If $\phi (n) =f(n) g(n),$ where $f'(n) g'(n) =C$ and $\large\frac {\phi ''}{\phi} =\frac {f''}{f} +\frac{g''}{g}+\frac{Kc}{fg}$ then value of K.

$(a)\;1 \\ (b)\;0 \\ (c)\;2 \\ (d)\;-1$

1 Answer

$\phi(x)=f(x)g(x)$
$\phi'(x) =f'(x)g(x) +f(x) g'(x)$
$\phi''(x) =f'(x)g(x) +f''(x) g'(x)+f''(x) g'(x) +f(x)g'(x)$
$\phi''(x) =2c+f''(x) g(x)+f(n) g''(x)$
$\large\frac{\phi ''(x)}{\phi (n)}=\frac{2c}{f(n) g(n) } +\frac{f''(x)}{f(x)}+\frac{g''(x)}{g(x)}$
$K=2$
Hence c is the correct answer.
answered Jan 31, 2014 by

1 answer

1 answer

1 answer

1 answer

1 answer