The function g x -f x -x-x- 0 has an extreme value when

The correct option is C g ( x )=f' ( x ) g(x)'=1x^2[xf'(x) f(x)] =0 Hence f'(x)=f(x)x Which is nothing but g(x) Hence f'(x)=g(x) ...

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Standard XII

Mathematics

Sufficient Condition for an Extrema

The function ...

Question

The function $g (x) = \frac{f (x)}{x}, x \neq 0$ has an extreme value when

g^{'} (x) = f (x)

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f (x) = 0

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x g^{'} (x) = f (x)

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g (x) = f^{'} (x)

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Solution

The correct option is C $g (x) = f^{'} (x)$
$g (x)^{'} = \frac{1}{x^{2}} [x f^{'} (x) - f (x)]$
$= 0$
Hence
$f^{'} (x) = \frac{f (x)}{x}$
Which is nothing but $g (x)$
Hence
$f^{'} (x) = g (x)$

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The function g(x)=f(x)x,x≠0 has an extreme value when

The correct option is C g(x)=f′(x)g(x)′=1x2[xf′(x)−f(x)]=0Hencef′(x)=f(x)xWhich is nothing but g(x)Hencef′(x)=g(x)

The function $g (x) = \frac{f (x)}{x}, x \neq 0$ has an extreme value when

The correct option is C $g (x) = f^{'} (x)$
$g (x)^{'} = \frac{1}{x^{2}} [x f^{'} (x) - f (x)]$
$= 0$
Hence
$f^{'} (x) = \frac{f (x)}{x}$
Which is nothing but $g (x)$
Hence
$f^{'} (x) = g (x)$