \over f} - {{g''} \over f}} \right)$

✓

Answer

Correct option: B.

$3{\rm{ }}\left( {{{f''} \over f} - {{g''} \over g}} \right)$

b
(b) We have $f(x)g(x) = 1$

Differentiating with respect to $ x$ , we get

$f'g + fg' = 0$…..$(i)$

Differentiating $(i)$ w.r.t. $x,$ we get

$f''g + 2f'g' + fg'' = 0$…..$(ii)$

Differentiating $(ii)$ w.r.t. $x,$ we get

$f'''g + g'''\,f + 3f''g' + 3g''f' = 0$

==> $\frac{{f'''}}{{f'}}(f'g) + \frac{{g'''}}{{g'}}(fg') + \frac{{3f''}}{f}(fg') + \frac{{3g''}}{g}(gf') = 0$

==> $\left( {\frac{{f'''}}{{f'}} + \frac{{3g''}}{g}} \right)\,(f'g) = - \left( {\frac{{g'''}}{{g'}} + \frac{{3f''}}{f}} \right)\,(fg')$

==> $ - \left( {\frac{{f'''}}{{f'}} + \frac{{3g''}}{g}} \right)\,(fg') = - \left( {\frac{{g'''}}{{g'}} + \frac{{3f''}}{g}} \right)fg'$ , [using $(i)$]

==> $\frac{{f'''}}{{f'}} + \frac{{3g''}}{g} = \frac{{g'''}}{{g'}} + \frac{{3f''}}{f} $

$\Rightarrow \frac{{f'''}}{{f'}} - \frac{{g'''}}{{g'}} = 3\left( {\frac{{f''}}{f} - \frac{{g''}}{g}} \right)$.

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